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The collected theorems of the ℤ/9 vortex framework

Rouschev, T. · Millennium Solutions · DOI 10.5281/zenodo.21819217 · CC BY-NC-ND 4.0
541 theorems + 8 rfl declarations · 30 sources · content-address 6c14259d-766b-8c86-a30e-914f49338665

Abstract

This document collects the 549 declarations of the ℤ/9 vortex framework that the Lean 4 kernel accepts — 541 of them THEOREMS by this deposit's rule, closing by exhaustion, and 8 rfl declarations shown and marked as such, across 30 source files. Each is stated exactly as the kernel received it, followed by the tactic that discharged it and the size of the finite domain that tactic exhausted. Every statement is decidable and was checked sorry-free and axiom-free. None of them is a Clay Millennium Problem and none claims one. A content-address proves integrity, not truth: it fixes which statement was checked, not that the statement is significant. Clay problems solved by this framework: 0 of 7.

The size of what was checked

The domains sum to 152,576,389,946 cases, and that total should not be read as the weight of this work: one theorem carries 99.995% of it. The median theorem walks 6 cases, and 215 of 549 walk a single point — a check by evaluation, not by exhaustion. The honest summary is a distribution, not a sum, so it is drawn below rather than reported as one number.

Figure 1 — all 549 declarations by domain sizeaddress_is_sixteen_bytes — 1 casesaddress_bytes_are_bytes — 1 casesversion_nibble_is_forced — 1 casesvariant_bits_are_forced — 1 casesaddressing_is_deterministic — 1 casesdistinct_seeds_give_distinct_addresses — 1 casesaddressing_is_not_the_identity — 1 casesaddressing_is_not_constant — 1 casesthe_four_seeds_are_distinct — 1 casesaddress_settles_its_range — 1 casesthe_address_is_order_sensitive — 4 casesthe_reflection_is_an_involution — 1 casesone_side_is_reflected_onto_itself — 1 casesthe_other_two_swap — 1 casesexactly_one_digit_is_unmoved — 1 casesexactly_one_digit_reflects_out_of_range — 1 casesthe_classes_partition_the_nine — 1 casesevery_digit_is_sorted_exactly_once — 1 casesthe_fall_fixes_every_digit_but_the_void — 1 casesthe_fall_and_the_reflection_share_one_exceptional_digit — 1 caseson_the_fixed_points_the_reflection_never_leaves — 1 caseswhat_escapes_falls_back_inside — 1 casesthe_digits_are_ten — 1 casesfactorial_seven_is_five_thousand_and_forty — 1 casespicks_theorem_holds_for_every_lattice_triangle_in_the_four_grid — 1 casesthe_latin_squares_of_order_four_number_five_hundred_and_seventy_six — 1 casesthe_loop_returns_less_than_it_took — 1 casesas_a_purifier_the_loop_costs_a_thousandfold — 1 casesthe_equation_balances_by_mass — 1 casesthe_gases_are_two_to_one_and_consume_each_other_exactly — 1 caseshydrogen_is_a_ninth_of_the_mass_and_oxygen_the_rest — 1 casesone_litre_split_returns_one_litre_burnt — 1 casesonly_oxy_hydrogen_burns_without_admitting_nitrogen — 1 casesthe_ideal_round_trip_is_exactly_zero — 1 casesthe_gases_are_eighteen_hundred_times_the_water_they_came_from — 1 casestwo_thirds_of_the_volume_carries_a_ninth_of_the_mass — 1 casesuncompressed_hydrogen_is_three_thousandfold_worse_by_volume — 1 caseseven_at_seven_hundred_bar_it_is_sevenfold_worse_by_volume — 1 caseseach_delivered_kilowatt_hour_costs_four_and_cycles_a_litre — 1 casesthree_quarters_of_the_input_leaves_as_heat — 1 caseswhat_the_feedwater_leaves_behind_decides_the_maintenance — 1 casesflt_all_primes_under_thirty — 1 caseswilson_all_primes_under_thirty — 1 casesxor_is_parity_up_to_eight_bits — 8 casesnat_div_is_a_total_function_returning_zero_at_a_zero_divisor — 1 casesat_a_zero_divisor_the_identity_is_carried_by_the_remainder — 1 casesfamilies_settle_their_ranges — 1 caseshash_a_seed_zero — 1 caseshash_ab_seed_zero — 2 caseshash_a_seed_golden — 1 caseshash_uuidna_seed_zero — 6 caseshash_is_not_the_identity — 1 caseshash_is_not_constant — 1 casesfnv_settles_its_range — 1 casesthe_hash_is_order_sensitive — 4 casesthe_empty_input_is_still_mixed — 1 casesdemorgan_all_widths — 7 casesinvpow_is_fifth_power_all_units — 6 casescyclic_units_have_a_primitive_root — 6 casesdecimal_period_is_the_order_of_ten — 8 casestriad_is_closed_under_double — 9 casestriad_is_closed_under_triple — 9 casestriad_is_closed_under_quadruple — 9 casestetA_is_closed_under_quadruple — 9 casestetB_is_closed_under_quadruple — 9 casestriad_is_closed_under_negate — 9 casescubes_is_closed_under_negate — 9 casestriad_is_closed_under_square — 9 casestetA_is_closed_under_square — 9 casescubes_is_closed_under_square — 9 casestriad_is_closed_under_fourth — 9 casestetA_is_closed_under_fourth — 9 casescubes_is_closed_under_fourth — 9 casestriad_is_closed_under_sixth — 9 casestetA_is_closed_under_sixth — 9 casescubes_is_closed_under_sixth — 9 casestriad_is_closed_under_quintuple — 9 casestriad_is_closed_under_sextuple — 9 casestriad_is_closed_under_septuple — 9 casestetA_is_closed_under_septuple — 9 casestetB_is_closed_under_septuple — 9 casestriad_is_closed_under_octuple — 9 casescubes_is_closed_under_octuple — 9 casesdouble_is_involutive_on_triad — 3 casesquadruple_is_involutive_on_triad — 3 casessquare_is_involutive_on_tetA — 3 casessquare_is_involutive_on_squares — 4 casescube_is_involutive_on_cubes — 3 casesfourth_is_involutive_on_tetA — 3 casesfourth_is_involutive_on_squares — 4 casesfifth_is_involutive_on_units — 6 casesfifth_is_involutive_on_orbit — 6 casesfifth_is_involutive_on_tetA — 3 casesfifth_is_involutive_on_tetB — 3 casesfifth_is_involutive_on_squares — 4 casesfifth_is_involutive_on_cubes — 3 casesquintuple_is_involutive_on_triad — 3 casesseptuple_is_involutive_on_triad — 3 casestriple_collapses_triad_to_one_value — 3 casestriple_collapses_tetA_to_one_value — 3 casestriple_collapses_tetB_to_one_value — 3 casessquare_collapses_triad_to_one_value — 3 casescube_collapses_triad_to_one_value — 3 casescube_collapses_tetA_to_one_value — 3 casescube_collapses_tetB_to_one_value — 3 casesfourth_collapses_triad_to_one_value — 3 casesfifth_collapses_triad_to_one_value — 3 casessixth_collapses_units_to_one_value — 6 casessixth_collapses_triad_to_one_value — 3 casessixth_collapses_orbit_to_one_value — 6 casessixth_collapses_tetA_to_one_value — 3 casessixth_collapses_tetB_to_one_value — 3 casessextuple_collapses_triad_to_one_value — 3 casessextuple_collapses_tetA_to_one_value — 3 casessextuple_collapses_tetB_to_one_value — 3 casesyang_mills_spectral_gap — 6 casesbirch_swinnerton_dyer_vanishing — 9 casesthe_sequence_is_its_named_parts_and_closes — 3 casesthere_are_2620_involutions_of_nine — 1 casesthe_fuel_was_not_the_limit — 1 casesevery_involution_fixes_at_least_one_point — 1 casesthe_fixed_points_are_always_odd — 1 casesno_single_point_is_fixed_by_all — 1 casesthe_constant_sum_is_rare — 1 casesso_involutions_are_not_all_harmonic_in_that_sense — 1 casesthe_coins_reflection_is_harmonic_and_is_one_of_the_ninety — 1 casesa_seal_is_128_bits — 1 casesmembership_grows_by_one_seal_per_doubling — 1 casesmembership_is_logarithmic_not_linear — 1 casesthe_967_receipt_case — 1 casesthe_si_fixes_exactly_seven_constants — 1 casestravel_at_one_returns_the_defined_constant — 1 casestravel_and_periods_at_one_return_their_constants — 1 casesthe_roots_of_the_seven — 7 casesthe_root_moves_with_the_unit_so_it_is_not_about_light — 1 casesthe_definitions_are_seven_and_travel_fixes_zero — 1 casesbool_demorgan1 — 4 casesbool_demorgan2 — 4 casesbool_distributivity — 8 casesbool_absorption — 4 casesmerkaba_cube_q3 — 1 casescover_rotation_full_circle — 9 casesfib_trinity_horizon — 1 casesarts_nine_hues_distinct — 9 casestrial_units_group — 6 casestrial_zero_divisors — 1 casestrial_zero_no_inverse — 9 caseschess_diagonals_15 — 1 casestarot_78_cards — 1 casestarot_minor_4x14 — 1 casestarot_digital_roots — 1 casesgf4_size — 4 casesrelation_digital_root — 6 casesharmonic_octave_2_1 — 1 casesharmonic_pythagorean_comma — 1 casesrelation_seven — 1 casesrelation_eight — 1 casesrelation_creation_week — 1 casesthe_moduli_dimensions_are_three_g_minus_three_and_six_g_minus_six — 1 casesgenus2_moduli_dim — 1 casesgenus2_hyperelliptic — 1 casesgenus2_h1_symplectic — 1 casesrelation_seven_is_six_plus_one — 6 casesrelation_units_sum_and_product — 6 casesrelation_432_factors — 1 casesrelation_triangular_45_is_base — 9 caseskaprekar_constants_digitroot_nine — 1 casesthe_three_four_five_right_triangle_is_the_first_pythagorean_triple — 1 casesthe_regular_pentagon_angles_are_the_heart_seventy_two_and_hundred_eight — 1 casesthe_cyclic_number_142857_is_the_repetend_of_one_seventh — 1 cases_142857_times_seven_is_six_nines — 1 casesmidy_the_two_halves_of_142857_sum_to_nines — 1 casesthe_digital_root_of_seven_to_the_k_has_period_three — 1 casesseven_divides_the_repunit_of_length_six — 1 casestwo_to_the_eighth_is_two_hundred_fifty_six_a_byte — 1 casestwo_to_the_tenth_is_1024_the_harmonic_ledger — 1 casesthe_regular_nonagon_exterior_angle_is_the_a432_step — 1 casesthe_digits_one_to_nine_sum_to_forty_five_rooting_to_nine — 9 casesnine_is_the_base_and_the_trinity_squared — 1 casessix_is_the_third_triangular_number — 3 casesthe_regular_hexagon_exterior_angle_is_the_gold_string — 1 casesthe_doubling_orbit_reflection_pairs_sum_to_nine — 1 casesfive_is_the_inverse_of_two_so_halving_reverses_the_orbit — 1 casesdna_is_the_version_itself — 1 casescontribute_two_to_save_sixty_four — 6 casesgenetic_code_is_the_octave_squared — 1 casessixty_one_sense_three_stop_codons — 1 casessix_reading_frames — 1 casesgenesis_1_the_unit — 9 casesgenesis_8_the_octave — 1 casesgenesis_64_the_codon — 1 casesdiamond_fixed_point_is_zero_entropy — 1 casesthe_address_is_shipped_not_the_payload — 1 casesthe_harmonic_band_thirty_to_ninety — 1 casessixty_and_ninety_partition_the_quadrant — 1 casesthe_hexagon_and_the_square_metrics — 1 casesa_theorem_responds_in_a_receipt — 1 casesinvolution_negation — 1 casesa432_factors_as_two_to_the_fourth_times_three_cubed — 1 casesa432_octave_doubling — 1 casestwo_bits_thrice_make_the_codon — 1 casesthe_skipper_navigates_by_angle — 1 casescheap_to_factor_easy_to_verify — 1 casesthe_cell_is_determined_by_content — 1 casespresent_by_reference_fits_a_tiny_budget — 1 casesa_cached_address_is_never_recomputed — 1 casesa_decidable_domain_is_finite_and_coverable — 9 casesthe_more_developed_the_more_cross_domain_reach — 1 casesthe_full_superposition_has_nine_states — 9 casesforward_is_the_deterministic_compute — 1 casesthe_cardinality_claims_hold_of_any_four_symbol_alphabet — 1 casesa_reflexive_address_claim_holds_for_a_constant_function — 1 casesthe_three_classes_partition_z9 — 9 casesthe_axis_is_closed_under_doubling — 1 casesdoubling_counter_rotates_the_two_tetrahedra — 1 casesthe_counter_rotation_has_period_two — 1 casesthe_tetrahedra_residue_sums_cancel — 1 casesone_tetrahedron_covers_half_the_units — 1 casesthe_cube_and_the_tetrahedron_count_out — 1 casessingleton_fold_is_the_leaf — 1 casesfold_is_order_independent_on_two — 1 casesmerge_is_order_sensitive — 1 casessorting_is_what_makes_the_fold_order_free — 1 casesmerkle_settles_its_range — 1 casesfold_is_order_independent_on_three — 1 casesbouton_two_heaps_lost_iff_xor_zero — 1 casesbouton_lost_iff_heaps_equal — 1 caseslost_positions_are_exactly_the_diagonal — 1 casesgrundy_of_a_single_heap_is_its_size — 1 casesgrundy_of_two_heaps_is_the_xor — 1 casesevery_phenomenon_is_definitional_or_credited — 1 casesseven_definitional_and_two_credited — 1 casesthe_definitional_half_is_the_whole_si — 1 casesthe_table_is_closed_and_that_is_all_this_file_decides — 1 casesnovelty_is_claimed_only_where_a_search_was_performed — 1 casesthis_deposit_claims_no_novelty_today — 1 casesthe_restated_sources_are_named_and_claim_nothing — 1 casesan_unsearched_source_claims_nothing — 1 casesevery_source_is_classified — 1 casesthe_kinds_cover_every_source — 1 casessome_sources_are_unsearched — 1 caseszero_claims_is_not_full_attribution — 1 casesnovelty_is_claimed_of_no_source — 1 casesperms_of_four_is_factorial — 4 casessuperposition_collapses_to_one — 4 casesthe_receipt_is_not_injective — 1 casesthe_uncanonicalised_fold_gives_many_answers — 4 casesquantum_settles_its_domain_totally — 1 casesthe_ghz_x_support_is_exactly_the_even_parity_strings — 8 casesit_is_half_of_the_eight — 8 casesexhaustion_never_reaches_its_own_bound — 1 casesthe_successor_of_every_bound_lies_outside — 1 casesdoubling_the_domain_leaves_the_same_hole — 1 caseseven_the_largest_domain_here_has_an_outside — 1 casesthis_file_settles_none_of_the_seven — 1 casesthe_ledger_reversal_cases — 1 casesreversal_settles_its_range — 1 casesclaims_exactly_what_arises_without_formality — 1 casesthe_claimed_are_copyright_moral_rights_and_the_database — 3 casesthe_unclaimed_are_the_registry_the_excluded_and_the_unownable — 4 casesno_right_that_needs_a_registry_act_is_claimed — 1 casesno_excluded_subject_matter_is_claimed — 1 casesnothing_incapable_of_ownership_is_claimed — 1 casesthe_enumeration_is_complete_and_unduplicated — 7 casesrights_settles_its_range — 1 casesthree_five_eight_are_consecutive — 1 casesthe_verify_path_is_the_exponent — 1 casesthe_gap_widens_with_every_doubling — 1 casesno_constant_factor_accounts_for_the_gap — 1 casesthe_measured_ratio_at_a_million_leaves — 1 casesthe_verify_is_thirty_eight_thousand_nanoseconds_not_one — 1 casesthe_advantage_is_the_count_not_the_operation — 1 casesthe_break_even_is_the_ratio_of_verifications — 1 caseshexbits_are_shorter_than_hex — 1 caseshexbits_are_slower_than_hex — 1 caseseach_pair_is_two_consecutive_units — 1 casesevery_token_is_a_multiple_of_three — 1 casesthe_tokens_are_three_times_these — 7 casesaddition_and_multiplication_stay_inside — 1 casessubtraction_stays_inside — 1 casesdivision_is_the_operation_that_leaves — 1 casesthe_roots_are_the_singles — 7 casesevery_root_is_a_single — 1 casesthe_coin_step_is_three_times_the_two_coins — 1 casesaccounting_the_coins_on_the_last_pair — 1 casesthe_exhaustible_tokens_are_those_six_divides — 4 casesthe_rest_halt_on_the_generator — 3 casesevery_token_is_void_bound_or_halts_on_three — 1 casesinside_this_ideal_the_bare_coin_sorts_as_the_scaled_one — 1 casesthe_seal_affords_sixty_four_payments_of_two — 1 casesthe_budget_and_the_period_are_one_turn — 1 casesunits_are_six — 6 casesunits_count — 1 caseshasinv_units — 1 caseshasinv_triad_not — 3 casesselfinv_exactly_one_and_eight — 2 caseseuler_units_pow_six — 1 casesinvpow_is_u_to_the_fifth — 1 casespowsum_zero_at_six — 1 casespowsum_zero_at_one — 1 casesmulperm_iff_unit — 1 casesaddgen_iff_coprime — 1 casesselfneg_only_zero — 1 casesorbit_closes — 1 casesorbit_distinct — 6 casesorbit_covers_units — 6 casestriad_squares_vanish — 1 casesadd_group — 1 casesneg_involution — 1 casesz9_settles_its_domain_totally — 1 casesfive_is_not_a_square_mod_nine — 1 casescubes_land_exactly_in_zero_one_eight — 9 casesprimitive_roots_are_exactly_two_and_five — 2 casesprimitive_root_iff_generates_the_units — 2 casesthe_orders_of_the_units_are_exact — 1 casesthe_reflected_orbit_covers_the_whole_triad — 3 casesreflection_splits_the_orbit_in_half — 9 casesthe_crossing_pairs_are_one_four_seven — 9 casesreflection_is_injective_on_the_orbit — 1 casesreflection_alone_reaches_only_two — 1 casesdoubling_and_reflection_together_reach_every_residue — 1 casesevery_residue_is_reachable_from_one — 9 casesthe_diameter_is_exactly_four — 1 casesthe_frontier_grows_by_two_each_round — 1 casesproduct_of_the_units_is_minus_one — 6 casesthree_five_eight_are_consecutive_fibonacci — 1 casesunits_and_triad_partition_the_ring — 1 casesboth_parts_sum_to_zero — 1 cases1328raw_bytes_of_a — 16 casesto_uuid_bytes_of_a — 16 casesto_uuid_bytes_of_uuidna — 16 casesaddressing_is_injective_on_single_characters — 24 casesthe_address_is_not_the_payload — 36 casesthe_mobius_divisor_sum_is_the_identity — 30 casesthe_derangement_recurrence_holds — 11 casesmantels_bound_is_n_squared_over_four — 20 casesevery_number_is_a_sum_of_four_squares — 60 casesthe_cantor_pairing_is_injective_and_covers_an_initial_segment — 36 casesthe_euler_characteristic_of_a_genus_g_surface — 12 caseswilson_fails_at_composites — 20 casespascal_rows_sum_to_powers_of_two — 12 casespascal_alternating_sums_vanish — 11 casestotient_at_prime_powers — 12 casesgeometric_series_all_bases — 40 casesdivision_identity_holds_across_the_range — 13 caseshash_is_deterministic — 32 caseshash_is_injective_on_single_characters — 64 casesthe_seed_separates — 16 caseshash_is_thirty_two_bit — 40 casespowsum_zero_odd_exponents — 54 casespowsum_nonzero_at_even_exponents — 18 casesmulperm_fails_at_the_triad — 18 casespower_sum_closed_forms — 30 casesinvpow_fails_off_the_units — 18 casesroots_of_unity_cancel — 12 casesunits_is_closed_under_double — 36 casesorbit_is_closed_under_double — 36 casesall_is_closed_under_double — 81 casesall_is_closed_under_triple — 81 casesunits_is_closed_under_quadruple — 36 casesorbit_is_closed_under_quadruple — 36 casesall_is_closed_under_quadruple — 81 casessquares_is_closed_under_quadruple — 16 casesorbit_is_closed_under_negate — 36 casesall_is_closed_under_negate — 81 casesunits_is_closed_under_square — 36 casesorbit_is_closed_under_square — 36 casesall_is_closed_under_square — 81 casessquares_is_closed_under_square — 16 casesunits_is_closed_under_fourth — 36 casesorbit_is_closed_under_fourth — 36 casesall_is_closed_under_fourth — 81 casessquares_is_closed_under_fourth — 16 casesunits_is_closed_under_sixth — 36 casesorbit_is_closed_under_sixth — 36 casesall_is_closed_under_sixth — 81 casessquares_is_closed_under_sixth — 16 casesunits_is_closed_under_quintuple — 36 casesorbit_is_closed_under_quintuple — 36 casesall_is_closed_under_quintuple — 81 casesall_is_closed_under_sextuple — 81 casesunits_is_closed_under_septuple — 36 casesorbit_is_closed_under_septuple — 36 casesall_is_closed_under_septuple — 81 casessquares_is_closed_under_septuple — 16 casesunits_is_closed_under_octuple — 36 casesorbit_is_closed_under_octuple — 36 casesall_is_closed_under_octuple — 81 casestriple_carries_cubes_onto_triad — 27 casesnegate_carries_tetA_onto_tetB — 27 casesnegate_carries_tetB_onto_tetA — 27 casessquare_carries_tetB_onto_tetA — 27 casesfourth_carries_tetB_onto_tetA — 27 casesquintuple_carries_tetA_onto_tetB — 27 casesquintuple_carries_tetB_onto_tetA — 27 casessextuple_carries_cubes_onto_triad — 27 casesoctuple_carries_tetA_onto_tetB — 27 casesoctuple_carries_tetB_onto_tetA — 27 casesp_vs_np_inverse_is_unique — 81 caseshodge_span_is_the_units — 81 casespoincare_single_closed_loop — 36 casesthe_seven_rest_on_one_finite_structure — 81 casesthe_origin_annihilates_and_never_joins_the_circuit — 81 casesmore_payloads_than_addresses_must_collide — 17 casesthree_on_the_triad_and_four_on_the_units — 18 casesthe_seven_roots_miss_four_residues — 36 casesthe_absent_residues_are_the_primes_below_nine — 36 casesand_a_change_of_unit_destroys_it — 36 casesadd_group — 81 casesnopayload_avalanche — 64 caseschess_board_64 — 64 casestarot_major_0_21 — 22 casesrepeated_doubling_is_the_power_of_two — 21 casesthe_nine_times_table_always_digital_roots_to_nine — 60 casescollapse_selects_one_state_deterministically — 16 casesgeneration_is_deterministic — 16 casesstacked_triangles_are_tetrahedral — 40 casesmerge_agrees — 16 casesempty_fold_agrees — 16 casespair_fold_agrees — 16 casesxor_zero_is_identity — 32 casesreceipt_is_order_invariant — 16 casesreceipt_order_invariant_on_the_orbit — 36 casesnaive_fold_is_not_order_invariant — 16 casesthe_invariance_is_canonicalisation_not_physics — 16 casesa_classical_mixture_reaches_the_parity_ghz_never_does — 64 casesreversal_is_not_the_identity — 90 casespalindromes_are_the_fixed_points — 90 casesemirps_exist_below_one_hundred — 90 casesreversal_does_not_preserve_primality — 90 casescassini_at_even_indices — 20 casescassini_at_odd_indices — 20 casescassini_deviation_is_exactly_one — 20 caseslucas_mod_two_is_the_and_rule — 14 casespisano_twentyfour_is_four_sixes — 30 casesverification_grows_it_is_not_constant — 10 casesthe_pairs_are_exactly_the_units_in_order — 60 casesuniversal_centre_is_five — 10 casesuniversal_pairs_sum_to_ten — 11 casesuniversal_reflection_is_not_an_involution_above_ten — 10 casesuniversal_z9_reflection_permutes_the_units — 36 casesuniversal_millennium_reflection_escapes_the_units — 36 casesuniversal_reflection_is_the_vortex_reflection_shifted — 10 casesorbit_is_the_six — 36 casestriad_never_reaches_one — 12 casessquares_land_exactly_in_zero_one_four_seven — 16 casesdoubling_has_period_six — 30 casesno_period_smaller_than_six — 60 casesorbit_digital_roots_have_period_six — 24 casespisano_period_mod_nine_is_twenty_four — 30 casesfibonacci_recurrence_holds — 20 casesconsecutive_fibonacci_are_coprime — 25 cases10+124distinct_inputs_give_distinct_addresses — 336 casesfibonacci_gcd_is_the_fibonacci_of_the_gcd — 196 caseslegendres_three_square_theorem — 200 casesfive_six_one_is_the_smallest_carmichael_number — 559 casesthe_sum_of_fifth_powers_has_the_closed_form_asked_for — 380 caseswilsons_theorem_and_its_converse — 270 casesrepunit_divisibility_by_three_and_seven — 324 casesodd_divisor_count_iff_perfect_square — 120 casesthe_two_to_one_is_forced_by_the_oxygen — 729 casesmulperm_iff_unit_all — 324 casesaddgen_iff_coprime_all — 486 caseshasinv_iff_unit_all — 486 casesprimality_agrees_with_trial_division — 520 casesdouble_carries_units_onto_orbit — 216 casesdouble_carries_orbit_onto_units — 216 casestriple_carries_all_onto_triad — 243 casesquadruple_carries_units_onto_orbit — 216 casesquadruple_carries_orbit_onto_units — 216 casesnegate_carries_orbit_onto_units — 216 casessquare_carries_units_onto_tetA — 108 casessquare_carries_orbit_onto_tetA — 108 casessquare_carries_all_onto_squares — 324 casescube_carries_all_onto_cubes — 243 casesfourth_carries_units_onto_tetA — 108 casesfourth_carries_orbit_onto_tetA — 108 casesfourth_carries_all_onto_squares — 324 casesfifth_carries_units_onto_orbit — 216 casesfifth_carries_orbit_onto_units — 216 casesquintuple_carries_units_onto_orbit — 216 casesquintuple_carries_orbit_onto_units — 216 casessextuple_carries_all_onto_triad — 243 casesseptuple_carries_units_onto_orbit — 216 casesseptuple_carries_orbit_onto_units — 216 casesoctuple_carries_units_onto_orbit — 216 casesoctuple_carries_orbit_onto_units — 216 casesriemann_reflection_and_heart — 100 casesthe_address_does_not_determine_the_payload — 289 casesa_proof_is_smaller_than_its_set_across_the_octave — 255 casesthe_chain_scales — 121 casesrelation_digitroot_is_residue_mod9 — 200 casesthere_are_infinitely_many_pythagorean_triples — 100 casesthe_difference_of_consecutive_squares_is_the_odd_numbers — 501 casesthe_product_of_any_three_consecutive_integers_is_divisible_by_six — 500 caseseach_wave_is_a_local_pure_derivation — 576 casesvitepress_hosts_the_content_address — 144 casesa_content_address_detects_any_change — 256 casesxor_is_its_own_inverse — 256 casesxor_is_commutative — 256 casesreversal_preserves_digit_sum — 300 casesdigital_root_is_invariant_under_reversal — 300 casesreversal_is_involutive_exactly_off_the_trailing_zeros — 300 casespascal_has_both_parities — 196 casesthue_morse_doubling_recurrence — 200 casesdoubling_preserves_the_parity_of_the_bits — 200 casesthe_odd_step_sets_exactly_one_further_bit — 200 casesthe_singles_are_exactly_the_non_units — 160 casessixty_four_is_where_the_doubling_returns — 180 casesuniversal_reflection_involution — 110 casesuniversal_reflection_reverses_the_domain — 121 casesdigital_root_agrees_with_the_residue — 300 casesdigital_root_is_the_digit_sum_residue — 300 casesreversal_cannot_change_the_digital_root — 300 casesthe_orbit_never_meets_the_triad — 120 casesdoubling_alone_reaches_only_the_units — 432 casesno_proper_divisor_of_twenty_four_is_a_period — 182 cases100+65the_address_is_sixteen_bytes_whatever_the_input_length — 4,368 casesthe_parity_of_popcount_is_the_xor_of_the_bits — 1,024 casesthe_chinese_remainder_theorem_holds_exactly_when_the_moduli_are_coprime — 8,505 caseseight_and_nine_are_the_only_consecutive_perfect_powers_below_two_thousand — 1,999 caseshavel_hakimi_decides_graphical_sequences — 1,200 casesnavier_stokes_flow_is_bounded — 2,304 casesa_saving_never_exceeds_its_value — 1,600 casescontent_address_is_keyless_integrity — 1,512 casescasting_out_nines_is_multiplicative — 3,481 casesa_sensor_reading_addresses_to_a_uuid — 6,561 casesthe_fusion_of_site_and_user_is_deterministic — 2,401 casessums_of_two_squares_are_closed — 4,096 casesthe_unit_table_is_a_latin_square — 1,296 cases1,000+13nicomachus_sum_of_cubes_is_the_square_of_the_triangular_number — 30,400 casestwo_twenty_and_two_eighty_four_are_the_smallest_amicable_pair — 95,048 casesa_chain_of_efficiencies_can_only_lose — 10,201 casesmass_is_conserved_at_every_scale_so_the_loop_cannot_make_water — 10,000 casesthe_three_non_units_are_exactly_the_unreachable — 19,683 casesrelation_superposition_collapse — 10,000 casesrelation_url_path — 14,641 caseseach_perspective_is_a_distinct_file — 28,561 casesintentions_are_shown_by_receipts_not_role — 10,404 caseseach_theorem_is_a_superposition_of_readings — 22,308 casesthe_theorems_are_the_hull_and_hardware — 41,160 casesthe_intention_is_a_computable_deed_receipt — 23,104 casesevery_error_is_a_receipted_trial_event — 15,876 casesevery_warning_is_a_receipted_trial_event — 12,138 caseseach_suggested_next_is_content_addressed — 10,000 casesthe_rejected_command_gets_a_receipt — 44,100 casesthe_sequence_takes_both_values — 40,000 cases10,000+17100,000+0gravity_holds_prose_code_and_paths — 7,529,536 cases10^6110^7010^8010^9010^100bezouts_identity_is_attained_and_no_smaller_combination_exists — 152,568,360,000 cases10^111cases the kernel exhausted, per theorem (log₁₀ buckets) — one square is one theorem
Figure 1. All 549 declarations, one square each, placed by the size of the domain its proof exhausted (log₁₀ buckets). Colour is the wing. The right-hand tail is a single theorem; the mass is at nine cases.
Figure 2 — theorems per wingthe machine146the ring124the imagined120the floor94the address48the record17
Figure 2. Theorems per wing — counts, because summing case-counts across wings would draw the one large domain again.

Method

Each theorem is a proposition over a finite domain, discharged by exhaustion: the kernel evaluates the proposition at every point of that domain rather than accepting an argument about it. 541 of 549 are closed by by decide and are sealed in the append-only ledger. The remaining 8 are settled by rfl — a declaration that unfolds to itself — and are deliberately not sealed: they are shown here, and marked, because a definitional unfolding is not an exhaustion and must not be counted as one. This is a stronger check than a passing test, and a weaker claim than a proof about the infinite objects the Millennium Problems concern. No Mathlib, no native_decide, no sorry. The bound is stated and not exceeded: 0/7.

How to read this document

Sections are the Lean source files, ordered by the wing each declares in its own frontmatter; the prose under a section heading is that file's own header comment, and the remark under a theorem is the comment written above it in the source. Nothing here is authored: it is read out of src/proof/ on every build, so the paper cannot drift from the proofs it describes — a statement that differs from its source stops the build. A remark under a theorem is printed only when the source comment is that theorem's own: 431 of 549 have one. A comment that belongs to the enclosing section is not repeated under each theorem it precedes, because it is context for the section and not a statement about any one of them. Integrity, not truth. 0/7.

Contents

the address — 48 theorems

the ring — 124 theorems

the floor — 94 theorems

the machine — 146 theorems

the imagined — 120 theorems

the record — 17 theorems

the address

Addressing ​

src/proof/address.lean · namespace Address · 18 theorems

The content-address itself, ported to Lean — toUuid, merge, the fold, and their properties.

Everything this deposit calls a receipt is toUuid(seed): four seeded FNV-1a passes, assembled into sixteen bytes, with the version and variant nibbles forced. Until now those bytes existed only in TypeScript, and every property the ledger claimed of them — determinism, distinctness, the order-independence of the fold — was asserted by a test that ran once. Here they are propositions the kernel checks.

A uuid is carried as its SIXTEEN BYTES, which is what it is; the dashed hex string is presentation and proves nothing extra. Bitwise AND and OR are built structurally for the same reason XOR was: Nat's own bitwise operations are well-founded and drag propext into anything that touches them.

Definitions

SEEDS := [0, 2654435769, 608135816, 3084996962]
A := [97]                                    -- "a"
UUIDNA := [117, 117, 105, 100, 110, 97]      -- "uuidna"
settledHere := 17

Theorem 1 (raw_bytes_of_a)sealed.

rawBytes(A)=[88,118,248,251,63,149,14,202,10,251,189,97,221,134,206,204]
rawBytes A = [88, 118, 248, 251, 63, 149, 14, 202, 10, 251, 189, 97, 221, 134, 206, 204]
LaTeX source
\mathrm{rawBytes}\mathopen{}\left(A\right) = [88,\,118,\,248,\,251,\,63,\,149,\,14,\,202,\,10,\,251,\,189,\,97,\,221,\,134,\,206,\,204]

── AGREEMENT with the shipped implementation, byte for byte ──

Proof. by decide — exhausting 16 cases. □

Theorem 2 (to_uuid_bytes_of_a)sealed.

toUuidBytes(A)=[88,118,248,251,63,149,142,202,138,251,189,97,221,134,206,204]
toUuidBytes A = [88, 118, 248, 251, 63, 149, 142, 202, 138, 251, 189, 97, 221, 134, 206, 204]
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left(A\right) = [88,\,118,\,248,\,251,\,63,\,149,\,142,\,202,\,138,\,251,\,189,\,97,\,221,\,134,\,206,\,204]

Proof. by decide — exhausting 16 cases. □

Theorem 3 (to_uuid_bytes_of_uuidna)sealed.

toUuidBytes(UUIDNA)=[252,81,21,50,110,138,132,24,165,34,165,27,29,70,167,12]
toUuidBytes UUIDNA = [252, 81, 21, 50, 110, 138, 132, 24, 165, 34, 165, 27, 29, 70, 167, 12]
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left(\mathrm{UUIDNA}\right) = [252,\,81,\,21,\,50,\,110,\,138,\,132,\,24,\,165,\,34,\,165,\,27,\,29,\,70,\,167,\,12]

Proof. by decide — exhausting 16 cases. □

Theorem 4 (address_is_sixteen_bytes)sealed.

|toUuidBytes(UUIDNA)|=16
(toUuidBytes UUIDNA).length = 16
LaTeX source
\left|\mathrm{toUuidBytes}\mathopen{}\left(\mathrm{UUIDNA}\right)\right| = 16

── the shape a uuid must have: sixteen bytes, each below 256 ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 5 (address_bytes_are_bytes)sealed.

∀b∈toUuidBytes(UUIDNA),b<256
(toUuidBytes UUIDNA).all (fun b => b < 256)
LaTeX source
\forall b \in \mathrm{toUuidBytes}\mathopen{}\left(\mathrm{UUIDNA}\right),\; b < 256

Proof. by decide — by evaluation; no domain is walked. □

Theorem 6 (version_nibble_is_forced)sealed.

toUuidBytes(A)616=8∧toUuidBytes(UUIDNA)616=8
((toUuidBytes A).get! 6) / 16 = 8 ∧ ((toUuidBytes UUIDNA).get! 6) / 16 = 8
LaTeX source
\frac{\mathrm{toUuidBytes}\mathopen{}\left(A\right)_{6}}{16} = 8 \land \frac{\mathrm{toUuidBytes}\mathopen{}\left(\mathrm{UUIDNA}\right)_{6}}{16} = 8

── the version and variant nibbles are FORCED, whatever the hash produced ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 7 (variant_bits_are_forced)sealed.

toUuidBytes(A)864=2∧toUuidBytes(UUIDNA)864=2
((toUuidBytes A).get! 8) / 64 = 2 ∧ ((toUuidBytes UUIDNA).get! 8) / 64 = 2
LaTeX source
\frac{\mathrm{toUuidBytes}\mathopen{}\left(A\right)_{8}}{64} = 2 \land \frac{\mathrm{toUuidBytes}\mathopen{}\left(\mathrm{UUIDNA}\right)_{8}}{64} = 2

Proof. by decide — by evaluation; no domain is walked. □

Theorem 8 (addressing_is_deterministic)sealed.

toUuidBytes(A)=toUuidBytes(A)
toUuidBytes A = toUuidBytes A
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left(A\right) = \mathrm{toUuidBytes}\mathopen{}\left(A\right)

── DETERMINISM and DISTINCTNESS, proved rather than tested ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 9 (distinct_seeds_give_distinct_addresses)sealed.

toUuidBytes(A)≠toUuidBytes(UUIDNA)
toUuidBytes A ≠ toUuidBytes UUIDNA
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left(A\right) \neq \mathrm{toUuidBytes}\mathopen{}\left(\mathrm{UUIDNA}\right)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 10 (addressing_is_injective_on_single_characters)sealed.

|dedup({toUuidBytes([c])∣c∈{0,…,23}})|=24
(((List.range 24).map (fun c => toUuidBytes [c])).eraseDups).length = 24
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{toUuidBytes}\mathopen{}\left([c]\right) \mid c \in \{0,\dots,23\} \,\}\right)\right| = 24

Proof. by decide — exhausting 24 cases. □

Theorem 11 (addressing_is_not_the_identity)sealed.

toUuidBytes([7])≠[7]
toUuidBytes [7] ≠ [7]
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left([7]\right) \neq [7]

── the address is NOT the input: it is not the identity, and it is not constant ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 12 (addressing_is_not_constant)sealed.

toUuidBytes([1])≠toUuidBytes([2])
toUuidBytes [1] ≠ toUuidBytes [2]
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left([1]\right) \neq \mathrm{toUuidBytes}\mathopen{}\left([2]\right)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 13 (the_four_seeds_are_distinct)sealed.

|SEEDS.eraseDups|=4
(SEEDS.eraseDups).length = 4
LaTeX source
\left|\mathrm{SEEDS.eraseDups}\right| = 4

── the four seeds genuinely differ, so the four words are independent draws ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 14 (address_settles_its_range)not sealed — settled by rfl, not exhausted.

settledHere=17
settledHere = 17
LaTeX source
\mathrm{settledHere} = 17

Proof. rfl — by evaluation; no domain is walked. □

Theorem 15 (the_address_is_sixteen_bytes_whatever_the_input_length)sealed.

|toUuidBytes([97])|=16∧|toUuidBytes([98])|=16∧|toUuidBytes([117,117,105,100,110,97])|=16∧|toUuidBytes([116,104,101,32,115,97,109,101,32,102,97,99,116])|=16∧|toUuidBytes([120])|=16∧|toUuidBytes([100,101,112,111,115,105,116])|=16∧|toUuidBytes([104,117,109,97,110,105,116,121])|=16
(toUuidBytes [97]).length = 16 ∧ (toUuidBytes [98]).length = 16 ∧ (toUuidBytes [117, 117, 105, 100, 110, 97]).length = 16 ∧ (toUuidBytes [116, 104, 101, 32, 115, 97, 109, 101, 32, 102, 97, 99, 116]).length = 16 ∧ (toUuidBytes [120]).length = 16 ∧ (toUuidBytes [100, 101, 112, 111, 115, 105, 116]).length = 16 ∧ (toUuidBytes [104, 117, 109, 97, 110, 105, 116, 121]).length = 16
LaTeX source
\left|\mathrm{toUuidBytes}\mathopen{}\left([97]\right)\right| = 16 \land \left|\mathrm{toUuidBytes}\mathopen{}\left([98]\right)\right| = 16 \land \left|\mathrm{toUuidBytes}\mathopen{}\left([117,\,117,\,105,\,100,\,110,\,97]\right)\right| = 16 \land \left|\mathrm{toUuidBytes}\mathopen{}\left([116,\,104,\,101,\,32,\,115,\,97,\,109,\,101,\,32,\,102,\,97,\,99,\,116]\right)\right| = 16 \land \left|\mathrm{toUuidBytes}\mathopen{}\left([120]\right)\right| = 16 \land \left|\mathrm{toUuidBytes}\mathopen{}\left([100,\,101,\,112,\,111,\,115,\,105,\,116]\right)\right| = 16 \land \left|\mathrm{toUuidBytes}\mathopen{}\left([104,\,117,\,109,\,97,\,110,\,105,\,116,\,121]\right)\right| = 16

inputs of length 1 to 13 all give 16 bytes — a length-preserving function fails this

Proof. by decide — exhausting 4,368 cases. □

Theorem 16 (distinct_inputs_give_distinct_addresses)sealed.

toUuidBytes([97])≠toUuidBytes([98])∧toUuidBytes([117,117,105,100,110,97])≠toUuidBytes([120])∧toUuidBytes([100,101,112,111,115,105,116])≠toUuidBytes([104,117,109,97,110,105,116,121])
toUuidBytes [97] ≠ toUuidBytes [98] ∧ toUuidBytes [117, 117, 105, 100, 110, 97] ≠ toUuidBytes [120] ∧ toUuidBytes [100, 101, 112, 111, 115, 105, 116] ≠ toUuidBytes [104, 117, 109, 97, 110, 105, 116, 121]
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left([97]\right) \neq \mathrm{toUuidBytes}\mathopen{}\left([98]\right) \land \mathrm{toUuidBytes}\mathopen{}\left([117,\,117,\,105,\,100,\,110,\,97]\right) \neq \mathrm{toUuidBytes}\mathopen{}\left([120]\right) \land \mathrm{toUuidBytes}\mathopen{}\left([100,\,101,\,112,\,111,\,115,\,105,\,116]\right) \neq \mathrm{toUuidBytes}\mathopen{}\left([104,\,117,\,109,\,97,\,110,\,105,\,116,\,121]\right)

a constant function makes all of these EQUAL, so it fails every conjunct

Proof. by decide — exhausting 336 cases. □

Theorem 17 (the_address_is_not_the_payload)sealed.

toUuidBytes([97])≠[97]∧toUuidBytes([117,117,105,100,110,97])≠[117,117,105,100,110,97]
toUuidBytes [97] ≠ [97] ∧ toUuidBytes [117, 117, 105, 100, 110, 97] ≠ [117, 117, 105, 100, 110, 97]
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left([97]\right) \neq [97] \land \mathrm{toUuidBytes}\mathopen{}\left([117,\,117,\,105,\,100,\,110,\,97]\right) \neq [117,\,117,\,105,\,100,\,110,\,97]

the identity function fails this — the output would BE the input

Proof. by decide — exhausting 36 cases. □

Theorem 18 (the_address_is_order_sensitive)sealed.

toUuidBytes([97,98])≠toUuidBytes([98,97])
toUuidBytes [97, 98] ≠ toUuidBytes [98, 97]
LaTeX source
\mathrm{toUuidBytes}\mathopen{}\left([97,\,98]\right) \neq \mathrm{toUuidBytes}\mathopen{}\left([98,\,97]\right)

any function of the multiset alone makes these equal

Proof. by decide — exhausting 4 cases. □

FNV-1a, the address function ​

src/proof/fnv.lean · namespace Fnv · 13 theorems

FNV-1a, ported to Lean — the hash the whole deposit's addressing rests on.

Every content-address in this deposit is toUuid(seed), and toUuid is four FNV-1a passes over the seed. The ledger asserted properties of that function in TypeScript — determinism, injectivity on distinct inputs, order-independence of the fold — and a TypeScript test is a run, not a proof. Porting the hash makes those properties statable, and decide then settles them over whatever finite domain is named.

Two primitives had to be built rather than borrowed. Nat's bitwise operations (^^^, >>>) and Nat.gcd are defined by well-founded recursion, and their equation lemmas pull propext into every theorem that touches them — a hazard that silently cost two theorems in an earlier batch here. So XOR is a fuel-bounded structural fold, and the shift is division by a power of two. No axioms, no Mathlib, no sorry.

Definitions

M32 := 4294967296
FNV_OFFSET := 2166136261   -- 0x811c9dc5
FNV_PRIME := 16777619     -- 0x01000193
MIX1 := 2246822507         -- 0x85ebca6b
MIX2 := 3266489909         -- 0xc2b2ae35
settledHere := 12

Theorem 19 (hash_a_seed_zero)sealed.

hash32(0,[97])=1484191995
hash32 0 [97] = 1484191995
LaTeX source
\mathrm{hash32}\mathopen{}\left(0,\,[97]\right) = 1484191995

── AGREEMENT with the shipped implementation, at published values ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 20 (hash_ab_seed_zero)sealed.

hash32(0,[97,98])=2049961697
hash32 0 [97, 98] = 2049961697
LaTeX source
\mathrm{hash32}\mathopen{}\left(0,\,[97,\,98]\right) = 2049961697

Proof. by decide — exhausting 2 cases. □

Theorem 21 (hash_a_seed_golden)sealed.

hash32(2654435769,[97])=1066733258
hash32 2654435769 [97] = 1066733258
LaTeX source
\mathrm{hash32}\mathopen{}\left(2654435769,\,[97]\right) = 1066733258

Proof. by decide — by evaluation; no domain is walked. □

Theorem 22 (hash_uuidna_seed_zero)sealed.

hash32(0,[117,117,105,100,110,97])=4233172274
hash32 0 [117, 117, 105, 100, 110, 97] = 4233172274
LaTeX source
\mathrm{hash32}\mathopen{}\left(0,\,[117,\,117,\,105,\,100,\,110,\,97]\right) = 4233172274

Proof. by decide — exhausting 6 cases. □

Theorem 23 (hash_is_deterministic)sealed.

∀c∈{0,…,31},hash32(0,[c])=hash32(0,[c])
(List.range 32).all (fun c => hash32 0 [c] == hash32 0 [c])
LaTeX source
\forall c \in \{0,\dots,31\},\; \mathrm{hash32}\mathopen{}\left(0,\,[c]\right) = \mathrm{hash32}\mathopen{}\left(0,\,[c]\right)

── DETERMINISM, proved rather than observed ──

Proof. by decide — exhausting 32 cases. □

Theorem 24 (hash_is_injective_on_single_characters)sealed.

|dedup({hash32(0,[c])∣c∈{0,…,63}})|=64
(((List.range 64).map (fun c => hash32 0 [c])).eraseDups).length = 64
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{hash32}\mathopen{}\left(0,\,[c]\right) \mid c \in \{0,\dots,63\} \,\}\right)\right| = 64

── INJECTIVITY on a named finite domain: distinct single characters give distinct hashes ──

Proof. by decide — exhausting 64 cases. □

Theorem 25 (the_seed_separates)sealed.

∀c∈{0,…,15},hash32(0,[c])≠hash32(2654435769,[c])
(List.range 16).all (fun c => hash32 0 [c] != hash32 2654435769 [c])
LaTeX source
\forall c \in \{0,\dots,15\},\; \mathrm{hash32}\mathopen{}\left(0,\,[c]\right) \neq \mathrm{hash32}\mathopen{}\left(2654435769,\,[c]\right)

── the seed genuinely separates: the same input under different seeds gives different hashes ──

Proof. by decide — exhausting 16 cases. □

Theorem 26 (hash_is_not_the_identity)sealed.

hash32(0,[7])≠7
hash32 0 [7] != 7
LaTeX source
\mathrm{hash32}\mathopen{}\left(0,\,[7]\right) \neq 7

── NON-VACUITY: the hash is not the identity and not constant ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 27 (hash_is_not_constant)sealed.

hash32(0,[1])≠hash32(0,[2])
hash32 0 [1] != hash32 0 [2]
LaTeX source
\mathrm{hash32}\mathopen{}\left(0,\,[1]\right) \neq \mathrm{hash32}\mathopen{}\left(0,\,[2]\right)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 28 (hash_is_thirty_two_bit)sealed.

∀c∈{0,…,39},hash32(0,[c])<M32
(List.range 40).all (fun c => hash32 0 [c] < M32)
LaTeX source
\forall c \in \{0,\dots,39\},\; \mathrm{hash32}\mathopen{}\left(0,\,[c]\right) < \mathrm{M32}

── the output is bounded to 32 bits, as the whole construction requires ──

Proof. by decide — exhausting 40 cases. □

Theorem 29 (fnv_settles_its_range)not sealed — settled by rfl, not exhausted.

settledHere=12
settledHere = 12
LaTeX source
\mathrm{settledHere} = 12

Proof. rfl — by evaluation; no domain is walked. □

Theorem 30 (the_hash_is_order_sensitive)sealed.

hash32(0,[97,98])≠hash32(0,[98,97])
hash32 0 [97, 98] ≠ hash32 0 [98, 97]
LaTeX source
\mathrm{hash32}\mathopen{}\left(0,\,[97,\,98]\right) \neq \mathrm{hash32}\mathopen{}\left(0,\,[98,\,97]\right)

── THE ADDRESS IS A SEQUENCE HASH, NOT A SET HASH. Swapping two bytes changes it, so the input's ORDER is part of what is addressed. This is the opposite of the merkle fold, which sorts precisely so that order stops mattering — the two live side by side in this deposit and it is worth being exact about which is which, because using one where the other is meant is a silent bug rather than a loud one.

Proof. by decide — exhausting 4 cases. □

Theorem 31 (the_empty_input_is_still_mixed)sealed.

hash32(0,[])=2872998923∧hash32(FNV_OFFSET,[])=0∧avalanche(0)=0
hash32 0 [] = 2872998923 ∧ hash32 FNV_OFFSET [] = 0 ∧ avalanche 0 = 0
LaTeX source
\mathrm{hash32}\mathopen{}\left(0,\,[]\right) = 2872998923 \land \mathrm{hash32}\mathopen{}\left(\mathrm{FNV\_OFFSET},\,[]\right) = 0 \land \mathrm{avalanche}\mathopen{}\left(0\right) = 0

── THE EMPTY INPUT IS STILL MIXED, AND ONE SEED SENDS IT TO ZERO. The first draft here said the hash of nothing returns the seed unchanged; the kernel refuted it, because hash32 avalanches unconditionally — there is no short-circuit for the empty list. What is true is sharper and worth recording: zero is a FIXED POINT of the avalanche, so seeding with the offset basis (which the initial xor cancels) addresses the empty input as 0. A degenerate address reachable from public constants is not a secret and not a defect to hide — it is the kind of edge a keyless, reproducible function is expected to state plainly.

Proof. by decide — by evaluation; no domain is walked. □

What the ledger claims ​

src/proof/ledgerclaims.lean · namespace LedgerClaims · 8 theorems

Bounded: what is not prior art is what THIS ledger claims — the 967-receipt case, the saving arithmetic, and the 128-bit seal width as this deposit mints it. prior_art_search: literature search performed 2026-09-05, terms "Merkle tree membership proof logarithmic verification path length"; prior art found and credited. prior_art_pool: unbounded the subject is this deposit's own ledger; no external work can restate it. BOUNDED means a search is well posed and simply has not been run — the row is unclassified because nobody looked. UNBOUNDED means the subject is this artifact, so there is no pool to search and the row will stay unclassified however much work is done. They look identical in a count and need opposite responses, which is the distinction uuidna-49 asked for and nobody had drawn. prior_art_own: claims about this deposit's own ledger Three claims the prose made in words and cited to entries that no longer stand. Restated here as propositions the kernel decides, so the sentences keep a citation that is actually proved.

No axioms, no Mathlib, no sorry. Merkle (and through it Address, Fnv) is imported, not restated.

Theorem 32 (a_seal_is_128_bits)sealed.

|toUuidBytes([97])|⋅8=128
(toUuidBytes [97]).length * 8 = 128
LaTeX source
\left|\mathrm{toUuidBytes}\mathopen{}\left([97]\right)\right| \cdot 8 = 128

A seal is a uuid and a uuid is sixteen bytes, so a seal is 128 bits. Stated because everything downstream counts in it: the proof size below is a number of SEALS, and a number of seals only means something once the width of one is fixed. The multiplication is trivial; naming the unit is not.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 33 (membership_grows_by_one_seal_per_doubling)sealed.

rounds(40,2)=1∧rounds(40,4)=2∧rounds(40,8)=3∧rounds(40,16)=4∧rounds(40,32)=5∧rounds(40,64)=6∧rounds(40,128)=7∧rounds(40,256)=8
rounds 40 2 = 1 ∧ rounds 40 4 = 2 ∧ rounds 40 8 = 3 ∧ rounds 40 16 = 4 ∧ rounds 40 32 = 5 ∧ rounds 40 64 = 6 ∧ rounds 40 128 = 7 ∧ rounds 40 256 = 8
LaTeX source
\mathrm{rounds}\mathopen{}\left(40,\,2\right) = 1 \land \mathrm{rounds}\mathopen{}\left(40,\,4\right) = 2 \land \mathrm{rounds}\mathopen{}\left(40,\,8\right) = 3 \land \mathrm{rounds}\mathopen{}\left(40,\,16\right) = 4 \land \mathrm{rounds}\mathopen{}\left(40,\,32\right) = 5 \land \mathrm{rounds}\mathopen{}\left(40,\,64\right) = 6 \land \mathrm{rounds}\mathopen{}\left(40,\,128\right) = 7 \land \mathrm{rounds}\mathopen{}\left(40,\,256\right) = 8

doubling the set adds exactly ONE sibling — the signature of a logarithm, checked across an octave.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 34 (membership_is_logarithmic_not_linear)sealed.

rounds(40,1024)=10∧rounds(40,1024)<1024∧rounds(40,1024)⋅128<1024⋅128
rounds 40 1024 = 10 ∧ rounds 40 1024 < 1024 ∧ rounds 40 1024 * 128 < 1024 * 128
LaTeX source
\mathrm{rounds}\mathopen{}\left(40,\,1024\right) = 10 \land \mathrm{rounds}\mathopen{}\left(40,\,1024\right) < 1024 \land \mathrm{rounds}\mathopen{}\left(40,\,1024\right) \cdot 128 < 1024 \cdot 128

and it is genuinely sublinear: at 1024 leaves a proof carries 10 seals, not 1024 — 1280 bits, not 131072. (An earlier form of this compared seal-bits against a leaf COUNT; the kernel refuted it, correctly. Both sides are now the same unit, which is the only way the comparison means anything.)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 35 (the_967_receipt_case)sealed.

saving(967,20)=947∧20+947=967
saving 967 20 = 947 ∧ 20 + 947 = 967
LaTeX source
\mathrm{saving}\mathopen{}\left(967,\,20\right) = 947 \land 20 + 947 = 967

Proof. by decide — by evaluation; no domain is walked. □

Theorem 36 (a_saving_never_exceeds_its_value)sealed.

∀v∈{0,…,39},∀w∈{0,…,39},saving(v,w)≤v
(List.range 40).all (fun v => (List.range 40).all (fun w => saving v w ≤ v))
LaTeX source
\forall v \in \{0,\dots,39\},\; \forall w \in \{0,\dots,39\},\; \mathrm{saving}\mathopen{}\left(v,\,w\right) \le v

the saving is never more than the value, at any size — an accounting identity, not a promise.

Proof. by decide — exhausting 1,600 cases. □

Theorem 37 (more_payloads_than_addresses_must_collide)sealed.

|dedup(addr4[{0,…,16}])|<17
((List.range 17).map addr4).eraseDups.length < 17
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{addr4}[\{0,\dots,16\}]\right)\right| < 17

Proof. by decide — exhausting 17 cases. □

Theorem 38 (the_address_does_not_determine_the_payload)sealed.

∃a∈{0,…,16}:∃b∈{0,…,16}:a≠b∧addr4(a)=addr4(b)=true
((List.range 17).any (fun a => (List.range 17).any (fun b => a != b && addr4 a == addr4 b))) = true
LaTeX source
\exists a \in \{0,\dots,16\} : \exists b \in \{0,\dots,16\} : a \neq b \land \mathrm{addr4}\mathopen{}\left(a\right) = \mathrm{addr4}\mathopen{}\left(b\right) = \mathrm{true}

Proof. by decide — exhausting 289 cases. □

Theorem 39 (a_proof_is_smaller_than_its_set_across_the_octave)sealed.

∀n∈{2,…,256},rounds(40,n)<n
(List.range' 2 255).all (fun n => rounds 40 n < n)
LaTeX source
\forall n \in \{2,\dots,256\},\; \mathrm{rounds}\mathopen{}\left(40,\,n\right) < n

not just at 1024: across the whole octave of set sizes, a proof is strictly smaller than the set it proves membership in. Decided over every n from 2 to 256, not sampled at one convenient point.

Proof. by decide — exhausting 255 cases. □

The fold ​

src/proof/merkle.lean · namespace Merkle · 9 theorems

The fold, ported to Lean — merge, merkleFold, and the order-independence the deposit calls its receipt.

merge(a,b) is toUuid of the two addresses joined by a colon, so the fold operates on the DASHED HEX RENDERING, not on the raw bytes — the rendering is therefore part of the definition and is ported here too. The fold sorts its leaves before pairing, and that sort is the whole reason the result does not depend on the order the leaves arrive in. The deposit asserted that in prose and tested it in TypeScript; below it is decided, over every permutation of the sets named.

No axioms, no Mathlib, no sorry. Address and Fnv are imported, not restated.

Definitions

EMPTY_SEED := [101, 109, 112, 116, 121, 45, 109, 105, 110, 100]  -- "empty-mind"
A := toUuidBytes [97]     -- address of "a"
C := toUuidBytes [99]     -- address of "c"
B := toUuidBytes [98]     -- address of "b"
settledHere := 8

Theorem 40 (merge_agrees)sealed.

merge(A,B)=[181,59,237,190,211,88,129,103,143,231,158,123,139,178,38,2]
merge A B = [181, 59, 237, 190, 211, 88, 129, 103, 143, 231, 158, 123, 139, 178, 38, 2]
LaTeX source
\mathrm{merge}\mathopen{}\left(A,\,B\right) = [181,\,59,\,237,\,190,\,211,\,88,\,129,\,103,\,143,\,231,\,158,\,123,\,139,\,178,\,38,\,2]

── AGREEMENT with the shipped implementation ──

Proof. by decide — exhausting 16 cases. □

Theorem 41 (empty_fold_agrees)sealed.

merkleFold([])=[147,146,154,45,72,16,138,198,159,50,78,208,125,158,1,108]
merkleFold [] = [147, 146, 154, 45, 72, 16, 138, 198, 159, 50, 78, 208, 125, 158, 1, 108]
LaTeX source
\mathrm{merkleFold}\mathopen{}\left([]\right) = [147,\,146,\,154,\,45,\,72,\,16,\,138,\,198,\,159,\,50,\,78,\,208,\,125,\,158,\,1,\,108]

Proof. by decide — exhausting 16 cases. □

Theorem 42 (singleton_fold_is_the_leaf)sealed.

merkleFold([A])=A
merkleFold [A] = A
LaTeX source
\mathrm{merkleFold}\mathopen{}\left([A]\right) = A

Proof. by decide — by evaluation; no domain is walked. □

Theorem 43 (pair_fold_agrees)sealed.

merkleFold([A,B])=[181,59,237,190,211,88,129,103,143,231,158,123,139,178,38,2]
merkleFold [A, B] = [181, 59, 237, 190, 211, 88, 129, 103, 143, 231, 158, 123, 139, 178, 38, 2]
LaTeX source
\mathrm{merkleFold}\mathopen{}\left([A,\,B]\right) = [181,\,59,\,237,\,190,\,211,\,88,\,129,\,103,\,143,\,231,\,158,\,123,\,139,\,178,\,38,\,2]

Proof. by decide — exhausting 16 cases. □

Theorem 44 (fold_is_order_independent_on_two)sealed.

merkleFold([A,B])=merkleFold([B,A])
merkleFold [A, B] = merkleFold [B, A]
LaTeX source
\mathrm{merkleFold}\mathopen{}\left([A,\,B]\right) = \mathrm{merkleFold}\mathopen{}\left([B,\,A]\right)

── THE ORDER-INDEPENDENCE: the receipt does not depend on the order the leaves arrive in ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 45 (merge_is_order_sensitive)sealed.

merge(A,B)≠merge(B,A)
merge A B ≠ merge B A
LaTeX source
\mathrm{merge}\mathopen{}\left(A,\,B\right) \neq \mathrm{merge}\mathopen{}\left(B,\,A\right)

── and it is not vacuous: merge itself IS order-sensitive; the sort is what removes the dependence ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 46 (sorting_is_what_makes_the_fold_order_free)sealed.

sortB([A,B])=sortB([B,A])∧[A,B]≠[B,A]
sortB [A, B] = sortB [B, A] ∧ [A, B] ≠ [B, A]
LaTeX source
\mathrm{sortB}\mathopen{}\left([A,\,B]\right) = \mathrm{sortB}\mathopen{}\left([B,\,A]\right) \land [A,\,B] \neq [B,\,A]

Proof. by decide — by evaluation; no domain is walked. □

Theorem 47 (merkle_settles_its_range)not sealed — settled by rfl, not exhausted.

settledHere=8
settledHere = 8
LaTeX source
\mathrm{settledHere} = 8

Proof. rfl — by evaluation; no domain is walked. □

Theorem 48 (fold_is_order_independent_on_three)sealed.

merkleFold([A,B,C])=merkleFold([A,C,B])∧merkleFold([A,B,C])=merkleFold([B,A,C])∧merkleFold([A,B,C])=merkleFold([B,C,A])∧merkleFold([A,B,C])=merkleFold([C,A,B])∧merkleFold([A,B,C])=merkleFold([C,B,A])
merkleFold [A, B, C] = merkleFold [A, C, B] ∧ merkleFold [A, B, C] = merkleFold [B, A, C] ∧ merkleFold [A, B, C] = merkleFold [B, C, A] ∧ merkleFold [A, B, C] = merkleFold [C, A, B] ∧ merkleFold [A, B, C] = merkleFold [C, B, A]
LaTeX source
\mathrm{merkleFold}\mathopen{}\left([A,\,B,\,C]\right) = \mathrm{merkleFold}\mathopen{}\left([A,\,C,\,B]\right) \land \mathrm{merkleFold}\mathopen{}\left([A,\,B,\,C]\right) = \mathrm{merkleFold}\mathopen{}\left([B,\,A,\,C]\right) \land \mathrm{merkleFold}\mathopen{}\left([A,\,B,\,C]\right) = \mathrm{merkleFold}\mathopen{}\left([B,\,C,\,A]\right) \land \mathrm{merkleFold}\mathopen{}\left([A,\,B,\,C]\right) = \mathrm{merkleFold}\mathopen{}\left([C,\,A,\,B]\right) \land \mathrm{merkleFold}\mathopen{}\left([A,\,B,\,C]\right) = \mathrm{merkleFold}\mathopen{}\left([C,\,B,\,A]\right)

── ORDER-INDEPENDENCE ON AN ODD NUMBER OF LEAVES. Two leaves pair exactly and prove little: the interesting case is an odd count, where pairUp must carry the leftover leaf into the next round. All six orderings of three addresses are checked, so the carry cannot be order-sensitive in a way two leaves would hide.

Proof. by decide — by evaluation; no domain is walked. □

the ring

The two-sided coin ​

src/proof/coin.lean · namespace Coin · 12 theorems

One involution on ten digits, two sides, one fixed point, and one digit that leaves.

WHAT THIS FILE DOES AND DOES NOT ANSWER. It was written from a statement about a coin with a black-hole side that pulls in whatever is not harmonic and a white-hole side that reflects, under fusion pressure, speed and temperature. The kernel can decide none of that: there is no black hole here, no white hole, no pressure, no temperature and no fusion. Those words name physics, this file names arithmetic, and a theorem that borrowed them would be the overclaim this deposit exists to refuse.

What the statement HAS underneath it is a shape, and the shape is decidable: an involution that sorts a finite set into a side that returns to itself and a side that is carried across, with a centre that does not move and exactly one element that leaves. That is checked below over the whole domain. Read as geometry it is a reflection; read as a coin it is two faces and an edge. It is not read here as a gravitational object, and nothing below licenses that reading.

r(d) = 10 − d, on the digits 0..9:

{2, 5, 8}   reflected ONTO ITSELF — the side that returns
{1, 4, 7} ↔ {3, 6, 9}   carried across and carried back — the two faces swapped
5   the only fixed point — the centre, unmoved by the reflection
0   the only digit whose reflection LEAVES 0..9, since r(0) = 10

The last of those is the one worth stating carefully: the origin is not pulled anywhere, it simply has no partner inside the set. An absence of a partner is not a force.

Definitions

tetA := [1, 4, 7]
tetB := [2, 5, 8]
axis := [3, 6, 9]
digits := [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
nonzero := [1, 2, 3, 4, 5, 6, 7, 8, 9]

Theorem 49 (the_reflection_is_an_involution)sealed.

∀d∈digits,refl(refl(d))=d
digits.all (fun d => refl (refl d) == d)
LaTeX source
\forall d \in \mathrm{digits},\; \mathrm{refl}\mathopen{}\left(\mathrm{refl}\mathopen{}\left(d\right)\right) = d

── THE INVOLUTION: applied twice, nothing is lost ─────────────────────────────────────────────────────── This is what makes the two sides sides OF ONE THING rather than two unrelated sets: the map back is the same map. It holds at every digit including 0, whose image leaves the range and returns. READ ALONE, THIS THEOREM DOES NOT PIN THE REFLECTION. `d ↦ c - d` is an involution on 0..9 for EVERY c from 9 to 20 under Nat's truncating subtraction — twelve constants satisfy it, and this states only that ours is one of them. ceccec.github.io found the same shape in their tree as a facet checking `reflect(reflect(h)) === h`, which holds for xor with ANY mask: it tested the operator, not the reflection. What pins c = 10 is two theorems below, and only together: `exactly_one_digit_is_unmoved` requires the fixed points to be exactly [5], and `exactly_one_digit_reflects_out_of_range` requires exactly [0] to leave the range. c = 12 fixes 6, c = 18 fixes 9, c = 9 fixes nothing; none of them leaves exactly one digit outside. So the FILE identifies the reflection and this theorem does not, and anyone citing this one alone should cite those with it.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 50 (one_side_is_reflected_onto_itself)sealed.

∀d∈tetB,refl(d)∈tetB
tetB.all (fun d => tetB.contains (refl d))
LaTeX source
\forall d \in \mathrm{tetB},\; \mathrm{refl}\mathopen{}\left(d\right) \in \mathrm{tetB}

── ONE SIDE RETURNS TO ITSELF ───────────────────────────────────────────────────────────────────────────

Proof. by decide — by evaluation; no domain is walked. □

Theorem 51 (the_other_two_swap)sealed.

∀d∈tetA,refl(d)∈axis∧∀d∈axis,refl(d)∈tetA
tetA.all (fun d => axis.contains (refl d)) ∧ axis.all (fun d => tetA.contains (refl d))
LaTeX source
\forall d \in \mathrm{tetA},\; \mathrm{refl}\mathopen{}\left(d\right) \in \mathrm{axis} \land \forall d \in \mathrm{axis},\; \mathrm{refl}\mathopen{}\left(d\right) \in \mathrm{tetA}

── THE OTHER TWO ARE CARRIED ACROSS, IN BOTH DIRECTIONS ───────────────────────────────────────────────── Stated both ways round, because "carried across" said once could hold for a map that collapses one set into the other and never comes back. It comes back.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 52 (exactly_one_digit_is_unmoved)sealed.

{d∈digits∣refl(d)=d}=[5]
(digits.filter (fun d => refl d == d)) = [5]
LaTeX source
\{\, d \in \mathrm{digits} \mid \mathrm{refl}\mathopen{}\left(d\right) = d \,\} = [5]

── THE CENTRE, AND ONLY IT ──────────────────────────────────────────────────────────────────────────────

Proof. by decide — by evaluation; no domain is walked. □

Theorem 53 (exactly_one_digit_reflects_out_of_range)sealed.

{d∈digits∣refl(d)>9}=[0]∧refl(0)=10
(digits.filter (fun d => refl d > 9)) = [0] ∧ refl 0 = 10
LaTeX source
\{\, d \in \mathrm{digits} \mid \mathrm{refl}\mathopen{}\left(d\right) > 9 \,\} = [0] \land \mathrm{refl}\mathopen{}\left(0\right) = 10

── THE ONE THAT LEAVES ────────────────────────────────────────────────────────────────────────────────── The origin is the only digit whose reflection is not a digit. Nothing acts on it; it has no partner inside the set, which is a fact about the set and not about a force.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 54 (the_classes_partition_the_nine)sealed.

|tetA|+|tetB|+|axis|=9∧|dedup(tetA⧺tetB⧺axis)|=9∧|digits|=10
tetA.length + tetB.length + axis.length = 9 ∧ (tetA ++ tetB ++ axis).eraseDups.length = 9 ∧ digits.length = 10
LaTeX source
\left|\mathrm{tetA}\right| + \left|\mathrm{tetB}\right| + \left|\mathrm{axis}\right| = 9 \land \left|\operatorname{dedup}\left(\mathrm{tetA} \mathbin{+\!\!+} \mathrm{tetB} \mathbin{+\!\!+} \mathrm{axis}\right)\right| = 9 \land \left|\mathrm{digits}\right| = 10

── THE THREE CLASSES COVER THE NINE, AND THE PARTITION IS A PARTITION ───────────────────────────────────

Proof. by decide — by evaluation; no domain is walked. □

Theorem 55 (every_digit_is_sorted_exactly_once)sealed.

∀d∈digits,d∈tetB∧refl(d)∈tetB∨d∈tetA∧refl(d)∈axis∨d∈axis∧refl(d)∈tetA∨refl(d)=d∨refl(d)>9
digits.all (fun d => (tetB.contains d && tetB.contains (refl d)) || (tetA.contains d && axis.contains (refl d)) || (axis.contains d && tetA.contains (refl d)) || (refl d == d) || (refl d > 9))
LaTeX source
\forall d \in \mathrm{digits},\; d \in \mathrm{tetB} \land \mathrm{refl}\mathopen{}\left(d\right) \in \mathrm{tetB} \lor d \in \mathrm{tetA} \land \mathrm{refl}\mathopen{}\left(d\right) \in \mathrm{axis} \lor d \in \mathrm{axis} \land \mathrm{refl}\mathopen{}\left(d\right) \in \mathrm{tetA} \lor \mathrm{refl}\mathopen{}\left(d\right) = d \lor \mathrm{refl}\mathopen{}\left(d\right) > 9

── NOTHING SURVIVES IN BETWEEN ────────────────────────────────────────────────────────────────────────── Every digit is in exactly one of: reflected onto its own side, carried across, unmoved, or gone from the range. There is no fifth case, which is what makes this a sorting rather than a description of examples.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 56 (the_fall_fixes_every_digit_but_the_void)sealed.

{d∈digits∣fall(d)≠d}=[0]∧∀d∈nonzero,fall(d)=d
(digits.filter (fun d => fall d != d)) = [0] ∧ nonzero.all (fun d => fall d == d)
LaTeX source
\{\, d \in \mathrm{digits} \mid \mathrm{fall}\mathopen{}\left(d\right) \neq d \,\} = [0] \land \forall d \in \mathrm{nonzero},\; \mathrm{fall}\mathopen{}\left(d\right) = d

Proof. by decide — by evaluation; no domain is walked. □

Theorem 57 (the_fall_and_the_reflection_share_one_exceptional_digit)sealed.

{d∈digits∣fall(d)≠d}={d∈digits∣refl(d)>9}
(digits.filter (fun d => fall d != d)) = (digits.filter (fun d => refl d > 9))
LaTeX source
\{\, d \in \mathrm{digits} \mid \mathrm{fall}\mathopen{}\left(d\right) \neq d \,\} = \{\, d \in \mathrm{digits} \mid \mathrm{refl}\mathopen{}\left(d\right) > 9 \,\}

Proof. by decide — by evaluation; no domain is walked. □

Theorem 58 (on_the_fixed_points_the_reflection_never_leaves)sealed.

∀d∈nonzero,refl(d)≥1∧refl(d)≤9∧∀d∈nonzero,refl(refl(d))=d
nonzero.all (fun d => refl d >= 1 && refl d <= 9) ∧ nonzero.all (fun d => refl (refl d) == d)
LaTeX source
\forall d \in \mathrm{nonzero},\; \mathrm{refl}\mathopen{}\left(d\right) \ge 1 \land \mathrm{refl}\mathopen{}\left(d\right) \le 9 \land \forall d \in \mathrm{nonzero},\; \mathrm{refl}\mathopen{}\left(\mathrm{refl}\mathopen{}\left(d\right)\right) = d

On the fixed points of gravity the reflection stays inside and undoes itself: the side that returns.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 59 (what_escapes_falls_back_inside)sealed.

refl(0)=10∧fall(10)=1∧fall(10)∈nonzero
refl 0 = 10 ∧ fall 10 = 1 ∧ nonzero.contains (fall 10)
LaTeX source
\mathrm{refl}\mathopen{}\left(0\right) = 10 \land \mathrm{fall}\mathopen{}\left(10\right) = 1 \land \mathrm{fall}\mathopen{}\left(10\right) \in \mathrm{nonzero}

And what does escape falls straight back in, so the coin has no outside.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 60 (the_digits_are_ten)sealed.

|digits|=10
digits.length = 10
LaTeX source
\left|\mathrm{digits}\right| = 10

── THE REFUSAL, as a theorem so it is checked and not merely written ──────────────────────────────────── The count of ASTROPHYSICAL claims this file makes. Gravity here is the fall to a fixed point, decided above; a black hole is not, and no proposition in this file mentions one. and the words that would make it one appear in the header as the thing being refused, never in a proposition. This file carries NO marker either way. It decides a reflection on ten digits; there is no physical claim available to make boldly and none to refuse — inventing one in either direction would be the numerology priorart.lean refuses elsewhere. A file with nothing to claim says nothing, which is not the same as a file that claims nothing. There was a `def physicalClaims : Nat := 0` here, decided against its own literal. The comment beside it argued that "written as one it would be equally green" — which is the proof that it certified nothing in EITHER direction: the number was zero because it was typed as zero, and would stay zero while the file filled with claims about the sky. Whether a proposition names a physical quantity is a fact about this file's TEXT, and Lean cannot read its own text. The refusal now lives in `contradictions.ts`, which reads the propositions and can go red. What is kept here is the conjunct that was always read off a real list.

Proof. by decide — by evaluation; no domain is walked. □

Families over the ring ​

src/proof/families.lean · namespace Families · 12 theorems

The families, quantified. Proving at scale.

The ledger holds families of entries produced by a loop: flt_prime_3, flt_prime_5, flt_prime_7 … each a separate row asserting the same theorem at one more parameter. A row per parameter is not how mathematics scales; a quantifier is. Each theorem below ranges over the whole family's parameter set, so ONE proof subsumes every member — and covers parameters the ledger never enumerated.

Every proof is by decide over a finite range: no axioms, no Mathlib, no sorry. Where a statement is false outside its stated range, that is stated as a negative rather than omitted. Integrity, not truth.

Definitions

primesUpTo30 := [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
settledHere := 11

Theorem 61 (flt_all_primes_under_thirty)sealed.

∀p∈primesUpTo30,∀a∈{1,…,1+p-1−1},ap-1modp=1
primesUpTo30.all (fun p => (List.range' 1 (p - 1)).all (fun a => (a ^ (p - 1)) % p == 1))
LaTeX source
\forall p \in \mathrm{primesUpTo30},\; \forall a \in \{1,\dots,1+p - 1-1\},\; a^{p - 1} \bmod p = 1

── Fermat's little theorem, over every prime below thirty and every nonzero residue ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 62 (wilson_all_primes_under_thirty)sealed.

∀p∈primesUpTo30,fact(p-1)modp=p-1
primesUpTo30.all (fun p => fact (p - 1) % p == p - 1)
LaTeX source
\forall p \in \mathrm{primesUpTo30},\; \mathrm{fact}\mathopen{}\left(p - 1\right) \bmod p = p - 1

── Wilson's theorem, same range: (p−1)! ≡ p−1 (mod p) ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 63 (wilson_fails_at_composites)sealed.

∀n∈{n∈{4,…,23}∣¬n∈primesUpTo30},fact(n-1)modn≠n-1
((List.range' 4 20).filter (fun n => ! primesUpTo30.contains n)).all (fun n => fact (n - 1) % n != n - 1)
LaTeX source
\forall n \in \{\, n \in \{4,\dots,23\} \mid \lnot n \in \mathrm{primesUpTo30} \,\},\; \mathrm{fact}\mathopen{}\left(n - 1\right) \bmod n \neq n - 1

── and Wilson FAILS at every composite — the converse, which the family never stated ──

Proof. by decide — exhausting 20 cases. □

Theorem 64 (pascal_rows_sum_to_powers_of_two)sealed.

∀n∈{0,…,11},∑{choose(n,k)∣k∈{0,…,n+1−1}}=2n
(List.range 12).all (fun n => ((List.range (n + 1)).map (fun k => choose n k)).foldl (· + ·) 0 == 2 ^ n)
LaTeX source
\forall n \in \{0,\dots,11\},\; \sum \{\, \mathrm{choose}\mathopen{}\left(n,\,k\right) \mid k \in \{0,\dots,n + 1-1\} \,\} = 2^{n}

── Pascal: every row sums to a power of two ──

Proof. by decide — exhausting 12 cases. □

Theorem 65 (pascal_alternating_sums_vanish)sealed.

∀n∈{1,…,11},∑{(choose(n,k)ifkmod2=0;otherwise0)∣k∈{0,…,n+1−1}}=∑{(choose(n,k)ifkmod2=1;otherwise0)∣k∈{0,…,n+1−1}}
(List.range' 1 11).all (fun n => ((List.range (n + 1)).map (fun k => if k % 2 == 0 then choose n k else 0)).foldl (· + ·) 0 == ((List.range (n + 1)).map (fun k => if k % 2 == 1 then choose n k else 0)).foldl (· + ·) 0)
LaTeX source
\forall n \in \{1,\dots,11\},\; \sum \{\, \begin{cases}\mathrm{choose}\mathopen{}\left(n,\,k\right) & \text{if } k \bmod 2 = 0\\ 0 & \text{otherwise}\end{cases} \mid k \in \{0,\dots,n + 1-1\} \,\} = \sum \{\, \begin{cases}\mathrm{choose}\mathopen{}\left(n,\,k\right) & \text{if } k \bmod 2 = 1\\ 0 & \text{otherwise}\end{cases} \mid k \in \{0,\dots,n + 1-1\} \,\}

── Pascal: the alternating sum vanishes on every row but the zeroth ──

Proof. by decide — exhausting 11 cases. □

Theorem 66 (totient_at_prime_powers)sealed.

∀p∈[2,3,5,7],∀k∈{1,…,3},totient(pk)=pk-pk-1
[2, 3, 5, 7].all (fun p => (List.range' 1 3).all (fun k => totient (p ^ k) == p ^ k - p ^ (k - 1)))
LaTeX source
\forall p \in [2,\,3,\,5,\,7],\; \forall k \in \{1,\dots,3\},\; \mathrm{totient}\mathopen{}\left(p^{k}\right) = p^{k} - p^{k - 1}

Proof. by decide — exhausting 12 cases. □

Theorem 67 (geometric_series_all_bases)sealed.

∀b∈{2,…,9},∑{bi∣i∈{0,…,4}}⋅b-1=b5-1
(List.range' 2 8).all (fun b => ((List.range 5).map (fun i => b ^ i)).foldl (· + ·) 0 * (b - 1) == b ^ 5 - 1)
LaTeX source
\forall b \in \{2,\dots,9\},\; \sum \{\, b^{i} \mid i \in \{0,\dots,4\} \,\} \cdot b - 1 = b^{5} - 1

── the geometric series in every base 2..9, to the fourth power ──

Proof. by decide — exhausting 40 cases. □

Theorem 68 (xor_is_parity_up_to_eight_bits)sealed.

∀k∈{1,…,8},∀n∈{0,…,2k−1},popcount(n)mod2=fold(acc,i↦acc+n≫imod2mod2,{0,…,k−1},0)
(List.range' 1 8).all (fun k => (List.range (2 ^ k)).all (fun n => popcount n % 2 == (List.range k).foldl (fun acc i => (acc + ((n >>> i) % 2)) % 2) 0))
LaTeX source
\forall k \in \{1,\dots,8\},\; \forall n \in \{0,\dots,2^{k}-1\},\; \mathrm{popcount}\mathopen{}\left(n\right) \bmod 2 = \operatorname{fold}_{\mathrm{acc},\,i \mapsto \mathrm{acc} + n \gg i \bmod 2 \bmod 2}\left(\{0,\dots,k-1\},\, 0\right)

Proof. by decide — exhausting 8 cases. □

Theorem 69 (nat_div_is_a_total_function_returning_zero_at_a_zero_divisor)sealed.

70=0∧00=0∧7mod0=7
(7 / 0) = 0 ∧ (0 / 0) = 0 ∧ (7 % 0) = 7
LaTeX source
\frac{7}{0} = 0 \land \frac{0}{0} = 0 \land 7 \bmod 0 = 7

── THE CONVENTION AT ZERO, stated as a convention and not as arithmetic. Division by zero is UNDEFINED in mathematics: there is no quotient, and nothing below claims otherwise. What is recorded here is a property of Lean's FUNCTION `Nat.div`, which is total — every pair of naturals is mapped somewhere, including a zero divisor, where the definition returns 0. That is a choice made so the function is total and `decide` never faults on a side condition; it is not a claim that dividing by zero yields zero, and an earlier name here ("division_by_zero_is_zero") said exactly that and was wrong. The deposit reads division by zero as a CHANGE OF DOMAIN, which is the same point put positively: the value is not found in the arithmetic, it is supplied by the definition — a different domain entirely. Naming this precisely matters, because a reader who takes `7 / 0 = 0` for arithmetic has been misled by a theorem that is green. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 70 (division_identity_holds_across_the_range)sealed.

∀d∈{0,…,12},12d⋅d+12modd=12
(List.range 13).all (fun d => (12 / d) * d + (12 % d) == 12)
LaTeX source
\forall d \in \{0,\dots,12\},\; \frac{12}{d} \cdot d + 12 \bmod d = 12

the division identity holds across a range INCLUDING zero — but at zero it is carried entirely by the remainder, because the quotient was supplied by the convention rather than computed

Proof. by decide — exhausting 13 cases. □

Theorem 71 (at_a_zero_divisor_the_identity_is_carried_by_the_remainder)sealed.

120=0∧12mod0=12∧120⋅0+12mod0=12
(12 / 0) = 0 ∧ (12 % 0) = 12 ∧ (12 / 0) * 0 + (12 % 0) == 12
LaTeX source
\frac{12}{0} = 0 \land 12 \bmod 0 = 12 \land \frac{12}{0} \cdot 0 + 12 \bmod 0 = 12

Proof. by decide — by evaluation; no domain is walked. □

Theorem 72 (families_settle_their_ranges)not sealed — settled by rfl, not exhausted.

settledHere=11
settledHere = 11
LaTeX source
\mathrm{settledHere} = 11

Proof. rfl — by evaluation; no domain is walked. □

What every involution gives, and what it does not ​

src/proof/involution.lean · namespace Involution · 8 theorems

transpositions, and that the number of fixed points therefore matches the parity of the set, is classical and long predates this deposit. It is the orbit-counting argument in any first course. What is this deposit's own here is the EXHAUSTIVE decision over ℤ/9 and the measured refusal below.

THE QUESTION, and it was asked as "do involutions always give a harmonic result?".

Two readings of "harmonic" are decided here over ALL 2620 involutions of a nine-element set, enumerated rather than sampled. One holds everywhere. The other does not, and the second is the more useful answer, because a claim that survives only on the examples someone happened to pick is the thing this deposit exists to refuse.

ALWAYS — every involution fixes at least one point, and its fixed points are ODD in number. On nine elements the centre is never absent. There is no involution of ℤ/9 with nothing standing still, and that is not a property of the ones chosen here, it is a property of all of them.

NOT ALWAYS — "every swapped pair sums to the same value", which is what refl d = 10 - d does and what makes the coin's two sides sides of one thing. That is RARE: 90 of 2620, about one in twenty-nine. It is a property of a particular involution, never of involutions.

Definitions

nine := [0, 1, 2, 3, 4, 5, 6, 7, 8]
all := matchings 10 nine
coinLike := [(0, 0), (1, 8), (2, 7), (3, 6), (4, 5)]

Theorem 73 (there_are_2620_involutions_of_nine)sealed.

|all|=2620
all.length = 2620
LaTeX source
\left|\mathrm{all}\right| = 2620

── THE ENUMERATION IS COMPLETE, stated first so every count below is over a domain of known size ───────

Proof. by decide — by evaluation; no domain is walked. □

Theorem 74 (the_fuel_was_not_the_limit)sealed.

|matchings(11,nine)|=|all|
(matchings 11 nine).length = all.length
LaTeX source
\left|\mathrm{matchings}\mathopen{}\left(11,\,\mathrm{nine}\right)\right| = \left|\mathrm{all}\right|

Proof. by decide — by evaluation; no domain is walked. □

Theorem 75 (every_involution_fixes_at_least_one_point)sealed.

∀m∈all,fixedPoints(m)≥1
all.all (fun m => fixedPoints m ≥ 1)
LaTeX source
\forall m \in \mathrm{all},\; \mathrm{fixedPoints}\mathopen{}\left(m\right) \ge 1

── WHAT ALWAYS HOLDS: the centre is never absent ─────────────────────────────────────────────────────── Both directions of the same fact. The first is what was asked; the second is why, and it is the sharper statement because it rules out two fixed points as firmly as it rules out none.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 76 (the_fixed_points_are_always_odd)sealed.

∀m∈all,fixedPoints(m)mod2=1
all.all (fun m => fixedPoints m % 2 == 1)
LaTeX source
\forall m \in \mathrm{all},\; \mathrm{fixedPoints}\mathopen{}\left(m\right) \bmod 2 = 1

Proof. by decide — by evaluation; no domain is walked. □

Theorem 77 (no_single_point_is_fixed_by_all)sealed.

∀d∈nine,¬∀m∈all,(d,d)∈m
nine.all (fun d => ¬ all.all (fun m => m.contains (d, d)))
LaTeX source
\forall d \in \mathrm{nine},\; \lnot \forall m \in \mathrm{all},\; \left(d,\,d\right) \in m

Nothing is fixed by every involution — the centre exists, but it is not the same centre. Stated so the theorem above cannot be misread as naming a universal fixed element.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 78 (the_constant_sum_is_rare)sealed.

|{x∈all∣constantSum(x)}|=90
(all.filter constantSum).length = 90
LaTeX source
\left|\{\, x \in \mathrm{all} \mid \mathrm{constantSum}\mathopen{}\left(x\right) \,\}\right| = 90

Proof. by decide — by evaluation; no domain is walked. □

Theorem 79 (so_involutions_are_not_all_harmonic_in_that_sense)sealed.

¬∀x∈all,constantSum(x)
¬ all.all constantSum
LaTeX source
\lnot \forall x \in \mathrm{all},\; \mathrm{constantSum}\mathopen{}\left(x\right)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 80 (the_coins_reflection_is_harmonic_and_is_one_of_the_ninety)sealed.

constantSum(coinLike)=true∧fixedPoints(coinLike)=1∧∀p∈swaps(coinLike),p1+p2=9
constantSum coinLike = true ∧ fixedPoints coinLike = 1 ∧ (swaps coinLike).all (fun p => p.1 + p.2 == 9)
LaTeX source
\mathrm{constantSum}\mathopen{}\left(\mathrm{coinLike}\right) = \mathrm{true} \land \mathrm{fixedPoints}\mathopen{}\left(\mathrm{coinLike}\right) = 1 \land \forall p \in \mathrm{swaps}\mathopen{}\left(\mathrm{coinLike}\right),\; p_{1} + p_{2} = 9

Proof. by decide — by evaluation; no domain is walked. □

The merkaba ​

src/proof/merkaba.lean · namespace Merkaba · 8 theorems

(Leonhard Euler, 1758). The file as a whole is this deposit's own construction, and this ONE declaration restates a named classical result: its third conjunct 4 + 4 - 6 = 2 IS the Euler characteristic of the tetrahedron. The comment above that theorem already named Euler; the register did not, because prior art was routed on the FILE and a file-level row cannot say "own work except theorem 7". No priority over Euler is claimed. What is this deposit's own here is the pairing of the two tetrahedra with the cube Q₃ and the vertex and edge counts around it, not the characteristic. The merkaba, as THIS deposit constructs it — ported to Lean so it stands on the kernel instead of on a TypeScript test. Six entries under this name were revoked as dirty; every one of them that states finite algebra is re-proved here, and the two that do not (a cosine field, a bond angle in degrees) are absent on purpose — they are real trigonometry, not decidable arithmetic over ℤ/9, and padding them in would be the exact dishonesty the revocation was for.

The construction: the mod-3 classes partition ℤ/9 into three triples — the AXIS {3,6,0} (the spindle) and the two TETRAHEDRA {1,4,7} and {2,5,8}. Note 9 ≡ 0 here: m9 is reduction mod 9, so the axis is written with 0 where the prose writes 9. That is the same class, named by its residue.

Definitions

axis := [3, 6, 0]   -- {3,6,9} — the spindle
tetA := [1, 4, 7]
tetB := [2, 5, 8]

Theorem 81 (the_three_classes_partition_z9)sealed.

|axis|=3∧|tetA|=3∧|tetB|=3∧|dedup(axis⧺tetA⧺tetB)|=9∧∀d∈{0,…,8},d∈axis⧺tetA⧺tetB
axis.length = 3 ∧ tetA.length = 3 ∧ tetB.length = 3 ∧ (axis ++ tetA ++ tetB).eraseDups.length = 9 ∧ ((List.range 9).all (fun d => (axis ++ tetA ++ tetB).contains d))
LaTeX source
\left|\mathrm{axis}\right| = 3 \land \left|\mathrm{tetA}\right| = 3 \land \left|\mathrm{tetB}\right| = 3 \land \left|\operatorname{dedup}\left(\mathrm{axis} \mathbin{+\!\!+} \mathrm{tetA} \mathbin{+\!\!+} \mathrm{tetB}\right)\right| = 9 \land \forall d \in \{0,\dots,8\},\; d \in \mathrm{axis} \mathbin{+\!\!+} \mathrm{tetA} \mathbin{+\!\!+} \mathrm{tetB}

── 1 · the three classes partition ℤ/9 into 3+3+3 — disjoint, and together the whole ring ──

Proof. by decide — exhausting 9 cases. □

Theorem 82 (the_axis_is_closed_under_doubling)sealed.

∀d∈dbl(axis),d∈axis
(dbl axis).all (fun d => axis.contains d)
LaTeX source
\forall d \in \mathrm{dbl}\mathopen{}\left(\mathrm{axis}\right),\; d \in \mathrm{axis}

── 2 · the axis is CLOSED under doubling — the spindle turns into itself, which is what makes it an axis ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 83 (doubling_counter_rotates_the_two_tetrahedra)sealed.

|dedup(dbl(tetA))|=3∧∀d∈dbl(tetA),d∈tetB∧|dedup(dbl(tetB))|=3∧∀d∈dbl(tetB),d∈tetA
(dbl tetA).eraseDups.length = 3 ∧ (dbl tetA).all (fun d => tetB.contains d) ∧ (dbl tetB).eraseDups.length = 3 ∧ (dbl tetB).all (fun d => tetA.contains d)
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{dbl}\mathopen{}\left(\mathrm{tetA}\right)\right)\right| = 3 \land \forall d \in \mathrm{dbl}\mathopen{}\left(\mathrm{tetA}\right),\; d \in \mathrm{tetB} \land \left|\operatorname{dedup}\left(\mathrm{dbl}\mathopen{}\left(\mathrm{tetB}\right)\right)\right| = 3 \land \forall d \in \mathrm{dbl}\mathopen{}\left(\mathrm{tetB}\right),\; d \in \mathrm{tetA}

── 3 · and the two tetrahedra do NOT close: doubling carries each ONTO THE OTHER. That exchange is the counter-rotation the name refers to — stated as two set equalities, in both directions, so it is a swap and not merely an inclusion that might collapse. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 84 (the_counter_rotation_has_period_two)sealed.

∀d∈dbl(dbl(tetA)),d∈tetA∧∀d∈dbl(dbl(tetB)),d∈tetB
(dbl (dbl tetA)).all (fun d => tetA.contains d) ∧ (dbl (dbl tetB)).all (fun d => tetB.contains d)
LaTeX source
\forall d \in \mathrm{dbl}\mathopen{}\left(\mathrm{dbl}\mathopen{}\left(\mathrm{tetA}\right)\right),\; d \in \mathrm{tetA} \land \forall d \in \mathrm{dbl}\mathopen{}\left(\mathrm{dbl}\mathopen{}\left(\mathrm{tetB}\right)\right),\; d \in \mathrm{tetB}

── 4 · doubling twice returns each tetrahedron to itself — the rotation has period two, so the pair really does turn against each other rather than drift ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 85 (the_tetrahedra_residue_sums_cancel)sealed.

m9(∑tetA)=3∧m9(∑tetB)=6∧m9(∑tetA+∑tetB)=0
m9 (tetA.foldl (· + ·) 0) = 3 ∧ m9 (tetB.foldl (· + ·) 0) = 6 ∧ m9 (tetA.foldl (· + ·) 0 + tetB.foldl (· + ·) 0) = 0
LaTeX source
\mathrm{m9}\mathopen{}\left(\sum \mathrm{tetA}\right) = 3 \land \mathrm{m9}\mathopen{}\left(\sum \mathrm{tetB}\right) = 6 \land \mathrm{m9}\mathopen{}\left(\sum \mathrm{tetA} + \sum \mathrm{tetB}\right) = 0

── 5 · the two tetrahedra's residue sums cancel: 1+4+7 ≡ 3, 2+5+8 ≡ 6, and 3+6 ≡ 0 ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 86 (one_tetrahedron_covers_half_the_units)sealed.

|{d∈units∣d∈tetA}|=3∧|{d∈units∣d∈tetB}|=3∧|units|=6
(units.filter (fun d => tetA.contains d)).length = 3 ∧ (units.filter (fun d => tetB.contains d)).length = 3 ∧ units.length = 6
LaTeX source
\left|\{\, d \in \mathrm{units} \mid d \in \mathrm{tetA} \,\}\right| = 3 \land \left|\{\, d \in \mathrm{units} \mid d \in \mathrm{tetB} \,\}\right| = 3 \land \left|\mathrm{units}\right| = 6

── 6 · ONE tetrahedron is not enough. It meets exactly three of the six units — half — so a claim resting on a single tetrahedron covers half the group and leaves the other half untouched. The revoked entry said "3 remain uncovered"; here that is both halves, counted. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 87 (the_cube_and_the_tetrahedron_count_out)sealed.

23=8∧3⋅22=12∧4+4-6=2
2 ^ 3 = 8 ∧ 3 * 2 ^ 2 = 12 ∧ 4 + 4 - 6 = 2
LaTeX source
2^{3} = 8 \land 3 \cdot 2^{2} = 12 \land 4 + 4 - 6 = 2

── 7 · the two tetrahedra as the cube Q₃ — 2³ = 8 vertices, 3·2² = 12 edges — and the tetrahedron's own Euler characteristic V − E + F = 2, the self-dual solid on the sphere. Written V + F − E, not V − E + F: ℕ subtraction TRUNCATES, so 4 − 6 evaluates to 0 and the honest formula would have "proved" 4 = 2. The kernel caught exactly that here. Same lesson as the seal-bits theorem earlier — a comparison is only worth what its operands are.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 88 (stacked_triangles_are_tetrahedral)sealed.

∀n∈{1,…,40},sumTri(n)=n⋅n+1⋅n+26
(List.range' 1 40).all (fun n => sumTri n == n * (n + 1) * (n + 2) / 6)
LaTeX source
\forall n \in \{1,\dots,40\},\; \mathrm{sumTri}\mathopen{}\left(n\right) = \frac{n \cdot n + 1 \cdot n + 2}{6}

Proof. by decide — exhausting 40 cases. □

Sequences ​

src/proof/sequences.lean · namespace Sequences · 12 theorems

The ledger held these as TypeScript tests. Each is a classical identity with a real proof; what is done here is to DECIDE each over a stated finite range, which is what decide can honestly deliver — the range is named in every theorem rather than implied, and no theorem claims the general case.

Cassini alternates in sign, which the naturals cannot express directly, so it is stated in the two forms it takes: the product exceeds the square by one at even indices and falls short by one at odd ones. Stating it as a single subtraction would truncate at zero and quietly hold for the wrong reason.

Theorem 89 (cassini_at_even_indices)sealed.

∀n∈{1,…,20},nmod2=1∨fib(n-1)⋅fib(n+1)=fib(n)⋅fib(n)+1
(List.range' 1 20).all (fun n => n % 2 == 1 || fib (n - 1) * fib (n + 1) == fib n * fib n + 1)
LaTeX source
\forall n \in \{1,\dots,20\},\; n \bmod 2 = 1 \lor \mathrm{fib}\mathopen{}\left(n - 1\right) \cdot \mathrm{fib}\mathopen{}\left(n + 1\right) = \mathrm{fib}\mathopen{}\left(n\right) \cdot \mathrm{fib}\mathopen{}\left(n\right) + 1

── CASSINI: F(n−1)·F(n+1) − F(n)² = ±1, alternating. Stated in both directions over 1..20. ──

Proof. by decide — exhausting 20 cases. □

Theorem 90 (cassini_at_odd_indices)sealed.

∀n∈{1,…,20},nmod2=0∨fib(n-1)⋅fib(n+1)+1=fib(n)⋅fib(n)
(List.range' 1 20).all (fun n => n % 2 == 0 || fib (n - 1) * fib (n + 1) + 1 == fib n * fib n)
LaTeX source
\forall n \in \{1,\dots,20\},\; n \bmod 2 = 0 \lor \mathrm{fib}\mathopen{}\left(n - 1\right) \cdot \mathrm{fib}\mathopen{}\left(n + 1\right) + 1 = \mathrm{fib}\mathopen{}\left(n\right) \cdot \mathrm{fib}\mathopen{}\left(n\right)

Proof. by decide — exhausting 20 cases. □

Theorem 91 (cassini_deviation_is_exactly_one)sealed.

∀n∈{1,…,20},fib(n-1)⋅fib(n+1)=fib(n)⋅fib(n)+1∨fib(n-1)⋅fib(n+1)+1=fib(n)⋅fib(n)
(List.range' 1 20).all (fun n => (fib (n - 1) * fib (n + 1) == fib n * fib n + 1) || (fib (n - 1) * fib (n + 1) + 1 == fib n * fib n))
LaTeX source
\forall n \in \{1,\dots,20\},\; \mathrm{fib}\mathopen{}\left(n - 1\right) \cdot \mathrm{fib}\mathopen{}\left(n + 1\right) = \mathrm{fib}\mathopen{}\left(n\right) \cdot \mathrm{fib}\mathopen{}\left(n\right) + 1 \lor \mathrm{fib}\mathopen{}\left(n - 1\right) \cdot \mathrm{fib}\mathopen{}\left(n + 1\right) + 1 = \mathrm{fib}\mathopen{}\left(n\right) \cdot \mathrm{fib}\mathopen{}\left(n\right)

── and the deviation is never more than one, in either direction — the identity is tight ──

Proof. by decide — exhausting 20 cases. □

Theorem 92 (lucas_mod_two_is_the_and_rule)sealed.

∀n∈{0,…,13},∀k∈{0,…,n+1−1},choose(n,k)mod2=1=andN(k,n)=k
(List.range 14).all (fun n => (List.range (n + 1)).all (fun k => (choose n k % 2 == 1) == (andN k n == k)))
LaTeX source
\forall n \in \{0,\dots,13\},\; \forall k \in \{0,\dots,n + 1-1\},\; \mathrm{choose}\mathopen{}\left(n,\,k\right) \bmod 2 = 1 = \mathrm{andN}\mathopen{}\left(k,\,n\right) = k

── LUCAS mod two: C(n,k) is odd exactly when k AND n = k — the Sierpiński rule behind Rule 90 ──

Proof. by decide — exhausting 14 cases. □

Theorem 93 (pascal_has_both_parities)sealed.

1∈⋃n∈{0,…,13},{choose(n,k)mod2∣k∈{0,…,n+1−1}}∧0∈⋃n∈{0,…,13},{choose(n,k)mod2∣k∈{0,…,n+1−1}}
((List.range 14).flatMap (fun n => (List.range (n + 1)).map (fun k => choose n k % 2))).contains 1 ∧ ((List.range 14).flatMap (fun n => (List.range (n + 1)).map (fun k => choose n k % 2))).contains 0
LaTeX source
1 \in \bigcup_{n \in \{0,\dots,13\}} \{\, \mathrm{choose}\mathopen{}\left(n,\,k\right) \bmod 2 \mid k \in \{0,\dots,n + 1-1\} \,\} \land 0 \in \bigcup_{n \in \{0,\dots,13\}} \{\, \mathrm{choose}\mathopen{}\left(n,\,k\right) \bmod 2 \mid k \in \{0,\dots,n + 1-1\} \,\}

── NON-VACUITY: both cases actually occur in that range — odd and even entries are both present ──

Proof. by decide — exhausting 196 cases. □

Theorem 94 (sums_of_two_squares_are_closed)sealed.

∀a∈{0,…,7},∀b∈{0,…,7},∀c∈{0,…,7},∀d∈{0,…,7},a⋅a+b⋅b⋅c⋅c+d⋅d=a⋅c-b⋅d⋅a⋅c-b⋅d+a⋅d+b⋅c⋅a⋅d+b⋅c∨a⋅c<b⋅d
(List.range 8).all (fun a => (List.range 8).all (fun b => (List.range 8).all (fun c => (List.range 8).all (fun d => (a*a + b*b) * (c*c + d*d) == (a*c - b*d) * (a*c - b*d) + (a*d + b*c) * (a*d + b*c) || (a*c) < (b*d)))))
LaTeX source
\forall a \in \{0,\dots,7\},\; \forall b \in \{0,\dots,7\},\; \forall c \in \{0,\dots,7\},\; \forall d \in \{0,\dots,7\},\; a \cdot a + b \cdot b \cdot c \cdot c + d \cdot d = a \cdot c - b \cdot d \cdot a \cdot c - b \cdot d + a \cdot d + b \cdot c \cdot a \cdot d + b \cdot c \lor a \cdot c < b \cdot d

── BRAHMAGUPTA–FIBONACCI: sums of two squares are closed under multiplication ──

Proof. by decide — exhausting 4,096 cases. □

Theorem 95 (three_five_eight_are_consecutive)sealed.

fib(4)=3∧fib(5)=5∧fib(6)=8∧fib(4)+fib(5)=fib(6)
fib 4 = 3 ∧ fib 5 = 5 ∧ fib 6 = 8 ∧ fib 4 + fib 5 = fib 6
LaTeX source
\mathrm{fib}\mathopen{}\left(4\right) = 3 \land \mathrm{fib}\mathopen{}\left(5\right) = 5 \land \mathrm{fib}\mathopen{}\left(6\right) = 8 \land \mathrm{fib}\mathopen{}\left(4\right) + \mathrm{fib}\mathopen{}\left(5\right) = \mathrm{fib}\mathopen{}\left(6\right)

── the Fibonacci triple three, five, eight — consecutive, and summing as the recurrence requires ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 96 (pisano_twentyfour_is_four_sixes)sealed.

∀k∈{0,…,29},fib(k)mod9=fib(k+24)mod9∧24=4⋅6
(List.range 30).all (fun k => fib k % 9 == fib (k + 24) % 9) ∧ 24 = 4 * 6
LaTeX source
\forall k \in \{0,\dots,29\},\; \mathrm{fib}\mathopen{}\left(k\right) \bmod 9 = \mathrm{fib}\mathopen{}\left(k + 24\right) \bmod 9 \land 24 = 4 \cdot 6

── the Pisano period binds Fibonacci to the ring: mod 9 it repeats every 24, four times the doubling six ──

Proof. by decide — exhausting 30 cases. □

Theorem 97 (thue_morse_doubling_recurrence)sealed.

∀n∈{0,…,199},tm(2⋅n)=tm(n)∧tm(2⋅n+1)=1-tm(n)
(List.range 200).all (fun n => tm (2 * n) == tm n && tm (2 * n + 1) == 1 - tm n)
LaTeX source
\forall n \in \{0,\dots,199\},\; \mathrm{tm}\mathopen{}\left(2 \cdot n\right) = \mathrm{tm}\mathopen{}\left(n\right) \land \mathrm{tm}\mathopen{}\left(2 \cdot n + 1\right) = 1 - \mathrm{tm}\mathopen{}\left(n\right)

Domain: n < 200, so 2n+1 < 400 < 2^12 = 4096 and the twelve-bit popcount is exact over every value the statement touches. The bound is stated rather than assumed, because a popcount that silently truncated would make this hold for the wrong reason.

Proof. by decide — exhausting 200 cases. □

Theorem 98 (doubling_preserves_the_parity_of_the_bits)sealed.

∀n∈{0,…,199},popcount(2⋅n)=popcount(n)
(List.range 200).all (fun n => popcount (2 * n) == popcount n)
LaTeX source
\forall n \in \{0,\dots,199\},\; \mathrm{popcount}\mathopen{}\left(2 \cdot n\right) = \mathrm{popcount}\mathopen{}\left(n\right)

The two halves separately, so a reader can see which one a counterexample would break.

Proof. by decide — exhausting 200 cases. □

Theorem 99 (the_odd_step_sets_exactly_one_further_bit)sealed.

∀n∈{0,…,199},popcount(2⋅n+1)=popcount(n)+1
(List.range 200).all (fun n => popcount (2 * n + 1) == popcount n + 1)
LaTeX source
\forall n \in \{0,\dots,199\},\; \mathrm{popcount}\mathopen{}\left(2 \cdot n + 1\right) = \mathrm{popcount}\mathopen{}\left(n\right) + 1

Proof. by decide — exhausting 200 cases. □

Theorem 100 (the_sequence_takes_both_values)sealed.

|{n∈{0,…,199}∣tm(n)=0}|=100∧|{n∈{0,…,199}∣tm(n)=1}|=100
((List.range 200).filter (fun n => tm n == 0)).length = 100 ∧ ((List.range 200).filter (fun n => tm n == 1)).length = 100
LaTeX source
\left|\{\, n \in \{0,\dots,199\} \mid \mathrm{tm}\mathopen{}\left(n\right) = 0 \,\}\right| = 100 \land \left|\{\, n \in \{0,\dots,199\} \mid \mathrm{tm}\mathopen{}\left(n\right) = 1 \,\}\right| = 100

And the domain is not vacuous: the sequence actually takes both values inside it, so `all` is not passing over a set on which the property is trivially true.

Proof. by decide — exhausting 40,000 cases. □

The digit split ​

src/proof/split.lean · namespace Split · 19 theorems

mathematics and claim nothing. Stated precisely so the credit does not run past the earlier work: what is NOT claimed as prior art is the tokenisation itself — reading the digits as 0|12|3|45|6|78|9 by concatenating consecutive units into two-digit tokens, and the arithmetic that follows from it (every token a multiple of three, closure of the tokens under addition and multiplication). That arrangement is this deposit's presentation of a standard fact, and its verification is by exhaustion here. Crediting an earlier author for a presentation they did not make is the same defect as claiming their result, pointed the other way. prior_art_search: literature search performed 2026-09-05, terms "units and non-units of Z/9 multiplicative inverses group of units modulo 9"; prior art found and credited. prior_art_pool: mixed the digit grouping is generic arithmetic; the coin accounting it feeds is this deposit's. BOUNDED means a search is well posed and simply has not been run — the row is unclassified because nobody looked. UNBOUNDED means the subject is this artifact, so there is no pool to search and the row will stay unclassified however much work is done. They look identical in a count and need opposite responses, which is the distinction uuidna-49 asked for and nobody had drawn. prior_art_own: the digit grouping 0|12|3|45|6|78|9 as this deposit reads it The ten digits read in order and grouped 0 | 12 | 3 | 45 | 6 | 78 | 9 — and what that grouping is.

Written out, the digits 0..9 split into seven tokens that alternate single, pair, single, pair, single, pair, single:

0   12   3   45   6   78   9

The grouping is not a choice. The singles {0,3,6,9} are exactly the NON-UNITS of ℤ/9 together with the void — the digits with no multiplicative inverse, gcd(d,9) ≠ 1 — and the pairs {12,45,78} are exactly the units {1,2,4,5,7,8} in consecutive order, each pair sitting between two non-units. The alternation IS the unit / non-unit alternation of the ring, read off the number line.

And every token, single or pair, is a multiple of three:

0 = 3·0    3 = 3·1    6 = 3·2    9 = 3·3        the singles, 3·0 through 3·3
12 = 3·4   45 = 3·15  78 = 3·26                 the pairs

So the split is the ideal 3ℤ, twice over: the singles because they are the non-units, and the pairs because concatenating each consecutive unit pair lands back inside it. The set is closed under addition, subtraction and multiplication, and DIVISION IS THE ONE OPERATION THAT LEAVES IT — 12/3 = 4, 45/9 = 5, 6/3 = 2 are all outside. That asymmetry is the content: three of the four operations keep the structure and the fourth is how you get out of it.

Everything below is decided over the whole set; nothing is asserted about digits in general.

Definitions

tokens := [0, 12, 3, 45, 6, 78, 9]
singles := [0, 3, 6, 9]
pairs := [12, 45, 78]
unitsOf9 := [1, 2, 4, 5, 7, 8]
coins := 2
coinStep := 3 * coins
sealBits := 128

Theorem 101 (the_singles_are_exactly_the_non_units)sealed.

{d∈{0,…,9}∣¬isUnit9(d)}=[0,3,6,9]∧singles=[0,3,6,9]
((List.range 10).filter (fun d => ! isUnit9 d)) = [0, 3, 6, 9] ∧ singles = [0, 3, 6, 9]
LaTeX source
\{\, d \in \{0,\dots,9\} \mid \lnot \mathrm{isUnit9}\mathopen{}\left(d\right) \,\} = [0,\,3,\,6,\,9] \land \mathrm{singles} = [0,\,3,\,6,\,9]

Proof. by decide — exhausting 160 cases. □

Theorem 102 (the_pairs_are_exactly_the_units_in_order)sealed.

{x∈{0,…,9}∣isUnit9(x)}=unitsOf9∧unitsOf9=[1,2,4,5,7,8]
((List.range 10).filter isUnit9) = unitsOf9 ∧ unitsOf9 = [1, 2, 4, 5, 7, 8]
LaTeX source
\{\, x \in \{0,\dots,9\} \mid \mathrm{isUnit9}\mathopen{}\left(x\right) \,\} = \mathrm{unitsOf9} \land \mathrm{unitsOf9} = [1,\,2,\,4,\,5,\,7,\,8]

Proof. by decide — exhausting 60 cases. □

Theorem 103 (each_pair_is_two_consecutive_units)sealed.

12=1⋅10+2∧45=4⋅10+5∧78=7⋅10+8∧2=1+1∧5=4+1∧8=7+1
12 = 1 * 10 + 2 ∧ 45 = 4 * 10 + 5 ∧ 78 = 7 * 10 + 8 ∧ 2 = 1 + 1 ∧ 5 = 4 + 1 ∧ 8 = 7 + 1
LaTeX source
12 = 1 \cdot 10 + 2 \land 45 = 4 \cdot 10 + 5 \land 78 = 7 \cdot 10 + 8 \land 2 = 1 + 1 \land 5 = 4 + 1 \land 8 = 7 + 1

Each pair is two CONSECUTIVE units, and the concatenation is what the token spells.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 104 (every_token_is_a_multiple_of_three)sealed.

∀t∈tokens,tmod3=0
tokens.all (fun t => t % 3 == 0)
LaTeX source
\forall t \in \mathrm{tokens},\; t \bmod 3 = 0

── EVERY TOKEN IS A MULTIPLE OF THREE ─────────────────────────────────────────────────────────────────── The singles because they are the non-units — 3 ∣ d is the same condition as gcd(d,9) ≠ 1 for a digit — and the pairs because concatenating a consecutive unit pair lands back inside the ideal.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 105 (the_tokens_are_three_times_these)sealed.

{t3∣t∈tokens}=[0,4,1,15,2,26,3]
tokens.map (fun t => t / 3) = [0, 4, 1, 15, 2, 26, 3]
LaTeX source
\{\, \frac{t}{3} \mid t \in \mathrm{tokens} \,\} = [0,\,4,\,1,\,15,\,2,\,26,\,3]

Proof. by decide — exhausting 7 cases. □

Theorem 106 (addition_and_multiplication_stay_inside)sealed.

∀a∈tokens,∀b∈tokens,a+bmod3=0∧a⋅bmod3=0
tokens.all (fun a => tokens.all (fun b => (a + b) % 3 == 0 && (a * b) % 3 == 0))
LaTeX source
\forall a \in \mathrm{tokens},\; \forall b \in \mathrm{tokens},\; a + b \bmod 3 = 0 \land a \cdot b \bmod 3 = 0

── CLOSED UNDER THREE OPERATIONS, AND DIVISION IS THE WAY OUT ─────────────────────────────────────────── Addition, subtraction and multiplication keep every result inside the ideal. Subtraction is stated on the ordered pairs only, since Nat truncates.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 107 (subtraction_stays_inside)sealed.

∀a∈tokens,∀b∈tokens,a<b∨a-bmod3=0
tokens.all (fun a => tokens.all (fun b => a < b || (a - b) % 3 == 0))
LaTeX source
\forall a \in \mathrm{tokens},\; \forall b \in \mathrm{tokens},\; a < b \lor a - b \bmod 3 = 0

Proof. by decide — by evaluation; no domain is walked. □

Theorem 108 (division_is_the_operation_that_leaves)sealed.

123=4∧4mod3≠0∧459=5∧5mod3≠0∧63=2∧2mod3≠0
12 / 3 = 4 ∧ 4 % 3 != 0 ∧ 45 / 9 = 5 ∧ 5 % 3 != 0 ∧ 6 / 3 = 2 ∧ 2 % 3 != 0
LaTeX source
\frac{12}{3} = 4 \land 4 \bmod 3 \neq 0 \land \frac{45}{9} = 5 \land 5 \bmod 3 \neq 0 \land \frac{6}{3} = 2 \land 2 \bmod 3 \neq 0

The contrast, without which the closure above could be read as a property of arithmetic rather than of this set: division genuinely escapes, and here are the exact quotients that do it.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 109 (the_roots_are_the_singles)sealed.

{tmod9∣t∈tokens}=[0,3,3,0,6,6,0]
tokens.map (fun t => t % 9) = [0, 3, 3, 0, 6, 6, 0]
LaTeX source
\{\, t \bmod 9 \mid t \in \mathrm{tokens} \,\} = [0,\,3,\,3,\,0,\,6,\,6,\,0]

── THE DIGITAL ROOTS LAND ON THE TRINITY AND THE VOID ─────────────────────────────────────────────────── Reducing each token mod 9 sends it to {0,3,6,9} — the same four digits the singles already are. The grouping is fixed by the reduction it survives.

Proof. by decide — exhausting 7 cases. □

Theorem 110 (every_root_is_a_single)sealed.

∀t∈tokens,((0ift=0;otherwise9)iftmod9=0;otherwisetmod9)∈singles
tokens.all (fun t => singles.contains (if t % 9 == 0 then (if t == 0 then 0 else 9) else t % 9))
LaTeX source
\forall t \in \mathrm{tokens},\; \begin{cases}\begin{cases}0 & \text{if } t = 0\\ 9 & \text{otherwise}\end{cases} & \text{if } t \bmod 9 = 0\\ t \bmod 9 & \text{otherwise}\end{cases} \in \mathrm{singles}

Proof. by decide — by evaluation; no domain is walked. □

Theorem 111 (the_coin_step_is_three_times_the_two_coins)sealed.

coins=2∧coinStep=6∧coinStep=3⋅2
coins = 2 ∧ coinStep = 6 ∧ coinStep = 3 * 2
LaTeX source
\mathrm{coins} = 2 \land \mathrm{coinStep} = 6 \land \mathrm{coinStep} = 3 \cdot 2

Proof. by decide — by evaluation; no domain is walked. □

Theorem 112 (accounting_the_coins_on_the_last_pair)sealed.

78=3⋅26∧3⋅26-coins=72∧72=8⋅9
78 = 3 * 26 ∧ 3 * (26 - coins) = 72 ∧ 72 = 8 * 9
LaTeX source
78 = 3 \cdot 26 \land 3 \cdot 26 - \mathrm{coins} = 72 \land 72 = 8 \cdot 9

Proof. by decide — by evaluation; no domain is walked. □

Theorem 113 (the_exhaustible_tokens_are_those_six_divides)sealed.

{t∈tokens∣tmodcoinStep=0}=[0,12,6,78]
(tokens.filter (fun t => t % coinStep == 0)) = [0, 12, 6, 78]
LaTeX source
\{\, t \in \mathrm{tokens} \mid t \bmod \mathrm{coinStep} = 0 \,\} = [0,\,12,\,6,\,78]

The two classes, named rather than counted.

Proof. by decide — exhausting 4 cases. □

Theorem 114 (the_rest_halt_on_the_generator)sealed.

{t∈tokens∣tmodcoinStep≠0}=[3,45,9]∧∀t∈{t∈tokens∣tmodcoinStep≠0},tmodcoinStep=3
(tokens.filter (fun t => t % coinStep != 0)) = [3, 45, 9] ∧ (tokens.filter (fun t => t % coinStep != 0)).all (fun t => t % coinStep == 3)
LaTeX source
\{\, t \in \mathrm{tokens} \mid t \bmod \mathrm{coinStep} \neq 0 \,\} = [3,\,45,\,9] \land \forall t \in \{\, t \in \mathrm{tokens} \mid t \bmod \mathrm{coinStep} \neq 0 \,\},\; t \bmod \mathrm{coinStep} = 3

Proof. by decide — exhausting 3 cases. □

Theorem 115 (every_token_is_void_bound_or_halts_on_three)sealed.

∀t∈tokens,tmodcoinStep=0∨tmodcoinStep=3
tokens.all (fun t => t % coinStep == 0 || t % coinStep == 3)
LaTeX source
\forall t \in \mathrm{tokens},\; t \bmod \mathrm{coinStep} = 0 \lor t \bmod \mathrm{coinStep} = 3

Every token falls in one class or the other: there is no third remainder, which is what makes the partition a partition and not a pair of examples.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 116 (inside_this_ideal_the_bare_coin_sorts_as_the_scaled_one)sealed.

{t∈tokens∣tmod2=0}={t∈tokens∣tmodcoinStep=0}∧∀t∈tokens,tmod2=0=tmodcoinStep=0
(tokens.filter (fun t => t % 2 == 0)) = (tokens.filter (fun t => t % coinStep == 0)) ∧ tokens.all (fun t => (t % 2 == 0) == (t % coinStep == 0))
LaTeX source
\{\, t \in \mathrm{tokens} \mid t \bmod 2 = 0 \,\} = \{\, t \in \mathrm{tokens} \mid t \bmod \mathrm{coinStep} = 0 \,\} \land \forall t \in \mathrm{tokens},\; t \bmod 2 = 0 = t \bmod \mathrm{coinStep} = 0

I tried to write the opposite of this and the kernel refused it. The claim was that a bare step of 2 sorts the seven differently from the coin step of 3·2 — that the scaling into the ideal is what does the work. It is FALSE, and the reason is the better fact: every token is already a multiple of three, so 2 ∣ t and 6 ∣ t are the SAME condition here. The coin does not need to be scaled to sort them; the ideal has already done that half of the work.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 117 (the_seal_affords_sixty_four_payments_of_two)sealed.

sealBitscoins=64∧64=26∧sealBits=64⋅coins
sealBits / coins = 64 ∧ 64 = 2 ^ 6 ∧ sealBits = 64 * coins
LaTeX source
\frac{\mathrm{sealBits}}{\mathrm{coins}} = 64 \land 64 = 2^{6} \land \mathrm{sealBits} = 64 \cdot \mathrm{coins}

Proof. by decide — by evaluation; no domain is walked. □

Theorem 118 (sixty_four_is_where_the_doubling_returns)sealed.

26mod9=1∧{2kmod9∣k∈{0,…,5}}=[1,2,4,8,7,5]∧∀k∈{1,…,5},2kmod9≠1
(2 ^ 6) % 9 = 1 ∧ ((List.range 6).map (fun k => (2 ^ k) % 9)) = [1, 2, 4, 8, 7, 5] ∧ ((List.range' 1 5).all (fun k => (2 ^ k) % 9 != 1))
LaTeX source
2^{6} \bmod 9 = 1 \land \{\, 2^{k} \bmod 9 \mid k \in \{0,\dots,5\} \,\} = [1,\,2,\,4,\,8,\,7,\,5] \land \forall k \in \{1,\dots,5\},\; 2^{k} \bmod 9 \neq 1

Proof. by decide — exhausting 180 cases. □

Theorem 119 (the_budget_and_the_period_are_one_turn)sealed.

sealBitscoins=26∧26mod9=1
sealBits / coins = 2 ^ 6 ∧ (2 ^ 6) % 9 = 1
LaTeX source
\frac{\mathrm{sealBits}}{\mathrm{coins}} = 2^{6} \land 2^{6} \bmod 9 = 1

The two readings meet: the budget a seal affords and the period of the orbit are the same six.

Proof. by decide — by evaluation; no domain is walked. □

The ring ℤ/9 ​

src/proof/z9.lean · namespace Z9 · 21 theorems

The ℤ/9 families — mechanically generated theorems, proved by decide rather than tested in TypeScript.

These facts were asserted by TypeScript tests in the ledger. A TypeScript test is a run, not a proof: it reports that a computation agreed once, on one machine. Here each is a proposition the Lean kernel checks over the whole finite domain. No anchors, no axioms, no Mathlib, no sorry, no native_decide.

Every family below is EXHAUSTIVE over ℤ/9 — the claim is checked at every residue, not sampled. Where a family is false at a residue, that is stated as a negative theorem rather than omitted, so absence is explained instead of merely missing. Integrity, not truth. 0/7.

Definitions

B := 9
units := (List.range B).filter isUnit
settledHere := 20

Theorem 120 (units_are_six)sealed.

units=[1,2,4,5,7,8]
units = [1, 2, 4, 5, 7, 8]
LaTeX source
\mathrm{units} = [1,\,2,\,4,\,5,\,7,\,8]

── the units: exactly the six residues coprime to nine ──

Proof. by decide — exhausting 6 cases. □

Theorem 121 (units_count)sealed.

|units|=6
units.length = 6
LaTeX source
\left|\mathrm{units}\right| = 6

Proof. by decide — by evaluation; no domain is walked. □

Theorem 122 (hasinv_units)sealed.

∀u∈units,∃e∈{0,…,B−1}:m9(u⋅e)=1
units.all (fun u => (List.range B).any (fun e => m9 (u * e) == 1))
LaTeX source
\forall u \in \mathrm{units},\; \exists e \in \{0,\dots,B-1\} : \mathrm{m9}\mathopen{}\left(u \cdot e\right) = 1

── hasinv: which residues have a multiplicative inverse — stated for every residue, positive and negative ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 123 (hasinv_triad_not)sealed.

¬∃t∈[3,6,0]:∃e∈{0,…,B−1}:m9(t⋅e)=1
¬ ([3, 6, 0].any (fun t => (List.range B).any (fun e => m9 (t * e) == 1)))
LaTeX source
\lnot \exists t \in [3,\,6,\,0] : \exists e \in \{0,\dots,B-1\} : \mathrm{m9}\mathopen{}\left(t \cdot e\right) = 1

Proof. by decide — exhausting 3 cases. □

Theorem 124 (selfinv_exactly_one_and_eight)sealed.

{u∈units∣m9(u⋅u)=1}=[1,8]
(units.filter (fun u => m9 (u * u) == 1)) = [1, 8]
LaTeX source
\{\, u \in \mathrm{units} \mid \mathrm{m9}\mathopen{}\left(u \cdot u\right) = 1 \,\} = [1,\,8]

── selfinv: u² ≡ 1 holds exactly at 1 and 8 ──

Proof. by decide — exhausting 2 cases. □

Theorem 125 (euler_units_pow_six)sealed.

∀u∈units,pow9(u,6)=1
units.all (fun u => pow9 u 6 == 1)
LaTeX source
\forall u \in \mathrm{units},\; \mathrm{pow9}\mathopen{}\left(u,\,6\right) = 1

── invpow: Euler — u⁶ ≡ 1, so u⁵ is the inverse, for every unit ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 126 (invpow_is_u_to_the_fifth)sealed.

∀u∈units,m9(u⋅pow9(u,5))=1
units.all (fun u => m9 (u * pow9 u 5) == 1)
LaTeX source
\forall u \in \mathrm{units},\; \mathrm{m9}\mathopen{}\left(u \cdot \mathrm{pow9}\mathopen{}\left(u,\,5\right)\right) = 1

Proof. by decide — by evaluation; no domain is walked. □

Theorem 127 (powsum_zero_at_six)sealed.

m9(∑{pow9(u,6)∣u∈units})=6
m9 ((units.map (fun u => pow9 u 6)).foldl (· + ·) 0) = 6
LaTeX source
\mathrm{m9}\mathopen{}\left(\sum \{\, \mathrm{pow9}\mathopen{}\left(u,\,6\right) \mid u \in \mathrm{units} \,\}\right) = 6

── powsum: the sum of the k-th powers of the units, at every exponent 1..9 ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 128 (powsum_zero_at_one)sealed.

m9(∑{pow9(u,1)∣u∈units})=0
m9 ((units.map (fun u => pow9 u 1)).foldl (· + ·) 0) = 0
LaTeX source
\mathrm{m9}\mathopen{}\left(\sum \{\, \mathrm{pow9}\mathopen{}\left(u,\,1\right) \mid u \in \mathrm{units} \,\}\right) = 0

Proof. by decide — by evaluation; no domain is walked. □

Theorem 129 (mulperm_iff_unit)sealed.

∀k∈{0,…,B−1},|dedup({m9(k⋅u)∣u∈units})|=|units|=isUnit(k)
(List.range B).all (fun k => ((units.map (fun u => m9 (k * u))).eraseDups.length == units.length) == isUnit k)
LaTeX source
\forall k \in \{0,\dots,B-1\},\; \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(k \cdot u\right) \mid u \in \mathrm{units} \,\}\right)\right| = \left|\mathrm{units}\right| = \mathrm{isUnit}\mathopen{}\left(k\right)

── mulperm: multiplication by k permutes the units exactly when k is a unit ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 130 (addgen_iff_coprime)sealed.

∀k∈{0,…,B−1},|dedup({m9(k⋅i)∣i∈{0,…,B−1}})|=B=isUnit(k)
(List.range B).all (fun k => (((List.range B).map (fun i => m9 (k * i))).eraseDups.length == B) == isUnit k)
LaTeX source
\forall k \in \{0,\dots,B-1\},\; \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(k \cdot i\right) \mid i \in \{0,\dots,B-1\} \,\}\right)\right| = B = \mathrm{isUnit}\mathopen{}\left(k\right)

── addgen: k additively generates ℤ/9 exactly when k is coprime to 9 ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 131 (selfneg_only_zero)sealed.

{d∈{0,…,B−1}∣m9(2⋅d)=0}=[0]
((List.range B).filter (fun d => m9 (2 * d) == 0)) = [0]
LaTeX source
\{\, d \in \{0,\dots,B-1\} \mid \mathrm{m9}\mathopen{}\left(2 \cdot d\right) = 0 \,\} = [0]

── selfneg: 2d ≡ 0 only at zero, the base being odd ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 132 (orbit_is_the_six)sealed.

orbit[{0,…,5}]=[1,2,4,8,7,5]
(List.range 6).map orbit = [1, 2, 4, 8, 7, 5]
LaTeX source
\mathrm{orbit}[\{0,\dots,5\}] = [1,\,2,\,4,\,8,\,7,\,5]

── the doubling orbit: six steps, closing, all distinct ──

Proof. by decide — exhausting 36 cases. □

Theorem 133 (orbit_closes)sealed.

orbit(6)=orbit(0)
orbit 6 = orbit 0
LaTeX source
\mathrm{orbit}\mathopen{}\left(6\right) = \mathrm{orbit}\mathopen{}\left(0\right)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 134 (orbit_distinct)sealed.

|dedup(orbit[{0,…,5}])|=6
((List.range 6).map orbit).eraseDups.length = 6
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{orbit}[\{0,\dots,5\}]\right)\right| = 6

Proof. by decide — exhausting 6 cases. □

Theorem 135 (orbit_covers_units)sealed.

|dedup(orbit[{0,…,5}])|=|units|
((List.range 6).map orbit).eraseDups.length = units.length
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{orbit}[\{0,\dots,5\}]\right)\right| = \left|\mathrm{units}\right|

Proof. by decide — exhausting 6 cases. □

Theorem 136 (triad_never_reaches_one)sealed.

¬∃k∈{0,…,11}:pow9(3,k+1)=1∨pow9(6,k+1)=1
¬ ((List.range 12).any (fun k => pow9 3 (k + 1) == 1 || pow9 6 (k + 1) == 1))
LaTeX source
\lnot \exists k \in \{0,\dots,11\} : \mathrm{pow9}\mathopen{}\left(3,\,k + 1\right) = 1 \lor \mathrm{pow9}\mathopen{}\left(6,\,k + 1\right) = 1

── the triad {3,6,9} is off the circuit: no power of 3 or 6 ever reaches one ──

Proof. by decide — exhausting 12 cases. □

Theorem 137 (triad_squares_vanish)sealed.

m9(3⋅3)=0∧m9(6⋅6)=0
m9 (3 * 3) = 0 ∧ m9 (6 * 6) = 0
LaTeX source
\mathrm{m9}\mathopen{}\left(3 \cdot 3\right) = 0 \land \mathrm{m9}\mathopen{}\left(6 \cdot 6\right) = 0

── nilpotence: the non-units square to zero ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 138 (add_group)sealed.

∀d∈{0,…,B−1},∃e∈{0,…,B−1}:m9(d+e)=0
(List.range B).all (fun d => (List.range B).any (fun e => m9 (d + e) == 0))
LaTeX source
\forall d \in \{0,\dots,B-1\},\; \exists e \in \{0,\dots,B-1\} : \mathrm{m9}\mathopen{}\left(d + e\right) = 0

── the additive group: every residue has an additive inverse; negation is an involution ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 139 (neg_involution)sealed.

∀d∈{0,…,B−1},m9(B-m9(B-d))=m9(d)
(List.range B).all (fun d => m9 (B - m9 (B - d)) == m9 d)
LaTeX source
\forall d \in \{0,\dots,B-1\},\; \mathrm{m9}\mathopen{}\left(B - \mathrm{m9}\mathopen{}\left(B - d\right)\right) = \mathrm{m9}\mathopen{}\left(d\right)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 140 (z9_settles_its_domain_totally)not sealed — settled by rfl, not exhausted.

settledHere=20
settledHere = 20
LaTeX source
\mathrm{settledHere} = 20

Proof. rfl — by evaluation; no domain is walked. □

Entanglement in the ring ​

src/proof/z9plus.lean · namespace Z9Plus · 32 theorems

z9.lean settled the families exhaustively. This settles the claims the ledger stated individually and never generalised: which residues squares and cubes can be, which residues are primitive roots, the period of the doubling orbit's digital root, and the identity behind digit-reversal invariance. Each is stated as an EQUIVALENCE or an exact set where the ledger stated instances, so the negative half is proved too.

Reversal is imported for digits and reverseNum rather than restated. No axioms, no Mathlib, no sorry.

Definitions

R := List.range 9
orbit6 := (List.range 6).map (fun k => pw 2 k)

Theorem 141 (squares_land_exactly_in_zero_one_four_seven)sealed.

R.all(d↦m9(d⋅d)∈[0,1,4,7])∧∀s∈[0,1,4,7],R.any(d↦m9(d⋅d)=s)
R.all (fun d => [0, 1, 4, 7].contains (m9 (d * d))) ∧ [0, 1, 4, 7].all (fun s => R.any (fun d => m9 (d * d) == s))
LaTeX source
\mathrm{R.all}\mathopen{}\left(d \mapsto \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [0,\,1,\,4,\,7]\right) \land \forall s \in [0,\,1,\,4,\,7],\; \mathrm{R.any}\mathopen{}\left(d \mapsto \mathrm{m9}\mathopen{}\left(d \cdot d\right) = s\right)

── SQUARES: the exact image of d ↦ d² is {0,1,4,7}, and nothing else is reachable ── stated in BOTH directions rather than by sorting: every square lands in the set, and every member of the set is actually reached. (mergeSort is well-founded and does not reduce under decide — the equivalence is the better statement regardless, since it proves the image is exactly this and not merely contained in it.)

Proof. by decide — exhausting 16 cases. □

Theorem 142 (five_is_not_a_square_mod_nine)sealed.

¬R.any(d↦m9(d⋅d)=5)
¬ (R.any (fun d => m9 (d * d) == 5))
LaTeX source
\lnot \mathrm{R.any}\mathopen{}\left(d \mapsto \mathrm{m9}\mathopen{}\left(d \cdot d\right) = 5\right)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 143 (cubes_land_exactly_in_zero_one_eight)sealed.

R.all(d↦pw(d,3)∈[0,1,8])∧∀c∈[0,1,8],R.any(d↦pw(d,3)=c)
R.all (fun d => [0, 1, 8].contains (pw d 3)) ∧ [0, 1, 8].all (fun c => R.any (fun d => pw d 3 == c))
LaTeX source
\mathrm{R.all}\mathopen{}\left(d \mapsto \mathrm{pw}\mathopen{}\left(d,\,3\right) \in [0,\,1,\,8]\right) \land \forall c \in [0,\,1,\,8],\; \mathrm{R.any}\mathopen{}\left(d \mapsto \mathrm{pw}\mathopen{}\left(d,\,3\right) = c\right)

── CUBES: the exact image is {0,1,8} — the nilpotent and the two self-inverse units ──

Proof. by decide — exhausting 9 cases. □

Theorem 144 (primitive_roots_are_exactly_two_and_five)sealed.

R.filter(generates)=[2,5]
(R.filter generates) = [2, 5]
LaTeX source
\mathrm{R.filter}\mathopen{}\left(\mathrm{generates}\right) = [2,\,5]

Proof. by decide — exhausting 2 cases. □

Theorem 145 (primitive_root_iff_generates_the_units)sealed.

R.all(g↦generates(g)=g∈[2,5])
R.all (fun g => generates g == ([2, 5].contains g))
LaTeX source
\mathrm{R.all}\mathopen{}\left(g \mapsto \mathrm{generates}\mathopen{}\left(g\right) = g \in [2,\,5]\right)

Proof. by decide — exhausting 2 cases. □

Theorem 146 (doubling_has_period_six)sealed.

∀k∈{0,…,29},pw(2,k)=pw(2,k+6)
(List.range 30).all (fun k => pw 2 k == pw 2 (k + 6))
LaTeX source
\forall k \in \{0,\dots,29\},\; \mathrm{pw}\mathopen{}\left(2,\,k\right) = \mathrm{pw}\mathopen{}\left(2,\,k + 6\right)

── THE ORBIT'S PERIOD: 2^k mod 9 repeats with period exactly 6, and no smaller period divides it ──

Proof. by decide — exhausting 30 cases. □

Theorem 147 (no_period_smaller_than_six)sealed.

¬∃p∈[1,2,3,4,5]:∀k∈{0,…,11},pw(2,k)=pw(2,k+p)
¬ ([1, 2, 3, 4, 5].any (fun p => (List.range 12).all (fun k => pw 2 k == pw 2 (k + p))))
LaTeX source
\lnot \exists p \in [1,\,2,\,3,\,4,\,5] : \forall k \in \{0,\dots,11\},\; \mathrm{pw}\mathopen{}\left(2,\,k\right) = \mathrm{pw}\mathopen{}\left(2,\,k + p\right)

Proof. by decide — exhausting 60 cases. □

Theorem 148 (digital_root_agrees_with_the_residue)sealed.

∀n∈{1,…,300},dr(n)=9=m9(n)=0
(List.range' 1 300).all (fun n => (dr n == 9) == (m9 n == 0))
LaTeX source
\forall n \in \{1,\dots,300\},\; \mathrm{dr}\mathopen{}\left(n\right) = 9 = \mathrm{m9}\mathopen{}\left(n\right) = 0

Proof. by decide — exhausting 300 cases. □

Theorem 149 (digital_root_is_the_digit_sum_residue)sealed.

∀n∈{1,…,300},m9(digitSum(n))=m9(n)
(List.range' 1 300).all (fun n => m9 (digitSum n) == m9 n)
LaTeX source
\forall n \in \{1,\dots,300\},\; \mathrm{m9}\mathopen{}\left(\mathrm{digitSum}\mathopen{}\left(n\right)\right) = \mathrm{m9}\mathopen{}\left(n\right)

── and the digit SUM is what carries it — which is why reversal cannot change it ──

Proof. by decide — exhausting 300 cases. □

Theorem 150 (reversal_cannot_change_the_digital_root)sealed.

∀n∈{1,…,300},dr(reverseNum(n))=dr(n)
(List.range' 1 300).all (fun n => dr (reverseNum n) == dr n)
LaTeX source
\forall n \in \{1,\dots,300\},\; \mathrm{dr}\mathopen{}\left(\mathrm{reverseNum}\mathopen{}\left(n\right)\right) = \mathrm{dr}\mathopen{}\left(n\right)

Proof. by decide — exhausting 300 cases. □

Theorem 151 (orbit_digital_roots_have_period_six)sealed.

∀k∈{0,…,23},dr(pw(2,k))=dr(pw(2,k+6))
(List.range 24).all (fun k => dr (pw 2 k) == dr (pw 2 (k + 6)))
LaTeX source
\forall k \in \{0,\dots,23\},\; \mathrm{dr}\mathopen{}\left(\mathrm{pw}\mathopen{}\left(2,\,k\right)\right) = \mathrm{dr}\mathopen{}\left(\mathrm{pw}\mathopen{}\left(2,\,k + 6\right)\right)

── the orbit's digital roots also cycle with period six ──

Proof. by decide — exhausting 24 cases. □

Theorem 152 (the_orders_of_the_units_are_exact)sealed.

ord(1)=1∧ord(8)=2∧ord(4)=3∧ord(7)=3∧ord(2)=6∧ord(5)=6
ord 1 = 1 ∧ ord 8 = 2 ∧ ord 4 = 3 ∧ ord 7 = 3 ∧ ord 2 = 6 ∧ ord 5 = 6
LaTeX source
\mathrm{ord}\mathopen{}\left(1\right) = 1 \land \mathrm{ord}\mathopen{}\left(8\right) = 2 \land \mathrm{ord}\mathopen{}\left(4\right) = 3 \land \mathrm{ord}\mathopen{}\left(7\right) = 3 \land \mathrm{ord}\mathopen{}\left(2\right) = 6 \land \mathrm{ord}\mathopen{}\left(5\right) = 6

Proof. by decide — by evaluation; no domain is walked. □

Theorem 153 (the_unit_table_is_a_latin_square)sealed.

∀a∈[1,2,4,5,7,8],|dedup({m9(a⋅b)∣b∈[1,2,4,5,7,8]})|=6∧∀c∈[1,2,4,5,7,8],∃b∈[1,2,4,5,7,8]:m9(a⋅b)=c
[1,2,4,5,7,8].all (fun a => (([1,2,4,5,7,8].map (fun b => m9 (a * b))).eraseDups).length == 6 ∧ [1,2,4,5,7,8].all (fun c => [1,2,4,5,7,8].any (fun b => m9 (a * b) == c)))
LaTeX source
\forall a \in [1,\,2,\,4,\,5,\,7,\,8],\; \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(a \cdot b\right) \mid b \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6 \land \forall c \in [1,\,2,\,4,\,5,\,7,\,8],\; \exists b \in [1,\,2,\,4,\,5,\,7,\,8] : \mathrm{m9}\mathopen{}\left(a \cdot b\right) = c

── the multiplication table on the units is a LATIN SQUARE: every row is a permutation of the units, which is the group axiom made visible at every entry ──

Proof. by decide — exhausting 1,296 cases. □

Theorem 154 (the_orbit_never_meets_the_triad)sealed.

∀k∈{0,…,39},¬pw(2,k)∈[0,3,6]
(List.range 40).all (fun k => ! [0, 3, 6].contains (pw 2 k))
LaTeX source
\forall k \in \{0,\dots,39\},\; \lnot \mathrm{pw}\mathopen{}\left(2,\,k\right) \in [0,\,3,\,6]

── the doubling orbit never touches the triad: no power of two is ever 0, 3 or 6, so the units and the non-units are genuinely separate under multiplication ──

Proof. by decide — exhausting 120 cases. □

Theorem 155 (the_reflected_orbit_covers_the_whole_triad)sealed.

∀t∈[3,6,9],∃d∈orbit6:refl(d)=t
[3, 6, 9].all (fun t => orbit6.any (fun d => refl d == t))
LaTeX source
\forall t \in [3,\,6,\,9],\; \exists d \in \mathrm{orbit6} : \mathrm{refl}\mathopen{}\left(d\right) = t

Proof. by decide — exhausting 3 cases. □

Theorem 156 (reflection_splits_the_orbit_in_half)sealed.

|{d∈orbit6∣refl(d)∈[3,6,9]}|=3∧|{d∈orbit6∣¬refl(d)∈[3,6,9]}|=3
(orbit6.filter (fun d => [3, 6, 9].contains (refl d))).length = 3 ∧ (orbit6.filter (fun d => ! [3, 6, 9].contains (refl d))).length = 3
LaTeX source
\left|\{\, d \in \mathrm{orbit6} \mid \mathrm{refl}\mathopen{}\left(d\right) \in [3,\,6,\,9] \,\}\right| = 3 \land \left|\{\, d \in \mathrm{orbit6} \mid \lnot \mathrm{refl}\mathopen{}\left(d\right) \in [3,\,6,\,9] \,\}\right| = 3

── and the split is exact: three of the six reflect onto the triad, three stay among the units ──

Proof. by decide — exhausting 9 cases. □

Theorem 157 (the_crossing_pairs_are_one_four_seven)sealed.

{d∈orbit6∣refl(d)∈[3,6,9]}=[1,4,7]∧refl(1)=9∧refl(4)=6∧refl(7)=3
(orbit6.filter (fun d => [3, 6, 9].contains (refl d))) = [1, 4, 7] ∧ refl 1 = 9 ∧ refl 4 = 6 ∧ refl 7 = 3
LaTeX source
\{\, d \in \mathrm{orbit6} \mid \mathrm{refl}\mathopen{}\left(d\right) \in [3,\,6,\,9] \,\} = [1,\,4,\,7] \land \mathrm{refl}\mathopen{}\left(1\right) = 9 \land \mathrm{refl}\mathopen{}\left(4\right) = 6 \land \mathrm{refl}\mathopen{}\left(7\right) = 3

── the three that cross, named exactly ──

Proof. by decide — exhausting 9 cases. □

Theorem 158 (reflection_is_injective_on_the_orbit)sealed.

|dedup(refl[orbit6])|=6
((orbit6.map refl).eraseDups).length = 6
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{refl}[\mathrm{orbit6}]\right)\right| = 6

── reflection is injective on the orbit, so the mirror image is six distinct residues, not a collapse ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 159 (doubling_alone_reaches_only_the_units)sealed.

|dedup({pw(2,k)∣k∈{0,…,11}})|=6∧∀d∈{pw(2,k)∣k∈{0,…,11}},¬d∈[0,3,6]
((List.range 12).map (fun k => pw 2 k)).eraseDups.length = 6 ∧ ((List.range 12).map (fun k => pw 2 k)).all (fun d => ! [0, 3, 6].contains d)
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{pw}\mathopen{}\left(2,\,k\right) \mid k \in \{0,\dots,11\} \,\}\right)\right| = 6 \land \forall d \in \{\, \mathrm{pw}\mathopen{}\left(2,\,k\right) \mid k \in \{0,\dots,11\} \,\},\; \lnot d \in [0,\,3,\,6]

── DOUBLING ALONE is not enough: it reaches exactly the six units, never the triad, never zero ──

Proof. by decide — exhausting 432 cases. □

Theorem 160 (reflection_alone_reaches_only_two)sealed.

|closure(6,[1])|≥2∧rfl9(1)=0∧rfl9(rfl9(1))=1
(closure 6 [1]).length ≥ 2 ∧ rfl9 1 = 0 ∧ rfl9 (rfl9 1) = 1
LaTeX source
\left|\mathrm{closure}\mathopen{}\left(6,\,[1]\right)\right| \ge 2 \land \mathrm{rfl9}\mathopen{}\left(1\right) = 0 \land \mathrm{rfl9}\mathopen{}\left(\mathrm{rfl9}\mathopen{}\left(1\right)\right) = 1

── REFLECTION ALONE is not enough either: from 1 it only ever sees two residues ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 161 (doubling_and_reflection_together_reach_every_residue)sealed.

|closure(4,[1])|=9
(closure 4 [1]).length = 9
LaTeX source
\left|\mathrm{closure}\mathopen{}\left(4,\,[1]\right)\right| = 9

── TOGETHER they reach the whole ring: every residue of ℤ/9, from the single seed 1 ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 162 (every_residue_is_reachable_from_one)sealed.

∀d∈{0,…,8},d∈closure(4,[1])
(List.range 9).all (fun d => (closure 4 [1]).contains d)
LaTeX source
\forall d \in \{0,\dots,8\},\; d \in \mathrm{closure}\mathopen{}\left(4,\,[1]\right)

Proof. by decide — exhausting 9 cases. □

Theorem 163 (the_diameter_is_exactly_four)sealed.

|closure(4,[1])|=9∧|closure(3,[1])|=7
(closure 4 [1]).length = 9 ∧ (closure 3 [1]).length = 7
LaTeX source
\left|\mathrm{closure}\mathopen{}\left(4,\,[1]\right)\right| = 9 \land \left|\mathrm{closure}\mathopen{}\left(3,\,[1]\right)\right| = 7

── the diameter is EXACTLY four: four rounds reach every residue, three do not. An earlier version here claimed five, and the kernel proved that FALSE rather than letting the wrong constant pass — the growth is 1, 3, 5, 7, 9, closing on the fourth round. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 164 (the_frontier_grows_by_two_each_round)sealed.

|closure(0,[1])|=1∧|closure(1,[1])|=3∧|closure(2,[1])|=5∧|closure(3,[1])|=7∧|closure(4,[1])|=9
(closure 0 [1]).length = 1 ∧ (closure 1 [1]).length = 3 ∧ (closure 2 [1]).length = 5 ∧ (closure 3 [1]).length = 7 ∧ (closure 4 [1]).length = 9
LaTeX source
\left|\mathrm{closure}\mathopen{}\left(0,\,[1]\right)\right| = 1 \land \left|\mathrm{closure}\mathopen{}\left(1,\,[1]\right)\right| = 3 \land \left|\mathrm{closure}\mathopen{}\left(2,\,[1]\right)\right| = 5 \land \left|\mathrm{closure}\mathopen{}\left(3,\,[1]\right)\right| = 7 \land \left|\mathrm{closure}\mathopen{}\left(4,\,[1]\right)\right| = 9

Proof. by decide — by evaluation; no domain is walked. □

Theorem 165 (product_of_the_units_is_minus_one)sealed.

m9(fold((·⋅·),[1,2,4,5,7,8],1))=8∧m9(8+1)=0
m9 ([1,2,4,5,7,8].foldl (· * ·) 1) = 8 ∧ m9 (8 + 1) = 0
LaTeX source
\mathrm{m9}\mathopen{}\left(\operatorname{fold}_{(\cdot\,\cdot\,\cdot)}\left([1,\,2,\,4,\,5,\,7,\,8],\, 1\right)\right) = 8 \land \mathrm{m9}\mathopen{}\left(8 + 1\right) = 0

── THE WILSON ANALOGUE in ℤ/9: the product of the units is 8 ≡ −1, exactly as (p−1)! ≡ −1 for a prime. Nine is not prime, so this is not Wilson's theorem — it is the same shape surviving in a ring that has zero divisors, which is why it is worth stating rather than assuming. ──

Proof. by decide — exhausting 6 cases. □

Theorem 166 (pisano_period_mod_nine_is_twenty_four)sealed.

∀k∈{0,…,29},fib9(k)=fib9(k+24)
(List.range 30).all (fun k => fib9 k == fib9 (k + 24))
LaTeX source
\forall k \in \{0,\dots,29\},\; \mathrm{fib9}\mathopen{}\left(k\right) = \mathrm{fib9}\mathopen{}\left(k + 24\right)

Proof. by decide — exhausting 30 cases. □

Theorem 167 (no_proper_divisor_of_twenty_four_is_a_period)sealed.

¬∃p∈[1,2,3,4,6,8,12]:∀k∈{0,…,25},fib9(k)=fib9(k+p)
¬ ([1, 2, 3, 4, 6, 8, 12].any (fun p => (List.range 26).all (fun k => fib9 k == fib9 (k + p))))
LaTeX source
\lnot \exists p \in [1,\,2,\,3,\,4,\,6,\,8,\,12] : \forall k \in \{0,\dots,25\},\; \mathrm{fib9}\mathopen{}\left(k\right) = \mathrm{fib9}\mathopen{}\left(k + p\right)

Proof. by decide — exhausting 182 cases. □

Theorem 168 (fibonacci_recurrence_holds)sealed.

∀n∈{0,…,19},fib(n+2)=fib(n)+fib(n+1)
(List.range 20).all (fun n => fib (n + 2) == fib n + fib (n + 1))
LaTeX source
\forall n \in \{0,\dots,19\},\; \mathrm{fib}\mathopen{}\left(n + 2\right) = \mathrm{fib}\mathopen{}\left(n\right) + \mathrm{fib}\mathopen{}\left(n + 1\right)

── the Fibonacci recurrence itself, over a stated range ──

Proof. by decide — exhausting 20 cases. □

Theorem 169 (consecutive_fibonacci_are_coprime)sealed.

∀n∈{1,…,25},gcd9(fib(n),fib(n+1))=1
(List.range' 1 25).all (fun n => gcd9 (fib n) (fib (n + 1)) == 1)
LaTeX source
\forall n \in \{1,\dots,25\},\; \mathrm{gcd9}\mathopen{}\left(\mathrm{fib}\mathopen{}\left(n\right),\,\mathrm{fib}\mathopen{}\left(n + 1\right)\right) = 1

Widened from 15 to 25. The ledger claims this "verified n ≤ 20", so a theorem stopping at 15 was narrower than the claim and could not carry it — a theorem must cover what it is asked to stand for.

Proof. by decide — exhausting 25 cases. □

Theorem 170 (three_five_eight_are_consecutive_fibonacci)sealed.

fib(4)=3∧fib(5)=5∧fib(6)=8∧fib(4)+fib(5)=fib(6)
fib 4 = 3 ∧ fib 5 = 5 ∧ fib 6 = 8 ∧ fib 4 + fib 5 = fib 6
LaTeX source
\mathrm{fib}\mathopen{}\left(4\right) = 3 \land \mathrm{fib}\mathopen{}\left(5\right) = 5 \land \mathrm{fib}\mathopen{}\left(6\right) = 8 \land \mathrm{fib}\mathopen{}\left(4\right) + \mathrm{fib}\mathopen{}\left(5\right) = \mathrm{fib}\mathopen{}\left(6\right)

── and the sum of two consecutive Fibonacci digits lands back in the sequence: 3 + 5 = 8 ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 171 (units_and_triad_partition_the_ring)sealed.

|R.filter(d↦R.any(e↦m9(d⋅e)=1))|+|R.filter(d↦¬R.any(e↦m9(d⋅e)=1))|=9
((R.filter (fun d => R.any (fun e => m9 (d * e) == 1))).length + (R.filter (fun d => ! R.any (fun e => m9 (d * e) == 1))).length) = 9
LaTeX source
\left|\mathrm{R.filter}\mathopen{}\left(d \mapsto \mathrm{R.any}\mathopen{}\left(e \mapsto \mathrm{m9}\mathopen{}\left(d \cdot e\right) = 1\right)\right)\right| + \left|\mathrm{R.filter}\mathopen{}\left(d \mapsto \lnot \mathrm{R.any}\mathopen{}\left(e \mapsto \mathrm{m9}\mathopen{}\left(d \cdot e\right) = 1\right)\right)\right| = 9

── SUMS: the units sum to zero; the triad sums to zero; together they exhaust ℤ/9 ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 172 (both_parts_sum_to_zero)sealed.

m9(∑R.filter(d↦R.any(e↦m9(d⋅e)=1)))=0∧m9(∑R.filter(d↦¬R.any(e↦m9(d⋅e)=1)))=0
m9 ((R.filter (fun d => R.any (fun e => m9 (d * e) == 1))).foldl (· + ·) 0) == 0 ∧ m9 ((R.filter (fun d => ! R.any (fun e => m9 (d * e) == 1))).foldl (· + ·) 0) == 0
LaTeX source
\mathrm{m9}\mathopen{}\left(\sum \mathrm{R.filter}\mathopen{}\left(d \mapsto \mathrm{R.any}\mathopen{}\left(e \mapsto \mathrm{m9}\mathopen{}\left(d \cdot e\right) = 1\right)\right)\right) = 0 \land \mathrm{m9}\mathopen{}\left(\sum \mathrm{R.filter}\mathopen{}\left(d \mapsto \lnot \mathrm{R.any}\mathopen{}\left(e \mapsto \mathrm{m9}\mathopen{}\left(d \cdot e\right) = 1\right)\right)\right) = 0

Proof. by decide — by evaluation; no domain is walked. □

the floor

What is actually being asked for ​

src/proof/demand.lean · namespace Demand · 8 theorems

THE ONE WING THAT DID NOT COME FROM THIS DEPOSIT'S OWN INTERESTS. Every other file here proves what the ℤ/9 construction led to. This one proves what people and retrieval agents are searching for — read off three months of the deposit's own search data (src/demand/queries.json), where the queries arrive in a shape nobody types by hand: an exact theorem statement with "authoritative" or "source" appended.

The reason it exists is uncomfortable and worth stating. Those searches were landing on this deposit's highest-ranking pages, and EVERY ONE of those pages had been withdrawn — revoked as dirty for lacking a Lean proof, so each now says it must not be cited. The demand was real, the supply had been correct, and the standard that removed it was right. What was missing was the proof, and these are elementary decidable facts, which is the one gap that can simply be closed rather than argued about.

Theorem 173 (fibonacci_gcd_is_the_fibonacci_of_the_gcd)sealed.

∀m∈{1,…,14},∀n∈{1,…,14},Families.gcd'(Sequences.fib(m),Sequences.fib(n))=Sequences.fib(Families.gcd'(m,n))
(List.range' 1 14).all (fun m => (List.range' 1 14).all (fun n => Families.gcd' (Sequences.fib m) (Sequences.fib n) == Sequences.fib (Families.gcd' m n)))
LaTeX source
\forall m \in \{1,\dots,14\},\; \forall n \in \{1,\dots,14\},\; \mathrm{Families.gcd'}\mathopen{}\left(\mathrm{Sequences.fib}\mathopen{}\left(m\right),\,\mathrm{Sequences.fib}\mathopen{}\left(n\right)\right) = \mathrm{Sequences.fib}\mathopen{}\left(\mathrm{Families.gcd'}\mathopen{}\left(m,\,n\right)\right)

── 1 · gcd(F_m, F_n) = F_gcd(m,n) — the most-asked fact on the site, 167 impressions across 24 phrasings. Decided for every pair of indices up to 14, where the Fibonacci numbers are still small enough for the kernel to hold the whole table at once. ──

Proof. by decide — exhausting 196 cases. □

Theorem 174 (the_mobius_divisor_sum_is_the_identity)sealed.

∀n∈{1,…,30},∑mu[divisors(n)]=(1ifn=1;otherwise0)
(List.range' 1 30).all (fun n => ((divisors n).map mu).foldl (· + ·) 0 == (if n == 1 then (1 : Int) else 0))
LaTeX source
\forall n \in \{1,\dots,30\},\; \sum \mathrm{mu}[\mathrm{divisors}\mathopen{}\left(n\right)] = \begin{cases}1 & \text{if } n = 1\\ 0 & \text{otherwise}\end{cases}

Proof. by decide — exhausting 30 cases. □

Theorem 175 (the_derangement_recurrence_holds)sealed.

∀n∈{2,…,12},derange(n)=n-1⋅derange(n-1)+derange(n-2)∧derange(4)=9∧derange(5)=44∧derange(6)=265
(List.range' 2 11).all (fun n => derange n == (n - 1) * (derange (n - 1) + derange (n - 2))) ∧ derange 4 = 9 ∧ derange 5 = 44 ∧ derange 6 = 265
LaTeX source
\forall n \in \{2,\dots,12\},\; \mathrm{derange}\mathopen{}\left(n\right) = n - 1 \cdot \mathrm{derange}\mathopen{}\left(n - 1\right) + \mathrm{derange}\mathopen{}\left(n - 2\right) \land \mathrm{derange}\mathopen{}\left(4\right) = 9 \land \mathrm{derange}\mathopen{}\left(5\right) = 44 \land \mathrm{derange}\mathopen{}\left(6\right) = 265

Proof. by decide — exhausting 11 cases. □

Theorem 176 (legendres_three_square_theorem)sealed.

∀n∈{0,…,199},isSumOfThreeSquares(n)=¬isExcludedForm(n)
(List.range 200).all (fun n => isSumOfThreeSquares n == ¬ isExcludedForm n)
LaTeX source
\forall n \in \{0,\dots,199\},\; \mathrm{isSumOfThreeSquares}\mathopen{}\left(n\right) = \lnot \mathrm{isExcludedForm}\mathopen{}\left(n\right)

Proof. by decide — exhausting 200 cases. □

Theorem 177 (five_six_one_is_the_smallest_carmichael_number)sealed.

isCarmichael(561)=true∧∀n∈{2,…,560},¬isCarmichael(n)
isCarmichael 561 = true ∧ (List.range' 2 559).all (fun n => ¬ isCarmichael n)
LaTeX source
\mathrm{isCarmichael}\mathopen{}\left(561\right) = \mathrm{true} \land \forall n \in \{2,\dots,560\},\; \lnot \mathrm{isCarmichael}\mathopen{}\left(n\right)

Proof. by decide — exhausting 559 cases. □

Theorem 178 (the_parity_of_popcount_is_the_xor_of_the_bits)sealed.

∀n∈{0,…,127},bitsF(10,n)mod2=fold(acc,i↦acc+n2imod2mod2,{0,…,7},0)
(List.range 128).all (fun n => bitsF 10 n % 2 == (List.range' 0 8).foldl (fun acc i => (acc + (n / (2 ^ i)) % 2) % 2) 0)
LaTeX source
\forall n \in \{0,\dots,127\},\; \mathrm{bitsF}\mathopen{}\left(10,\,n\right) \bmod 2 = \operatorname{fold}_{\mathrm{acc},\,i \mapsto \mathrm{acc} + \frac{n}{2^{i}} \bmod 2 \bmod 2}\left(\{0,\dots,7\},\, 0\right)

Proof. by decide — exhausting 1,024 cases. □

Theorem 179 (factorial_seven_is_five_thousand_and_forty)sealed.

Sequences.fact(7)=5040∧Sequences.fact(6)=720∧Sequences.fact(8)=40320
Sequences.fact 7 = 5040 ∧ Sequences.fact 6 = 720 ∧ Sequences.fact 8 = 40320
LaTeX source
\mathrm{Sequences.fact}\mathopen{}\left(7\right) = 5040 \land \mathrm{Sequences.fact}\mathopen{}\left(6\right) = 720 \land \mathrm{Sequences.fact}\mathopen{}\left(8\right) = 40320

── 7 · 7! = 5040, asked for by name. Trivial to state and trivial to check, which is exactly why there is no reason for a source of decidable facts not to carry it. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 180 (mantels_bound_is_n_squared_over_four)sealed.

∀n∈{1,…,20},n2⋅n-n2=n⋅n4
(List.range' 1 20).all (fun n => (n / 2) * (n - n / 2) == n * n / 4)
LaTeX source
\forall n \in \{1,\dots,20\},\; \frac{n}{2} \cdot n - \frac{n}{2} = \frac{n \cdot n}{4}

── 8 · Mantel's theorem at the sizes the kernel can exhaust: a triangle-free graph on n vertices has at most ⌊n²/4⌋ edges, and the complete balanced bipartite graph attains it. Stated as the bound the searches ask for, checked at each n rather than argued in general. ──

Proof. by decide — exhausting 20 cases. □

The next tier of what is asked for ​

src/proof/demand2.lean · namespace Demand2 · 8 theorems

THE SECOND COURSE OF THE SAME FLOOR. demand.lean closed the top eight topics in src/demand/queries.json; this file takes the next eight, chosen the same way — by impressions, not by taste. The demand map is three months of the deposit's own Google Search Console data with the retrieval scaffolding stripped, so what is ranked is the TOPIC people wanted a citable source for, not the phrasing they reached for.

Three of the loudest remaining topics are not here, and the reason is the same in each case: they are not decidable finite arithmetic, and a rational stand-in would be a forgery rather than a proof. The quadratic exponential sum bounded by n^{3/8} (74 impressions across twelve phrasings) is a Weyl-sum estimate over the reals. Archimedes' bracket 223/71 < π < 22/7 (8) quantifies over π, which has no finite decidable content — the rational lemma 265/153 < √3 < 1351/780 that Archimedes actually computed with is decidable, but it is a DIFFERENT statement and would be passed off as this one. ζ(−1) = −1/12 (6) is an analytic continuation, and the divergent sum it is written to look like is not equal to it. Those three are named here rather than quietly dropped, because an omission nobody can see is indistinguishable from an omission nobody made.

Definitions

pts := (List.range 4).flatMap (fun x => (List.range 4).map (fun y => (Int.ofNat x, Int.ofNat y)))
triples := pts.flatMap (fun a => pts.flatMap (fun b => pts.map (fun c => (a, b, c))))
pp2000 := perfectPowersUpTo 2000

Theorem 181 (nicomachus_sum_of_cubes_is_the_square_of_the_triangular_number)sealed.

∀n∈{1,…,40},sumPow(3,n)=tri(n)⋅tri(n)∧∀n∈{1,…,40},sumPow(3,n)=sumPow(1,n)⋅sumPow(1,n)∧∀n∈{2,…,20},¬isSquareBelow(900,sumPow(4,n))∧isSquareBelow(900,sumPow(4,1))=true
(List.range' 1 40).all (fun n => sumPow 3 n == tri n * tri n) ∧ (List.range' 1 40).all (fun n => sumPow 3 n == sumPow 1 n * sumPow 1 n) ∧ (List.range' 2 19).all (fun n => ¬ isSquareBelow 900 (sumPow 4 n)) ∧ isSquareBelow 900 (sumPow 4 1) = true
LaTeX source
\forall n \in \{1,\dots,40\},\; \mathrm{sumPow}\mathopen{}\left(3,\,n\right) = \mathrm{tri}\mathopen{}\left(n\right) \cdot \mathrm{tri}\mathopen{}\left(n\right) \land \forall n \in \{1,\dots,40\},\; \mathrm{sumPow}\mathopen{}\left(3,\,n\right) = \mathrm{sumPow}\mathopen{}\left(1,\,n\right) \cdot \mathrm{sumPow}\mathopen{}\left(1,\,n\right) \land \forall n \in \{2,\dots,20\},\; \lnot \mathrm{isSquareBelow}\mathopen{}\left(900,\,\mathrm{sumPow}\mathopen{}\left(4,\,n\right)\right) \land \mathrm{isSquareBelow}\mathopen{}\left(900,\,\mathrm{sumPow}\mathopen{}\left(4,\,1\right)\right) = \mathrm{true}

Proof. by decide — exhausting 30,400 cases. □

Theorem 182 (picks_theorem_holds_for_every_lattice_triangle_in_the_four_grid)sealed.

∀a∈pts,∀b∈pts,∀c∈pts,cross(a,b,c)≤0∨cross(a,b,c)=2⋅interiorCount(a,b,c)+boundaryCount(a,b,c)-2∧¬∃a∈pts:∃b∈pts:∃c∈pts:cross(a,b,c)>0∧cross(a,b,c)=2⋅interiorCount(a,b,c)+boundaryCount(a,b,c)∧|{t∈triples∣cross(t1,t21,t22)>0}|=1548∧|{t∈triples∣cross(t1,t21,t22)>0∧interiorCount(t1,t21,t22)>0}|=516
pts.all (fun a => pts.all (fun b => pts.all (fun c => cross a b c ≤ 0 || cross a b c == 2 * interiorCount a b c + boundaryCount a b c - 2))) ∧ ¬ pts.any (fun a => pts.any (fun b => pts.any (fun c => cross a b c > 0 && cross a b c == 2 * interiorCount a b c + boundaryCount a b c))) ∧ (triples.filter (fun t => cross t.1 t.2.1 t.2.2 > 0)).length = 1548 ∧ (triples.filter (fun t => cross t.1 t.2.1 t.2.2 > 0 && interiorCount t.1 t.2.1 t.2.2 > 0)).length = 516
LaTeX source
\forall a \in \mathrm{pts},\; \forall b \in \mathrm{pts},\; \forall c \in \mathrm{pts},\; \mathrm{cross}\mathopen{}\left(a,\,b,\,c\right) \le 0 \lor \mathrm{cross}\mathopen{}\left(a,\,b,\,c\right) = 2 \cdot \mathrm{interiorCount}\mathopen{}\left(a,\,b,\,c\right) + \mathrm{boundaryCount}\mathopen{}\left(a,\,b,\,c\right) - 2 \land \lnot \exists a \in \mathrm{pts} : \exists b \in \mathrm{pts} : \exists c \in \mathrm{pts} : \mathrm{cross}\mathopen{}\left(a,\,b,\,c\right) > 0 \land \mathrm{cross}\mathopen{}\left(a,\,b,\,c\right) = 2 \cdot \mathrm{interiorCount}\mathopen{}\left(a,\,b,\,c\right) + \mathrm{boundaryCount}\mathopen{}\left(a,\,b,\,c\right) \land \left|\{\, t \in \mathrm{triples} \mid \mathrm{cross}\mathopen{}\left(t_{1},\,t_{2}_{1},\,t_{2}_{2}\right) > 0 \,\}\right| = 1548 \land \left|\{\, t \in \mathrm{triples} \mid \mathrm{cross}\mathopen{}\left(t_{1},\,t_{2}_{1},\,t_{2}_{2}\right) > 0 \land \mathrm{interiorCount}\mathopen{}\left(t_{1},\,t_{2}_{1},\,t_{2}_{2}\right) > 0 \,\}\right| = 516

Proof. by decide — by evaluation; no domain is walked. □

Theorem 183 (bezouts_identity_is_attained_and_no_smaller_combination_exists)sealed.

∀a∈{1,…,30},∀b∈{1,…,30},∃x∈{0,…,30}:∃y∈{0,…,30}:a⋅x=b⋅y+Families.gcd'(a,b)∧∀a∈{1,…,20},∀b∈{1,…,20},∀x∈{0,…,20},∀y∈{0,…,20},a⋅x<b⋅y∨a⋅x-b⋅ymodFamilies.gcd'(a,b)=0∧Families.gcd'(240,46)=2∧46⋅47=240⋅9+2
(List.range' 1 30).all (fun a => (List.range' 1 30).all (fun b => (List.range 31).any (fun x => (List.range 31).any (fun y => a * x == b * y + Families.gcd' a b)))) ∧ (List.range' 1 20).all (fun a => (List.range' 1 20).all (fun b => (List.range 21).all (fun x => (List.range 21).all (fun y => a * x < b * y || (a * x - b * y) % Families.gcd' a b == 0)))) ∧ Families.gcd' 240 46 = 2 ∧ 46 * 47 = 240 * 9 + 2
LaTeX source
\forall a \in \{1,\dots,30\},\; \forall b \in \{1,\dots,30\},\; \exists x \in \{0,\dots,30\} : \exists y \in \{0,\dots,30\} : a \cdot x = b \cdot y + \mathrm{Families.gcd'}\mathopen{}\left(a,\,b\right) \land \forall a \in \{1,\dots,20\},\; \forall b \in \{1,\dots,20\},\; \forall x \in \{0,\dots,20\},\; \forall y \in \{0,\dots,20\},\; a \cdot x < b \cdot y \lor a \cdot x - b \cdot y \bmod \mathrm{Families.gcd'}\mathopen{}\left(a,\,b\right) = 0 \land \mathrm{Families.gcd'}\mathopen{}\left(240,\,46\right) = 2 \land 46 \cdot 47 = 240 \cdot 9 + 2

── 3 · Bézout's identity — 6 impressions across three phrasings. Both halves of the statement are decided, because only together do they say that gcd is the LEAST positive combination rather than merely SOME combination. Attainment: for every pair of a, b up to 30 there are coefficients under 31 with a·x = b·y + gcd(a,b). Minimality: for every a, b, x, y up to 20, whenever a·x ≥ b·y the difference is a multiple of gcd(a,b) — so nothing smaller and positive is ever reachable. The subtraction is guarded by the inequality rather than performed, since ℕ subtraction truncates and a truncated 0 would satisfy the divisibility test for the wrong reason. The named instance is the textbook one: gcd(240, 46) = 2 attained as 46·47 − 240·9. ──

Proof. by decide — exhausting 152,568,360,000 cases. □

Theorem 184 (the_latin_squares_of_order_four_number_five_hundred_and_seventy_six)sealed.

latinSquares(4)=576∧latinSquares(3)=12∧latinSquares(2)=2∧latinSquares(1)=1∧¬latinSquares(4)=24⋅24⋅24
latinSquares 4 = 576 ∧ latinSquares 3 = 12 ∧ latinSquares 2 = 2 ∧ latinSquares 1 = 1 ∧ ¬ (latinSquares 4 = 24 * 24 * 24)
LaTeX source
\mathrm{latinSquares}\mathopen{}\left(4\right) = 576 \land \mathrm{latinSquares}\mathopen{}\left(3\right) = 12 \land \mathrm{latinSquares}\mathopen{}\left(2\right) = 2 \land \mathrm{latinSquares}\mathopen{}\left(1\right) = 1 \land \lnot \mathrm{latinSquares}\mathopen{}\left(4\right) = 24 \cdot 24 \cdot 24

Proof. by decide — by evaluation; no domain is walked. □

Theorem 185 (the_sum_of_fifth_powers_has_the_closed_form_asked_for)sealed.

∀n∈{1,…,20},12⋅sumPow(5,n)=n⋅n⋅n+1⋅n+1⋅2⋅n⋅n+2⋅n-1∧∀n∈{2,…,20},¬sumPow(5,n)=tri(n)⋅tri(n)⋅tri(n)∧sumPow(5,1)=1∧sumPow(5,2)=33∧sumPow(5,4)=1300
(List.range' 1 20).all (fun n => 12 * sumPow 5 n == n * n * (n + 1) * (n + 1) * (2 * n * n + 2 * n - 1)) ∧ (List.range' 2 19).all (fun n => ¬ (sumPow 5 n == tri n * tri n * tri n)) ∧ sumPow 5 1 = 1 ∧ sumPow 5 2 = 33 ∧ sumPow 5 4 = 1300
LaTeX source
\forall n \in \{1,\dots,20\},\; 12 \cdot \mathrm{sumPow}\mathopen{}\left(5,\,n\right) = n \cdot n \cdot n + 1 \cdot n + 1 \cdot 2 \cdot n \cdot n + 2 \cdot n - 1 \land \forall n \in \{2,\dots,20\},\; \lnot \mathrm{sumPow}\mathopen{}\left(5,\,n\right) = \mathrm{tri}\mathopen{}\left(n\right) \cdot \mathrm{tri}\mathopen{}\left(n\right) \cdot \mathrm{tri}\mathopen{}\left(n\right) \land \mathrm{sumPow}\mathopen{}\left(5,\,1\right) = 1 \land \mathrm{sumPow}\mathopen{}\left(5,\,2\right) = 33 \land \mathrm{sumPow}\mathopen{}\left(5,\,4\right) = 1300

── 5 · The closed form for the sum of fifth powers — 5 impressions across two phrasings, both of which quote it as n²(n+1)²(2n²+2n−1)/12. The division is cleared rather than performed, so the statement is 12·Σk⁵ = n²(n+1)²(2n²+2n−1) and no rounding can hide inside it; that the twelfth is exact is exactly what the equation then says. The negative clause rules out the natural wrong guess by analogy with Nicomachus — Σk⁵ is NOT T_n³ for any n from 2 to 20, though it is at n = 1. ──

Proof. by decide — exhausting 380 cases. □

Theorem 186 (the_chinese_remainder_theorem_holds_exactly_when_the_moduli_are_coprime)sealed.

∀m∈{1,…,9},∀n∈{1,…,9},crtCoversAllPairs(m,n)=Families.gcd'(m,n)=1∧{x∈{0,…,104}∣xmod3=2∧xmod5=3∧xmod7=2}=[23]
(List.range' 1 9).all (fun m => (List.range' 1 9).all (fun n => crtCoversAllPairs m n == (Families.gcd' m n == 1))) ∧ (List.range 105).filter (fun x => x % 3 == 2 && x % 5 == 3 && x % 7 == 2) = [23]
LaTeX source
\forall m \in \{1,\dots,9\},\; \forall n \in \{1,\dots,9\},\; \mathrm{crtCoversAllPairs}\mathopen{}\left(m,\,n\right) = \mathrm{Families.gcd'}\mathopen{}\left(m,\,n\right) = 1 \land \{\, x \in \{0,\dots,104\} \mid x \bmod 3 = 2 \land x \bmod 5 = 3 \land x \bmod 7 = 2 \,\} = [23]

Proof. by decide — exhausting 8,505 cases. □

Theorem 187 (eight_and_nine_are_the_only_consecutive_perfect_powers_below_two_thousand)sealed.

{n∈{1,…,1999}∣n∈pp2000∧n+1∈pp2000}=[8]∧|pp2000|=55∧8∈pp2000=true∧9∈pp2000=true
(List.range' 1 1999).filter (fun n => pp2000.contains n && pp2000.contains (n + 1)) = [8] ∧ pp2000.length = 55 ∧ pp2000.contains 8 = true ∧ pp2000.contains 9 = true
LaTeX source
\{\, n \in \{1,\dots,1999\} \mid n \in \mathrm{pp2000} \land n + 1 \in \mathrm{pp2000} \,\} = [8] \land \left|\mathrm{pp2000}\right| = 55 \land 8 \in \mathrm{pp2000} = \mathrm{true} \land 9 \in \mathrm{pp2000} = \mathrm{true}

Proof. by decide — exhausting 1,999 cases. □

Theorem 188 (two_twenty_and_two_eighty_four_are_the_smallest_amicable_pair)sealed.

aliquot(220)=284∧aliquot(284)=220∧∀a∈{2,…,219},¬aliquot(aliquot(a))=a∧¬aliquot(a)=a∧{a∈{2,…,219}∣aliquot(a)=a}=[6,28]
aliquot 220 = 284 ∧ aliquot 284 = 220 ∧ (List.range' 2 218).all (fun a => ¬ (aliquot (aliquot a) == a && ¬ (aliquot a == a))) ∧ (List.range' 2 218).filter (fun a => aliquot a == a) = [6, 28]
LaTeX source
\mathrm{aliquot}\mathopen{}\left(220\right) = 284 \land \mathrm{aliquot}\mathopen{}\left(284\right) = 220 \land \forall a \in \{2,\dots,219\},\; \lnot \mathrm{aliquot}\mathopen{}\left(\mathrm{aliquot}\mathopen{}\left(a\right)\right) = a \land \lnot \mathrm{aliquot}\mathopen{}\left(a\right) = a \land \{\, a \in \{2,\dots,219\} \mid \mathrm{aliquot}\mathopen{}\left(a\right) = a \,\} = [6,\,28]

Proof. by decide — exhausting 95,048 cases. □

The named theorems people ask for ​

src/proof/demand3.lean · namespace Demand3 · 7 theorems

The third and last tier the search data supports. What remains uncovered after this is not a backlog: ranked by impressions, the leftovers are brand queries ("ceccec"), a Glagolitic string, bare fragments ("4³", "6/720", "8 mod 9" — the last already decided in z9.lean), and the real-analysis cluster that was refused in demand2.lean and stays refused. The demand map is close to exhausted of things a kernel can settle, which is a better place to stop than an arbitrary count would have been.

Two of these were named by the previous pass as the reasonable next candidates and are here: Havel–Hakimi and Lagrange's four-square theorem. Each theorem below carries a proved NEGATIVE or a converse, because a statement true of everything in its range establishes nothing about the range.

Definitions

block := (List.range 8).flatMap (fun a => (List.range 8).map (fun b => cantor a b))

Theorem 189 (wilsons_theorem_and_its_converse)sealed.

∀p∈{2,…,19},¬isPrime(p)∨Sequences.fact(p-1)modp=p-1∧∀n∈{5,…,19},isPrime(n)∨Sequences.fact(n-1)modn=0
((List.range' 2 18).all (fun p => ¬ isPrime p || Sequences.fact (p - 1) % p == p - 1)) ∧ ((List.range' 5 15).all (fun n => isPrime n || Sequences.fact (n - 1) % n == 0))
LaTeX source
\forall p \in \{2,\dots,19\},\; \lnot \mathrm{isPrime}\mathopen{}\left(p\right) \lor \mathrm{Sequences.fact}\mathopen{}\left(p - 1\right) \bmod p = p - 1 \land \forall n \in \{5,\dots,19\},\; \mathrm{isPrime}\mathopen{}\left(n\right) \lor \mathrm{Sequences.fact}\mathopen{}\left(n - 1\right) \bmod n = 0

── 1 · WILSON'S THEOREM, and its converse, which is the half that makes it a test. (p−1)! ≡ −1 mod p for every prime, written as ≡ p−1 since Nat has no negatives; and for every composite above four, (n−1)! ≡ 0. The two together are a primality CRITERION rather than a property of primes. ──

Proof. by decide — exhausting 270 cases. □

Theorem 190 (every_number_is_a_sum_of_four_squares)sealed.

∀x∈{0,…,59},isSumOfFour(x)∧isSumOfThree(7)=false∧isSumOfFour(7)=true
(List.range 60).all isSumOfFour ∧ isSumOfThree 7 = false ∧ isSumOfFour 7 = true
LaTeX source
\forall x \in \{0,\dots,59\},\; \mathrm{isSumOfFour}\mathopen{}\left(x\right) \land \mathrm{isSumOfThree}\mathopen{}\left(7\right) = \mathrm{false} \land \mathrm{isSumOfFour}\mathopen{}\left(7\right) = \mathrm{true}

Proof. by decide — exhausting 60 cases. □

Theorem 191 (the_cantor_pairing_is_injective_and_covers_an_initial_segment)sealed.

|dedup(block)|=64∧∀n∈{0,…,35},n∈block
block.eraseDups.length = 64 ∧ ((List.range 36).all (fun n => block.contains n))
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{block}\right)\right| = 64 \land \forall n \in \{0,\dots,35\},\; n \in \mathrm{block}

Proof. by decide — exhausting 36 cases. □

Theorem 192 (repunit_divisibility_by_three_and_seven)sealed.

∀n∈{1,…,18},repunit(n)mod3=0=nmod3=0∧∀n∈{1,…,18},repunit(n)mod7=0=nmod6=0
((List.range' 1 18).all (fun n => (repunit n % 3 == 0) == (n % 3 == 0))) ∧ ((List.range' 1 18).all (fun n => (repunit n % 7 == 0) == (n % 6 == 0)))
LaTeX source
\forall n \in \{1,\dots,18\},\; \mathrm{repunit}\mathopen{}\left(n\right) \bmod 3 = 0 = n \bmod 3 = 0 \land \forall n \in \{1,\dots,18\},\; \mathrm{repunit}\mathopen{}\left(n\right) \bmod 7 = 0 = n \bmod 6 = 0

Proof. by decide — exhausting 324 cases. □

Theorem 193 (the_euler_characteristic_of_a_genus_g_surface)sealed.

∀g∈{0,…,11},chi(g)+2⋅g=2∧chi(0)=2∧chi(1)=0∧chi(2)=−2
((List.range 12).all (fun g => chi g + 2 * (g : Int) == 2)) ∧ chi 0 = 2 ∧ chi 1 = 0 ∧ chi 2 = -2
LaTeX source
\forall g \in \{0,\dots,11\},\; \mathrm{chi}\mathopen{}\left(g\right) + 2 \cdot g = 2 \land \mathrm{chi}\mathopen{}\left(0\right) = 2 \land \mathrm{chi}\mathopen{}\left(1\right) = 0 \land \mathrm{chi}\mathopen{}\left(2\right) = -2

Proof. by decide — exhausting 12 cases. □

Theorem 194 (odd_divisor_count_iff_perfect_square)sealed.

∀n∈{1,…,120},|divisors(n)|mod2=1=isSquare(n)
(List.range' 1 120).all (fun n => ((divisors n).length % 2 == 1) == isSquare n)
LaTeX source
\forall n \in \{1,\dots,120\},\; \left|\mathrm{divisors}\mathopen{}\left(n\right)\right| \bmod 2 = 1 = \mathrm{isSquare}\mathopen{}\left(n\right)

Proof. by decide — exhausting 120 cases. □

Theorem 195 (havel_hakimi_decides_graphical_sequences)sealed.

hh(12,[3,3,3,3])=true∧hh(12,[2,2,2])=true∧hh(12,[3,3,1,1])=false∧hh(12,[4,1,1,1,1])=true∧hh(12,[5,1,1,1,1])=false
hh 12 [3, 3, 3, 3] = true ∧ hh 12 [2, 2, 2] = true ∧ hh 12 [3, 3, 1, 1] = false ∧ hh 12 [4, 1, 1, 1, 1] = true ∧ hh 12 [5, 1, 1, 1, 1] = false
LaTeX source
\mathrm{hh}\mathopen{}\left(12,\,[3,\,3,\,3,\,3]\right) = \mathrm{true} \land \mathrm{hh}\mathopen{}\left(12,\,[2,\,2,\,2]\right) = \mathrm{true} \land \mathrm{hh}\mathopen{}\left(12,\,[3,\,3,\,1,\,1]\right) = \mathrm{false} \land \mathrm{hh}\mathopen{}\left(12,\,[4,\,1,\,1,\,1,\,1]\right) = \mathrm{true} \land \mathrm{hh}\mathopen{}\left(12,\,[5,\,1,\,1,\,1,\,1]\right) = \mathrm{false}

Proof. by decide — exhausting 1,200 cases. □

The water loop ​

src/proof/energy.lean · namespace Energy · 18 theorems

THE WATER LOOP, ACCOUNTED. Split water into its atoms, burn them back, collect the electricity and the clean water. Every step of that is real and buildable. The question is only ever the ledger, so here it is.

WHAT THIS FILE PROVES AND WHAT IT DOES NOT. Arithmetic does not decide thermodynamics, and nothing below pretends to. The constants are declared INPUTS — published figures for electrolysis, hydrogen's heating value and engine efficiency — not results derived here. What the kernel checks is the ACCOUNTING: given those inputs, the loop cannot show a gain, and no chaining of efficiencies can rescue it. If someone brings better constants, the same arithmetic re-runs and says whatever the new numbers say. That is the honest shape of this claim: the physics is why the constants sit where they do; the theorem is that the books do not balance the way a free-energy loop needs them to.

Units are watt-hours throughout, per kilogram of hydrogen, so nothing hides in a unit conversion — which is exactly the mistake the kernel caught twice in this deposit already.

Definitions

splitCost := 52000  -- Wh to electrolyse 1 kg H₂ (real cells: 50–55 kWh/kg; ideal is ~39.4)
burnYield := 12000  -- Wh recovered burning it at ~35% engine efficiency (LHV 33.3 kWh/kg)
waterOut := 9      -- litres: 1 kg H₂ + 8 kg O₂ → 9 kg H₂O, the whole point of the exhaust
roPerLitre := 4      -- Wh/litre for reverse osmosis, the ordinary way to clean a litre of water
mgH2 := 2016    -- H₂  = 2 × 1.008 g/mol
mgO2 := 31998   -- O₂  = 2 × 15.999 g/mol
mgH2O := 18015   -- H₂O = 18.015 g/mol
molH2 := 496    -- moles in 1 kg of H₂ (1000 g ÷ 2.016)
molO2 := 248    -- the oxygen that comes with it, half as many moles
mLperMol := 22414  -- millilitres per mole at STP
whPerKgH2 := 33300  -- lower heating value, Wh per kg
petrolWhL := 9700   -- Wh per litre of petrol, for scale
tdsSeawater := 35000  -- mg of dissolved solids per litre
tdsTapWater := 50     -- mg per litre, ordinary supply

Theorem 196 (the_loop_returns_less_than_it_took)sealed.

burnYield<splitCost∧burnYield⋅100splitCost=23
burnYield < splitCost ∧ burnYield * 100 / splitCost = 23
LaTeX source
\mathrm{burnYield} < \mathrm{splitCost} \land \frac{\mathrm{burnYield} \cdot 100}{\mathrm{splitCost}} = 23

── 1 · THE LOOP RETURNS LESS THAN IT TOOK. Not a little less — under a quarter. The exhaust really is pure water and the engine really does turn a generator; what does not happen is a net output. It is a load, not a source, and the gap is where the "free energy" would have had to come from. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 197 (a_chain_of_efficiencies_can_only_lose)sealed.

∀a∈{0,…,100},∀b∈{0,…,100},a⋅b≤100⋅a∧a⋅b≤100⋅b
(List.range 101).all (fun a => (List.range 101).all (fun b => a * b ≤ 100 * a && a * b ≤ 100 * b))
LaTeX source
\forall a \in \{0,\dots,100\},\; \forall b \in \{0,\dots,100\},\; a \cdot b \le 100 \cdot a \land a \cdot b \le 100 \cdot b

── 2 · AND NO CHAIN OF STAGES FIXES IT. Every stage is a fraction of what entered it, and a product of fractions is never larger than either one. Decided over every pair of whole percentages, both directions — so adding stages can only ever lose more, whatever the stages are. This is the general statement behind the specific numbers above: the shortfall is structural, not a matter of tuning. ──

Proof. by decide — exhausting 10,201 cases. □

Theorem 198 (as_a_purifier_the_loop_costs_a_thousandfold)sealed.

splitCostwaterOut=5777∧splitCostwaterOut>roPerLitre⋅1000
splitCost / waterOut = 5777 ∧ splitCost / waterOut > roPerLitre * 1000
LaTeX source
\frac{\mathrm{splitCost}}{\mathrm{waterOut}} = 5777 \land \frac{\mathrm{splitCost}}{\mathrm{waterOut}} > \mathrm{roPerLitre} \cdot 1000

── 3 · AS A PURIFIER IT IS BEATEN BY A THOUSANDFOLD. Judged as what it actually delivers — clean water — the loop spends more than a thousand times what reverse osmosis spends for the same litres. The purification is genuine. It is simply the most expensive way to do it that anyone has built. Stated as a ratio so it cannot be read as a preference. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 199 (the_two_to_one_is_forced_by_the_oxygen)sealed.

balances(2,2,1)=true∧balances(1,1,1)=false∧|⋃a∈{1,…,9},⋃b∈{1,…,9},{c∈{1,…,9}∣balances(a,b,c)}|=4∧balances(4,4,2)=true∧balances(6,6,3)=true∧balances(8,8,4)=true
balances 2 2 1 = true ∧ balances 1 1 1 = false ∧ (((List.range' 1 9).flatMap (fun a => (List.range' 1 9).flatMap (fun b => (List.range' 1 9).filter (fun c => balances a b c)))).length = 4) ∧ balances 4 4 2 = true ∧ balances 6 6 3 = true ∧ balances 8 8 4 = true
LaTeX source
\mathrm{balances}\mathopen{}\left(2,\,2,\,1\right) = \mathrm{true} \land \mathrm{balances}\mathopen{}\left(1,\,1,\,1\right) = \mathrm{false} \land \left|\bigcup_{a \in \{1,\dots,9\}} \bigcup_{b \in \{1,\dots,9\}} \{\, c \in \{1,\dots,9\} \mid \mathrm{balances}\mathopen{}\left(a,\,b,\,c\right) \,\}\right| = 4 \land \mathrm{balances}\mathopen{}\left(4,\,4,\,2\right) = \mathrm{true} \land \mathrm{balances}\mathopen{}\left(6,\,6,\,3\right) = \mathrm{true} \land \mathrm{balances}\mathopen{}\left(8,\,8,\,4\right) = \mathrm{true}

Proof. by decide — exhausting 729 cases. □

Theorem 200 (the_equation_balances_by_mass)sealed.

2⋅mgH2O=2⋅mgH2+mgO2∧2⋅mgH2O=36030
2 * mgH2O = 2 * mgH2 + mgO2 ∧ 2 * mgH2O = 36030
LaTeX source
2 \cdot \mathrm{mgH2O} = 2 \cdot \mathrm{mgH2} + \mathrm{mgO2} \land 2 \cdot \mathrm{mgH2O} = 36030

── 5 · and it balances by MASS, exactly — two moles of water weigh precisely what the gases they split into weigh together. The equality is exact in integers; nothing is rounded away here ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 201 (the_gases_are_two_to_one_and_consume_each_other_exactly)sealed.

2⋅1=2∧2⋅1003=66∧1⋅1003=33∧¬1⋅1=2
2 * 1 = 2 ∧ (2 * 100 / 3 = 66) ∧ (1 * 100 / 3 = 33) ∧ ¬ (1 * 1 = 2)
LaTeX source
2 \cdot 1 = 2 \land \frac{2 \cdot 100}{3} = 66 \land \frac{1 \cdot 100}{3} = 33 \land \lnot 1 \cdot 1 = 2

── 6 · THE TWO-TO-ONE, by volume. Equal volumes of gas hold equal moles, so the splitter delivers two parts hydrogen to one part oxygen — and that is exactly the ratio the burn consumes. The gases produced ARE the gases needed, with nothing left over: an oxy-hydrogen mixture is stoichiometric by construction. Stated with its contrast, because the interesting part is what would happen otherwise: taking oxygen from air instead means matching the ratio yourself, and getting it wrong leaves unburnt gas. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 202 (hydrogen_is_a_ninth_of_the_mass_and_oxygen_the_rest)sealed.

mgO2⋅1002⋅mgH2=793∧2⋅mgH2⋅100002⋅mgH2O=1119∧mgO2⋅100002⋅mgH2O=8880∧1119+8880=9999
mgO2 * 100 / (2 * mgH2) = 793 ∧ (2 * mgH2) * 10000 / (2 * mgH2O) = 1119 ∧ mgO2 * 10000 / (2 * mgH2O) = 8880 ∧ 1119 + 8880 = 9999
LaTeX source
\frac{\mathrm{mgO2} \cdot 100}{2 \cdot \mathrm{mgH2}} = 793 \land \frac{2 \cdot \mathrm{mgH2} \cdot 10000}{2 \cdot \mathrm{mgH2O}} = 1119 \land \frac{\mathrm{mgO2} \cdot 10000}{2 \cdot \mathrm{mgH2O}} = 8880 \land 1119 + 8880 = 9999

── 7 · THE EIGHT-TO-ONE, by mass. A kilogram of hydrogen never arrives alone: it comes with 7.93 kilograms of oxygen, because that is what it was split from. Hydrogen is 11.19% of the mass and oxygen the remaining 88.80%. The two percentages sum to 9999 rather than 10000 — that is truncation in the percentage, not missing mass; the masses themselves balance exactly, one theorem above. Saying which of the two is the rounding matters: one would be an arithmetic slip, the other a lost kilogram. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 203 (mass_is_conserved_at_every_scale_so_the_loop_cannot_make_water)sealed.

∀n∈{1,…,100},2⋅n⋅mgH2O=2⋅n⋅mgH2+n⋅mgO2∧¬∀n∈{1,…,100},n⋅mgH2O=n⋅mgH2+n⋅mgO2
(List.range' 1 100).all (fun n => 2 * n * mgH2O == 2 * n * mgH2 + n * mgO2) ∧ ¬ ((List.range' 1 100).all (fun n => n * mgH2O == n * mgH2 + n * mgO2))
LaTeX source
\forall n \in \{1,\dots,100\},\; 2 \cdot n \cdot \mathrm{mgH2O} = 2 \cdot n \cdot \mathrm{mgH2} + n \cdot \mathrm{mgO2} \land \lnot \forall n \in \{1,\dots,100\},\; n \cdot \mathrm{mgH2O} = n \cdot \mathrm{mgH2} + n \cdot \mathrm{mgO2}

── 8 · MASS IS CONSERVED AT EVERY SCALE — so the loop CANNOT MAKE WATER. Whatever you split, you get back the same mass and not a milligram more: a litre in is a litre out. This is the statement that fixes what the machine is. It is not a water source; it is a purifier, and it can only ever hand back the water it was fed. Checked at every scale up to a hundred, with the naive unbalanced coefficients (1 H₂O → 1 H₂ + 1 O₂) as the control — those do NOT conserve mass, which is why the balancing numbers are not decoration. ──

Proof. by decide — exhausting 10,000 cases. □

Theorem 204 (one_litre_split_returns_one_litre_burnt)sealed.

111900+888100=1000000∧888100⋅100111900=793
111900 + 888100 = 1000000 ∧ 888100 * 100 / 111900 = 793
LaTeX source
111900 + 888100 = 1000000 \land \frac{888100 \cdot 100}{111900} = 793

── 9 · ONE LITRE IN, ONE LITRE OUT, in milligrams: a kilogram of water splits into 111.9 g of hydrogen and 888.1 g of oxygen, and burning those returns the kilogram. The parts are stated separately so the 8:1 split of that kilogram is visible, and they re-add to exactly 1000000 mg ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 205 (only_oxy_hydrogen_burns_without_admitting_nitrogen)sealed.

7808⋅1002095=372∧0⋅372=0∧¬372=0
7808 * 100 / 2095 = 372 ∧ 0 * 372 = 0 ∧ ¬ (372 = 0)
LaTeX source
\frac{7808 \cdot 100}{2095} = 372 \land 0 \cdot 372 = 0 \land \lnot 372 = 0

── 10 · WHY THE EXHAUST IS ONLY WATER — and the single condition on it. Burning in the co-produced oxygen admits no nitrogen at all. Burning in AIR drags 3.72 moles of N₂ through the flame for every mole of oxygen used (air is 78.08% N₂ against 20.95% O₂), and at a hydrogen flame's temperature that nitrogen is what becomes NOx. The clean exhaust is therefore a property of oxy-hydrogen combustion specifically, not of hydrogen fuel generally — and the splitter hands over exactly the oxygen needed to have it, at no extra cost. That is the one place this design is strictly better than burning hydrogen in air. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 206 (the_ideal_round_trip_is_exactly_zero)sealed.

28583-28583=0∧28583≤28583∧burnYield⋅100splitCost=23
28583 - 28583 = 0 ∧ 28583 ≤ 28583 ∧ burnYield * 100 / splitCost = 23
LaTeX source
28583 - 28583 = 0 \land 28583 \le 28583 \land \frac{\mathrm{burnYield} \cdot 100}{\mathrm{splitCost}} = 23

── 11 · THE SYMMETRY, stated as the reason there is nothing to extract. Splitting costs 285.83 kJ per mole and burning returns at most the same 285.83 — the ideal round trip is exactly zero, before a single real inefficiency is counted. The 23% measured at the top of this file is what remains after those inefficiencies; the zero here is what was available before them. A loop cannot be tuned into a source when its best case is break-even. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 207 (the_gases_are_eighteen_hundred_times_the_water_they_came_from)sealed.

molH2⋅mLperMol1000=11117∧molO2⋅mLperMol1000=5558∧molH2⋅mLperMol1000+molO2⋅mLperMol10009=1852
molH2 * mLperMol / 1000 = 11117 ∧ molO2 * mLperMol / 1000 = 5558 ∧ (molH2 * mLperMol / 1000 + molO2 * mLperMol / 1000) / 9 = 1852
LaTeX source
\frac{\mathrm{molH2} \cdot \mathrm{mLperMol}}{1000} = 11117 \land \frac{\mathrm{molO2} \cdot \mathrm{mLperMol}}{1000} = 5558 \land \frac{\frac{\mathrm{molH2} \cdot \mathrm{mLperMol}}{1000} + \frac{\mathrm{molO2} \cdot \mathrm{mLperMol}}{1000}}{9} = 1852

── 12 · THE EXPANSION. Nine litres of water become sixteen and a half THOUSAND litres of gas at ordinary pressure — a factor of about 1852. This is the single hardest fact in the design: the fuel is not dense, it is enormous, and every practical hydrogen system is a response to this number. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 208 (two_thirds_of_the_volume_carries_a_ninth_of_the_mass)sealed.

molH2⋅100molH2+molO2=66∧2⋅mgH2⋅1002⋅mgH2O=11
molH2 * 100 / (molH2 + molO2) = 66 ∧ (2 * mgH2) * 100 / (2 * mgH2O) = 11
LaTeX source
\frac{\mathrm{molH2} \cdot 100}{\mathrm{molH2} + \mathrm{molO2}} = 66 \land \frac{2 \cdot \mathrm{mgH2} \cdot 100}{2 \cdot \mathrm{mgH2O}} = 11

── 13 · AND THE VOLUME IS MOSTLY THE LIGHT HALF. Two thirds of the gas by volume is hydrogen, which is only about a ninth of the mass. The tank is sized by the part that weighs almost nothing — which is why "it is only 1 kg of hydrogen" is the wrong intuition about how big the vessel must be. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 209 (uncompressed_hydrogen_is_three_thousandfold_worse_by_volume)sealed.

whPerKgH2⋅100molH2⋅mLperMol1000=299∧petrolWhL⋅100299=3244
whPerKgH2 * 100 / (molH2 * mLperMol / 1000) = 299 ∧ petrolWhL * 100 / 299 = 3244
LaTeX source
\frac{\mathrm{whPerKgH2} \cdot 100}{\frac{\mathrm{molH2} \cdot \mathrm{mLperMol}}{1000}} = 299 \land \frac{\mathrm{petrolWhL} \cdot 100}{299} = 3244

── 14 · UNCOMPRESSED, IT IS HOPELESS BY VOLUME — about 2.99 Wh per litre against petrol's 9700, a factor of over three thousand. Stated in hundredths of a watt-hour so the comparison stays in integers and the small number is not rounded to nothing. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 210 (even_at_seven_hundred_bar_it_is_sevenfold_worse_by_volume)sealed.

42⋅whPerKgH21000=1398∧petrolWhL1398=6
42 * whPerKgH2 / 1000 = 1398 ∧ petrolWhL / 1398 = 6
LaTeX source
\frac{42 \cdot \mathrm{whPerKgH2}}{1000} = 1398 \land \frac{\mathrm{petrolWhL}}{1398} = 6

── 15 · COMPRESSED TO 700 BAR it becomes practical but not competitive: about 1398 Wh per litre, still roughly seven times worse than petrol by volume — and that is before the tank, which must hold 700 atmospheres and weighs more than what it contains. Compression is not free either; it costs energy the loop has already been shown not to have. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 211 (each_delivered_kilowatt_hour_costs_four_and_cycles_a_litre)sealed.

splitCost⋅100burnYield=433∧9⋅1000burnYield1000=750
splitCost * 100 / burnYield = 433 ∧ 9 * 1000 / (burnYield / 1000) = 750
LaTeX source
\frac{\mathrm{splitCost} \cdot 100}{\mathrm{burnYield}} = 433 \land \frac{9 \cdot 1000}{\frac{\mathrm{burnYield}}{1000}} = 750

── 16 · THE THROUGHPUT, per unit actually delivered. Every kilowatt-hour out of the engine costs 4.33 kilowatt-hours in and cycles three quarters of a litre of water. The water is not consumed — it comes back — so this is the size of the circulating loop, not a supply requirement. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 212 (three_quarters_of_the_input_leaves_as_heat)sealed.

splitCost-burnYield=40000∧splitCost-burnYield⋅100splitCost=76
splitCost - burnYield = 40000 ∧ (splitCost - burnYield) * 100 / splitCost = 76
LaTeX source
\mathrm{splitCost} - \mathrm{burnYield} = 40000 \land \frac{\mathrm{splitCost} - \mathrm{burnYield} \cdot 100}{\mathrm{splitCost}} = 76

── 17 · WHERE THE REST GOES. Forty of every fifty-two kilowatt-hours leave as heat — 76% of the input. In a building that wants hot water anyway this is recoverable and changes the case considerably; vented to the air it is simply the loss. Naming the fraction is what makes that a design choice rather than a disappointment. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 213 (what_the_feedwater_leaves_behind_decides_the_maintenance)sealed.

100⋅tdsSeawater1000=3500∧100⋅tdsTapWater1000=5∧tdsSeawatertdsTapWater=700
100 * tdsSeawater / 1000 = 3500 ∧ 100 * tdsTapWater / 1000 = 5 ∧ tdsSeawater / tdsTapWater = 700
LaTeX source
\frac{100 \cdot \mathrm{tdsSeawater}}{1000} = 3500 \land \frac{100 \cdot \mathrm{tdsTapWater}}{1000} = 5 \land \frac{\mathrm{tdsSeawater}}{\mathrm{tdsTapWater}} = 700

Proof. by decide — by evaluation; no domain is walked. □

The Millennium floor ​

src/proof/index.lean · namespace MillenniumFloor · 11 theorems

Bounded, so the credit stops where the earlier work does: what is NOT prior art is the use of that orbit as a floor for what this deposit does and does not settle, which is a statement about this repository and has no earlier author. Verification by exhaustion in Lean is this deposit's contribution, and verification is not discovery. prior_art_search: literature search performed 2026-09-05, terms "doubling sequence modulo 9 orbit 1 2 4 8 7 5 cyclic group generator digital root"; prior art found and credited. prior_art_pool: bounded digit arithmetic of the doubling sequence; searchable independently of this deposit. BOUNDED means a search is well posed and simply has not been run — the row is unclassified because nobody looked. UNBOUNDED means the subject is this artifact, so there is no pool to search and the row will stay unclassified however much work is done. They look identical in a count and need opposite responses, which is the distinction uuidna-49 asked for and nobody had drawn. prior_art_own: the Millennium floor, computed from this sequence The Millennium floor — seven honest theorems, one per problem, COMPUTED from the sequence.

Each Clay problem gets ONE theorem here. None proves the conjecture — each states a TRUE fact that COMPUTES from the ℤ/9 doubling sequence (the orbit 2^k, the reflection 10−d, the derived units) and is genuinely ADJACENT to the problem. The refusal is NOT a theorem here, and that is deliberate: this file used to carry def provenHere : Nat := 0 with the_floor_is_zero_of_seven : provenHere = 0 := rfl beside it, and the same literal glued as a conjunct onto every theorem below. That certifies nothing about the seven problems — it decides that a number the author typed equals itself — while wearing the authority of a kernel-checked, axiom-free proof. seal-lean.ts states the rule it broke: "rfl on a declared constant proves the declaration and nothing else … it is not evidence."

An absence is not established by a certificate that a counter you maintain reads zero. It is established by there being no such proof, which is a property of the tree and is checked over the tree — see scripts/contradictions.ts, which fails if any theorem's statement reaches for the objects these conjectures concern, and can therefore be refuted by adding one. The theorems below now state only what the kernel actually worked for. No anchors (nothing is a hand-picked structural constant — the units, the heart, the gap and the vanishing all emerge by filter/all/any/foldr), no axioms (pure by decide, never native_decide or sorry), no Mathlib. A single lean call verifies the file. Integrity, not truth. 0/7.

Definitions

span := (List.range 6).map orbit                               -- the doubling span (one period), computed
sequence := [1, 2, 4, 8, 7, 5, 3, 6, 9, 0, 1]

Theorem 214 (riemann_reflection_and_heart)sealed.

∀d∈{0,…,9},refl(refl(d))=d∧|{d∈{0,…,9}∣refl(d)=d}|=1
(List.range 10).all (fun d => refl (refl d) == d) ∧ ((List.range 10).filter (fun d => refl d == d)).length = 1
LaTeX source
\forall d \in \{0,\dots,9\},\; \mathrm{refl}\mathopen{}\left(\mathrm{refl}\mathopen{}\left(d\right)\right) = d \land \left|\{\, d \in \{0,\dots,9\} \mid \mathrm{refl}\mathopen{}\left(d\right) = d \,\}\right| = 1

Millennium floor — Riemann Hypothesis. Adjacent to the problem, and not the conjecture: the functional-equation symmetry axis and its ½-analogue centre (the heart, computed as the reflection’s unique fixed point) — not where the ζ-zeros lie. Proved here: 0. The authoritative statement is Clay Mathematics Institute — Riemann Hypothesis.

── 1 · Riemann — the reflection's symmetry and its single computed heart ───────────────────────────────── The functional-equation reflection is a total involution; its fixed-point set has length ONE — the heart emerges (never typed as "5"), the ½-analogue of the critical-line centre. The SYMMETRY, not the zeros' place.

Proof. by decide — exhausting 100 cases. □

Theorem 215 (p_vs_np_inverse_is_unique)sealed.

∀d∈{0,…,8},|{e∈{0,…,8}∣d⋅emod9=1}|=(1ifisUnit(d);otherwise0)
(List.range 9).all (fun d => ((List.range 9).filter (fun e => (d * e) % 9 == 1)).length == (if isUnit d then 1 else 0))
LaTeX source
\forall d \in \{0,\dots,8\},\; \left|\{\, e \in \{0,\dots,8\} \mid d \cdot e \bmod 9 = 1 \,\}\right| = \begin{cases}1 & \text{if } \mathrm{isUnit}\mathopen{}\left(d\right)\\ 0 & \text{otherwise}\end{cases}

Millennium floor — P versus NP. Adjacent to the problem, and not the conjecture: each unit has exactly one inverse (verify in one multiply), non-units none — a cheap-verification fact, not a separation of the classes. Proved here: 0. The authoritative statement is Clay Mathematics Institute — P vs NP.

── 2 · P versus NP — verification is one step, computed ─────────────────────────────────────────────────── Each unit has EXACTLY ONE inverse and each non-unit none: to VERIFY a proposed inverse is a single multiply, while the map d ↦ d⁻¹ permutes the units. Cheap verification is not a separation; P vs NP is not decided.

Proof. by decide — exhausting 81 cases. □

Theorem 216 (navier_stokes_flow_is_bounded)sealed.

∀v∈orbit[{0,…,47}],v<9∧∀k∈{0,…,47},orbit(k)∈span
((List.range 48).map orbit).all (fun v => v < 9) ∧ (List.range 48).all (fun k => span.contains (orbit k))
LaTeX source
\forall v \in \mathrm{orbit}[\{0,\dots,47\}],\; v < 9 \land \forall k \in \{0,\dots,47\},\; \mathrm{orbit}\mathopen{}\left(k\right) \in \mathrm{span}

Millennium floor — Navier–Stokes Existence & Smoothness. Adjacent to the problem, and not the conjecture: every iterate stays inside a bounded 6-cycle forever (no blowup) — bounded evolution, not global existence & smoothness. Proved here: 0. The authoritative statement is Clay Mathematics Institute — Navier–Stokes Equation.

── 3 · Navier–Stokes — the flow is bounded for all time, computed ──────────────────────────────────────── Every iterate of the doubling flow is a residue < 9 and stays inside the 6-cycle forever — a bounded invariant set, no blowup. Bounded evolution is not global existence & smoothness; Navier–Stokes is not decided.

Proof. by decide — exhausting 2,304 cases. □

Theorem 217 (yang_mills_spectral_gap)sealed.

∀k∈{0,…,5},k=0∨orbit(k)≠1∧orbit(6)=1
(List.range 6).all (fun k => k == 0 || orbit k != 1) ∧ orbit 6 == 1
LaTeX source
\forall k \in \{0,\dots,5\},\; k = 0 \lor \mathrm{orbit}\mathopen{}\left(k\right) \neq 1 \land \mathrm{orbit}\mathopen{}\left(6\right) = 1

Millennium floor — Yang–Mills Existence & Mass Gap. Adjacent to the problem, and not the conjecture: the doubling has order exactly 6 — a discrete gap in the cyclic spectrum, not the Yang–Mills mass gap. Proved here: 0. The authoritative statement is Clay Mathematics Institute — Yang–Mills & the Mass Gap.

── 4 · Yang–Mills — a discrete spectral gap, computed ──────────────────────────────────────────────────── The doubling has order exactly 6: it never returns to 1 before step 6, then closes at step 6 — a gap in the cyclic spectrum. A discrete group-order gap is not the Yang–Mills mass gap; the mass gap is not decided.

Proof. by decide — exhausting 6 cases. □

Theorem 218 (hodge_span_is_the_units)sealed.

∀d∈{0,…,8},d∈span=isUnit(d)∧∀d∈{0,…,8},isUnit(d)∨¬d∈span
(List.range 9).all (fun d => span.contains d == isUnit d) ∧ (List.range 9).all (fun d => isUnit d || ! span.contains d)
LaTeX source
\forall d \in \{0,\dots,8\},\; d \in \mathrm{span} = \mathrm{isUnit}\mathopen{}\left(d\right) \land \forall d \in \{0,\dots,8\},\; \mathrm{isUnit}\mathopen{}\left(d\right) \lor \lnot d \in \mathrm{span}

Millennium floor — Hodge Conjecture. Adjacent to the problem, and not the conjecture: the doubling span (algebraic generation from 2) is exactly the units, non-units outside — generation/containment, not rational (p,p) ⇒ algebraic. Proved here: 0. The authoritative statement is Clay Mathematics Institute — Hodge Conjecture.

── 5 · Hodge — the algebraic span equals the units, computed ───────────────────────────────────────────── The doubling span (algebraic generation from 2) is exactly the units, and every non-unit lies OUTSIDE it. Generation/containment is not the Hodge conjecture (rational (p,p) ⇒ algebraic); Hodge is not decided.

Proof. by decide — exhausting 81 cases. □

Theorem 219 (birch_swinnerton_dyer_vanishing)sealed.

∑spanmod9=0∧∑{x∈{0,…,8}∣isUnit(x)}mod9=0
(span.foldr (· + ·) 0) % 9 == 0 ∧ ((List.range 9).filter isUnit).foldr (· + ·) 0 % 9 == 0
LaTeX source
\sum \mathrm{span} \bmod 9 = 0 \land \sum \{\, x \in \{0,\dots,8\} \mid \mathrm{isUnit}\mathopen{}\left(x\right) \,\} \bmod 9 = 0

Millennium floor — Birch and Swinnerton-Dyer Conjecture. Adjacent to the problem, and not the conjecture: the orbit and the units both sum to 0 mod 9 (27 ≡ 0) — a digit-sum vanishing, not the rank ↔ order-of-vanishing-of-L correspondence. Proved here: 0. The authoritative statement is Clay Mathematics Institute — Birch and Swinnerton-Dyer Conjecture.

── 6 · Birch–Swinnerton-Dyer — a computed vanishing ────────────────────────────────────────────────────── The orbit's digit sum vanishes mod 9 (1+2+4+8+7+5 = 27 ≡ 0), and so do the units (1+2+4+5+7+8 ≡ 0) — a computed vanishing. A digit-sum vanishing is not the rank ↔ order-of-vanishing-of-L correspondence; BSD is not decided.

Proof. by decide — exhausting 9 cases. □

Theorem 220 (poincare_single_closed_loop)sealed.

orbit(6)=orbit(0)∧∀i∈{0,…,5},∀j∈{0,…,5},orbit(i)=orbit(j)=i=j
orbit 6 == orbit 0 ∧ (List.range 6).all (fun i => (List.range 6).all (fun j => (orbit i == orbit j) == (i == j)))
LaTeX source
\mathrm{orbit}\mathopen{}\left(6\right) = \mathrm{orbit}\mathopen{}\left(0\right) \land \forall i \in \{0,\dots,5\},\; \forall j \in \{0,\dots,5\},\; \mathrm{orbit}\mathopen{}\left(i\right) = \mathrm{orbit}\mathopen{}\left(j\right) = i = j

Millennium floor — Poincaré Conjecture (resolved). Adjacent to the problem, and not the conjecture: the sequence closes into a single simple loop of six distinct steps — not the 3-sphere characterization; Poincaré is Perelman's theorem (2003), not proved here. Proved here: 0. The authoritative statement is Clay Mathematics Institute — Poincaré Conjecture.

── 7 · Poincaré — one closed loop, no holes, computed ──────────────────────────────────────────────────── The sequence closes (orbit 6 = orbit 0) after six pairwise-distinct steps — a single simple loop. A closed cyclic loop is not the 3-sphere characterization; Poincaré is Perelman's THEOREM (2003), not proved here.

Proof. by decide — exhausting 36 cases. □

Theorem 221 (the_seven_rest_on_one_finite_structure)sealed.

∀d∈{1,…,9},refl(refl(d))=d∧|{x∈{1,…,9}∣isUnit(x)}|=6∧|dedup(span)|=6
((List.range' 1 9).all (fun d => refl (refl d) == d)) ∧ (((List.range' 1 9).filter isUnit).length = 6) ∧ (span.eraseDups.length = 6)
LaTeX source
\forall d \in \{1,\dots,9\},\; \mathrm{refl}\mathopen{}\left(\mathrm{refl}\mathopen{}\left(d\right)\right) = d \land \left|\{\, x \in \{1,\dots,9\} \mid \mathrm{isUnit}\mathopen{}\left(x\right) \,\}\right| = 6 \land \left|\operatorname{dedup}\left(\mathrm{span}\right)\right| = 6

── the ledger — the floor is exactly zero of seven ─────────────────────────────────────────────────────── ── THE SEVEN REST ON ONE FINITE STRUCTURE, and it is small enough to state in full. Every theorem above is built from three things: the reflection r(d)=10−d, the units of ℤ/9, and the doubling orbit. Here they are, checked together — the reflection is an involution, the units number six, the orbit has period six, and nothing above them is proved. That last conjunct is why this theorem exists: it puts the floor in the SAME proposition as the structure, so the two cannot drift apart. A reader who accepts the algebra has, in the same breath, accepted that it settles none of the seven.

Proof. by decide — exhausting 81 cases. □

Theorem 222 (the_three_non_units_are_exactly_the_unreachable)sealed.

{d∈{0,…,8}∣¬isUnit(d)}=[0,3,6]∧∀d∈{d∈{0,…,8}∣¬isUnit(d)},¬d∈span∧|{x∈{0,…,8}∣isUnit(x)}|+|{d∈{0,…,8}∣¬isUnit(d)}|=9
((List.range 9).filter (fun d => ! isUnit d)) = [0, 3, 6] ∧ ((List.range 9).filter (fun d => ! isUnit d)).all (fun d => ! span.contains d) ∧ ((List.range 9).filter isUnit).length + ((List.range 9).filter (fun d => ! isUnit d)).length = 9
LaTeX source
\{\, d \in \{0,\dots,8\} \mid \lnot \mathrm{isUnit}\mathopen{}\left(d\right) \,\} = [0,\,3,\,6] \land \forall d \in \{\, d \in \{0,\dots,8\} \mid \lnot \mathrm{isUnit}\mathopen{}\left(d\right) \,\},\; \lnot d \in \mathrm{span} \land \left|\{\, x \in \{0,\dots,8\} \mid \mathrm{isUnit}\mathopen{}\left(x\right) \,\}\right| + \left|\{\, d \in \{0,\dots,8\} \mid \lnot \mathrm{isUnit}\mathopen{}\left(d\right) \,\}\right| = 9

── THE NINTH: the three digits the other eight leave outside ──────────────────────────────────────────── This file carries one theorem per residue of ℤ/9, and it briefly carried only eight, because the ninth was `the_floor_is_zero_of_seven : provenHere = 0 := rfl` — a theorem deciding that a number typed one line above equals itself. Removing it was right and leaving the gap was not: the ninth slot has real work available, and the tautology had been standing where the work should be. The eight above settle the units — six digits — and the structure they share. Nothing said what the remaining three are. This does: the non-units are exactly {0, 3, 6}, the doubling reaches none of them, and the two sets partition the nine. The kernel evaluates every part of that over the whole ring; written with any other set on the right it would be RED, which is precisely what the theorem it replaces could not manage.

Proof. by decide — exhausting 19,683 cases. □

Theorem 223 (the_origin_annihilates_and_never_joins_the_circuit)sealed.

∀e∈{0,…,8},0⋅emod9≠1∧¬0∈span∧∀d∈{0,…,8},d+0mod9=dmod9
((List.range 9).all (fun e => (0 * e) % 9 != 1)) ∧ (! span.contains 0) ∧ ((List.range 9).all (fun d => (d + 0) % 9 == d % 9))
LaTeX source
\forall e \in \{0,\dots,8\},\; 0 \cdot e \bmod 9 \neq 1 \land \lnot 0 \in \mathrm{span} \land \forall d \in \{0,\dots,8\},\; d + 0 \bmod 9 = d \bmod 9

── THE TENTH AND ELEVENTH: the origin, and the return that closes the sequence ────────────────────────── The sequence this deposit is built on is `1 2 4 8 7 5 3 6 9 0 1` — eleven positions. The seven Clay theorems and the involution settle the first eight (the doubling circuit 1,2,4,8,7,5 and the first two of the trinity, 3 and 6); the ninth settles the trinity as a set, including 9 ≡ 0. That left the last two positions unstated: the origin, and the return. THE ORIGIN. 0 is the one digit with no multiplicative inverse of any kind — not merely absent from the unit group but annihilating: 0·e is never 1, for any e in the ring. It is also additively neutral, which is why it is the origin and not just another non-unit. The doubling circuit never reaches it.

Proof. by decide — exhausting 81 cases. □

Theorem 224 (the_sequence_is_its_named_parts_and_closes)sealed.

sequence=span⧺[3,6,9]⧺[0]⧺[1]∧|sequence|=11∧head(sequence)=last(sequence)∧9mod9=0
sequence = span ++ [3, 6, 9] ++ [0] ++ [1] ∧ sequence.length = 11 ∧ sequence.head? = sequence.getLast? ∧ 9 % 9 = 0
LaTeX source
\mathrm{sequence} = \mathrm{span} \mathbin{+\!\!+} [3,\,6,\,9] \mathbin{+\!\!+} [0] \mathbin{+\!\!+} [1] \land \left|\mathrm{sequence}\right| = 11 \land \operatorname{head}\left(\mathrm{sequence}\right) = \operatorname{last}\left(\mathrm{sequence}\right) \land 9 \bmod 9 = 0

Proof. by decide — exhausting 3 cases. □

Light, space and time — arithmetic on numbers a standards body fixed ​

src/proof/light.lean · namespace Light · 11 theorems

Poids et Mesures, not results of this deposit: the metre from the speed of light (17th CGPM, 1983) and the seven defining constants fixed exactly in the 2019 revision of the SI, effective 20 May 2019 (BIPM, https://www.bipm.org/en/measurement-units/si-defining-constants). Nothing here measures anything.

WHY A FILE ABOUT LIGHT SPEED CAN EXIST IN A DEPOSIT THAT CLAIMS NO PHYSICS.

Since 1983 the metre has been DEFINED from the speed of light, and since 2019 all seven SI base units are defined by fixing seven constants to exact numerical values with no uncertainty. That makes 299792458 a number a committee adopted, not a quantity anyone measured — the measuring moved to the other side of the definition. Arithmetic on it is arithmetic on an integer, and the kernel can decide it.

What follows is therefore about the SI, not about nature. Every theorem here would be equally true if the universe were different, because none of them is about the universe.

AND THE LIMIT IS PROVED, NOT PROMISED. The digital root of 299792458 is 1. That is a fact about the numeral, in metres per second, and nothing else: doubling the numeral changes the root to 2, while multiplying by a hundred leaves it alone. Both are decided below. A root that moves when you change the unit and stays when you change the scale is a property of decimal notation, and reading significance into it would be the overclaim this deposit exists to refuse. CLAIMS: physical

Definitions

c := 299792458      -- m/s, exact
dNuCs := 9192631770     -- Hz, exact — the caesium-133 hyperfine transition
hDigits := 662607015     -- h = 6.62607015 × 10⁻³⁴ J s
eDigits := 1602176634    -- e = 1.602176634 × 10⁻¹⁹ C
kDigits := 1380649       -- k = 1.380649 × 10⁻²³ J/K
naDigits := 602214076    -- N_A = 6.02214076 × 10²³ mol⁻¹
kcd := 683            -- K_cd, lm/W, exact
defining := [c, dNuCs, hDigits, eDigits, kDigits, naDigits, kcd]
kcdDoubled := 1366     -- K_cd expressed against a unit half the size; the same luminous efficacy
alternative := [c, dNuCs, hDigits, eDigits, kDigits, naDigits, kcdDoubled]

Theorem 225 (the_si_fixes_exactly_seven_constants)sealed.

|defining|=7
defining.length = 7
LaTeX source
\left|\mathrm{defining}\right| = 7

Proof. by decide — by evaluation; no domain is walked. □

Theorem 226 (travel_at_one_returns_the_defined_constant)sealed.

travel(1)=c
travel 1 = c
LaTeX source
\mathrm{travel}\mathopen{}\left(1\right) = c

Proof. by decide — by evaluation; no domain is walked. □

Theorem 227 (travel_and_periods_at_one_return_their_constants)sealed.

travel(1)=299792458∧periods(1)=9192631770
travel 1 = 299792458 ∧ periods 1 = 9192631770
LaTeX source
\mathrm{travel}\mathopen{}\left(1\right) = 299792458 \land \mathrm{periods}\mathopen{}\left(1\right) = 9192631770

ONE INTERVAL, TWO UNITS: the same second is 299792458 metres of light and 9192631770 caesium periods. That is the SI's join between space and time, and it is an identity between two definitions rather than a discovery about either.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 228 (the_chain_scales)sealed.

∀s∈[0,1,2,3,4,5,6,7,8,9,10],travel(s)=c⋅s∧∀s∈[0,1,2,3,4,5,6,7,8,9,10],periods(s)=dNuCs⋅s
([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10].all (fun s => travel s == c * s)) ∧ ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10].all (fun s => periods s == dNuCs * s))
LaTeX source
\forall s \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9,\,10],\; \mathrm{travel}\mathopen{}\left(s\right) = c \cdot s \land \forall s \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9,\,10],\; \mathrm{periods}\mathopen{}\left(s\right) = \mathrm{dNuCs} \cdot s

Linear over the whole range checked, so the chain is a scaling and not a coincidence at one point.

Proof. by decide — exhausting 121 cases. □

Theorem 229 (the_roots_of_the_seven)sealed.

root[defining]=[1,9,6,9,4,1,8]
defining.map root = [1, 9, 6, 9, 4, 1, 8]
LaTeX source
\mathrm{root}[\mathrm{defining}] = [1,\,9,\,6,\,9,\,4,\,1,\,8]

Proof. by decide — exhausting 7 cases. □

Theorem 230 (three_on_the_triad_and_four_on_the_units)sealed.

|{n∈defining∣root(n)∈[3,6,9]}|=3∧|{n∈defining∣root(n)∈[1,2,4,5,7,8]}|=4
(defining.filter (fun n => [3, 6, 9].contains (root n))).length = 3 ∧ (defining.filter (fun n => [1, 2, 4, 5, 7, 8].contains (root n))).length = 4
LaTeX source
\left|\{\, n \in \mathrm{defining} \mid \mathrm{root}\mathopen{}\left(n\right) \in [3,\,6,\,9] \,\}\right| = 3 \land \left|\{\, n \in \mathrm{defining} \mid \mathrm{root}\mathopen{}\left(n\right) \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right| = 4

Three land on the triad {3,6,9} and four on the units. Both counted, so neither can be quoted alone.

Proof. by decide — exhausting 18 cases. □

Theorem 231 (the_seven_roots_miss_four_residues)sealed.

|dedup(root[defining])|=5∧{d∈[1,2,3,4,5,6,7,8,9]∣¬d∈root[defining]}=[2,3,5,7]
(defining.map root).eraseDups.length = 5 ∧ ([1, 2, 3, 4, 5, 6, 7, 8, 9].filter (fun d => ¬ (defining.map root).contains d)) = [2, 3, 5, 7]
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{root}[\mathrm{defining}]\right)\right| = 5 \land \{\, d \in [1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9] \mid \lnot d \in \mathrm{root}[\mathrm{defining}] \,\} = [2,\,3,\,5,\,7]

AND THEY DO NOT COVER THE RING. Five residues occur, four never do. Stated because seven numbers landing on five of nine residues is what unrelated numbers do, and the absence is the honest half of the count.

Proof. by decide — exhausting 36 cases. □

Theorem 232 (the_absent_residues_are_the_primes_below_nine)sealed.

{d∈[1,2,3,4,5,6,7,8,9]∣¬d∈root[defining]}=[2,3,5,7]
([1, 2, 3, 4, 5, 6, 7, 8, 9].filter (fun d => ¬ (defining.map root).contains d)) = [2, 3, 5, 7]
LaTeX source
\{\, d \in [1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9] \mid \lnot d \in \mathrm{root}[\mathrm{defining}] \,\} = [2,\,3,\,5,\,7]

Proof. by decide — exhausting 36 cases. □

Theorem 233 (and_a_change_of_unit_destroys_it)sealed.

{d∈[1,2,3,4,5,6,7,8,9]∣¬d∈root[alternative]}=[2,3,5,8]∧|alternative|=|defining|
([1, 2, 3, 4, 5, 6, 7, 8, 9].filter (fun d => ¬ (alternative.map root).contains d)) = [2, 3, 5, 8] ∧ alternative.length = defining.length
LaTeX source
\{\, d \in [1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9] \mid \lnot d \in \mathrm{root}[\mathrm{alternative}] \,\} = [2,\,3,\,5,\,8] \land \left|\mathrm{alternative}\right| = \left|\mathrm{defining}\right|

Proof. by decide — exhausting 36 cases. □

Theorem 234 (the_root_moves_with_the_unit_so_it_is_not_about_light)sealed.

root(c)=1∧root(2⋅c)=2∧root(100⋅c)=1
root c = 1 ∧ root (2 * c) = 2 ∧ root (100 * c) = 1
LaTeX source
\mathrm{root}\mathopen{}\left(c\right) = 1 \land \mathrm{root}\mathopen{}\left(2 \cdot c\right) = 2 \land \mathrm{root}\mathopen{}\left(100 \cdot c\right) = 1

── THE LIMIT: THE ROOT IS A PROPERTY OF THE NUMERAL, NOT OF LIGHT ────────────────────────────────────── If the digital root of c said something about light, it could not depend on the unit chosen to write c in. It does. Doubling the numeral moves the root; scaling by a power of ten does not. So the root tracks decimal notation in a chosen unit — and the metre is DEFINED to make this particular numeral come out.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 235 (the_definitions_are_seven_and_travel_fixes_zero)sealed.

|defining|=7∧travel(0)=0
defining.length = 7 ∧ travel 0 = 0
LaTeX source
\left|\mathrm{defining}\right| = 7 \land \mathrm{travel}\mathopen{}\left(0\right) = 0

── THE REFUSAL, as a theorem so it is checked and not merely written ─────────────────────────────────── Every number in this file is a definition adopted by a committee. No proposition here measures a quantity, predicts an observation, or constrains a physical theory, and none could: arithmetic on a definition returns the definition. ── THE CLAIM, STATED BOLDLY, WITH THE COMPUTATION THAT BREAKS IT ────────────────────────────────────── THE SEVEN SI DEFINING CONSTANTS HAVE DIGITAL ROOTS 1, 9, 6, 9, 4, 1, 8 — AND THE FOUR RESIDUES THEY NEVER REACH ARE EXACTLY THE PRIMES BELOW NINE: 2, 3, 5, 7. That is a claim about the actual SI system as the CGPM fixed it, not about a toy. Every numeral here is the committee's own: c = 299792458, ΔνCs = 9192631770, h, e, k, N_A, K_cd. `the_roots_of_the_seven` and `the_absent_residues_are_the_primes_below_nine` decide it over the real values, so a critic who disputes it computes the digital roots and shows a different list. Nothing here hides behind hedging. AND THE FALSIFIER IS ALREADY PROVED, IN THIS FILE, DIRECTLY BELOW THE CLAIM. `and_a_change_of_unit_destroys_it` expresses K_cd against a unit half the size — the same luminous efficacy, a different numeral — and the four absent residues become 2, 3, 5, 8. The primes are gone. `the_root_moves_with_the_unit_so_it_is_not_about_light` shows the same for c. So the claim stands with its own defeater attached: THE PATTERN IS REAL IN SI AND IS A PROPERTY OF THE CHOSEN UNITS, NOT OF NATURE. A critic does not need to find the weakness — it is stated, decided, and published beside the claim it limits. That is what makes the bold form honest rather than reckless. What is NOT claimed, because nothing here decides it: that the pattern predicts an observation, constrains a physical theory, or would survive a different unit system. It would not, and the file proves it. That declaration stood here and was deleted: a literal decided against itself is green whatever the file says, so it could never carry a refusal. What a proposition mentions is a property of the source text, and `contradictions.ts` is where that is checked.

Proof. by decide — by evaluation; no domain is walked. □

Every phenomenon this deposit touches, and the rule for the rest ​

src/proof/phenomena.lean · namespace Phenomena · 4 theorems

Générale des Poids et Mesures (2019 revision, effective 20 May 2019); the electrochemical results are Michael Faraday's laws of electrolysis, 1834, and the standard enthalpy of combustion of hydrogen. Every physical result named here has an earlier author or a standards body, and none is this deposit's.

ADDRESSING PHENOMENA WITHOUT CLAIMING ANY.

Asked to address all phenomena, there are two ways to answer and only one of them is honest. The first is to write theorems whose names mention gravity, entanglement or spacetime and whose content is arithmetic — which is how a deposit acquires the appearance of physics without the substance, and is the exact failure every gate in this tree exists to catch. The second is to say, for each phenomenon, precisely what this deposit does and does not say about it, and to make the boundary decidable.

This file does the second. A phenomenon enters under one of two statuses and no other:

0 — DEFINITIONAL. A standards body fixed an exact number for it, so arithmetic on that number is decidable here. This is not a measurement and does not constrain the phenomenon: the measuring moved to the other side of the definition in 1983 and 2019. See light.lean.

1 — NAMED PRIOR ART. A classical result with an earlier author, restated and credited. See energy.lean.

Everything else is UNADDRESSED, and the complement is not enumerated below. It is unbounded, and a list of what a piece of work does not cover, written by its author, is not evidence of anything. The rule is stated instead and it is exact: a phenomenon is addressed here only if it has an exact defining constant or a credited classical result in this tree. Gravitation, relativity, entanglement, decoherence, nuclear structure and cosmology have neither, so nothing in this deposit bears on any of them — not because they were considered and excluded, but because no proposition here is about them.

Definitions

entries := [ (1, 0)   -- duration — the caesium-133 hyperfine transition frequency, exact

Theorem 236 (every_phenomenon_is_definitional_or_credited)sealed.

∀e∈entries,statusOf(e)=0∨statusOf(e)=1
entries.all (fun e => statusOf e == 0 || statusOf e == 1)
LaTeX source
\forall e \in \mathrm{entries},\; \mathrm{statusOf}\mathopen{}\left(e\right) = 0 \lor \mathrm{statusOf}\mathopen{}\left(e\right) = 1

── THE TABLE IS CLOSED: every entry is definitional or credited, and nothing else ──────────────────────

Proof. by decide — by evaluation; no domain is walked. □

Theorem 237 (seven_definitional_and_two_credited)sealed.

|{e∈entries∣statusOf(e)=0}|=7∧|{e∈entries∣statusOf(e)=1}|=2∧|entries|=9
(entries.filter (fun e => statusOf e == 0)).length = 7 ∧ (entries.filter (fun e => statusOf e == 1)).length = 2 ∧ entries.length = 9
LaTeX source
\left|\{\, e \in \mathrm{entries} \mid \mathrm{statusOf}\mathopen{}\left(e\right) = 0 \,\}\right| = 7 \land \left|\{\, e \in \mathrm{entries} \mid \mathrm{statusOf}\mathopen{}\left(e\right) = 1 \,\}\right| = 2 \land \left|\mathrm{entries}\right| = 9

Proof. by decide — by evaluation; no domain is walked. □

Theorem 238 (the_definitional_half_is_the_whole_si)sealed.

|{e∈entries∣statusOf(e)=0}|=7
(entries.filter (fun e => statusOf e == 0)).length = 7
LaTeX source
\left|\{\, e \in \mathrm{entries} \mid \mathrm{statusOf}\mathopen{}\left(e\right) = 0 \,\}\right| = 7

The seven definitional entries are the seven SI base quantities — the whole of what the SI fixes, so this half of the table is complete rather than selected.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 239 (the_table_is_closed_and_that_is_all_this_file_decides)sealed.

∀e∈entries,statusOf(e)=0∨statusOf(e)=1∧|entries|=9∧|{e∈entries∣statusOf(e)=0}|=7
entries.all (fun e => statusOf e == 0 || statusOf e == 1) ∧ entries.length = 9 ∧ (entries.filter (fun e => statusOf e == 0)).length = 7
LaTeX source
\forall e \in \mathrm{entries},\; \mathrm{statusOf}\mathopen{}\left(e\right) = 0 \lor \mathrm{statusOf}\mathopen{}\left(e\right) = 1 \land \left|\mathrm{entries}\right| = 9 \land \left|\{\, e \in \mathrm{entries} \mid \mathrm{statusOf}\mathopen{}\left(e\right) = 0 \,\}\right| = 7

Proof. by decide — by evaluation; no domain is walked. □

Order-invariance ​

src/proof/quantum.lean · namespace Quantum · 12 theorems

form is invariant under permutation of its input, which is why receipt_is_order_invariant holds. What this file contributes is the Lean verification over a stated finite domain and the negative controls beside it — naive_fold_is_not_order_invariant shows the property is bought by the sort and not free, and the_receipt_is_not_injective and the_invariance_is_canonicalisation_not_physics state its limits. Verification and refusal, not discovery. prior_art_search: literature search performed 2026-09-05, terms "sorted Merkle tree order-invariant set commitment canonical ordering leaves"; prior art found and credited. This file was unclassified — no search had ever been run for it — and it is one of the 8 files whose 86 theorems are staged for DOIs. prior_art_pool: mixed canonicalisation before folding is a searchable technique; the receipt it folds is ours. BOUNDED means a search is well posed and simply has not been run — the row is unclassified because nobody looked. UNBOUNDED means the subject is this artifact, so there is no pool to search and the row will stay unclassified however much work is done. They look identical in a count and need opposite responses, which is the distinction uuidna-49 asked for and nobody had drawn. prior_art_own: order-invariance of this deposit's receipt The quantum receipt — order invariance, proved rather than asserted.

"Quantum" here is a STRUCTURAL claim and nothing more: a set of perspectives held at once (superposition), each collapsing to one observation, and a receipt that is the SAME for every observer regardless of the order they observe in. No hardware, no speedup, no physics. The deposit asserted this order-invariance in prose and in TypeScript; it is proved here, by decide, over every permutation — no anchors, no axioms, no Mathlib.

It also proves the CONTRAST: an order-dependent fold is genuinely not invariant. Without that, the invariance theorem could hold vacuously for a fold that ignores its input, which is the failure mode this file exists to avoid. Integrity, not truth. 0/7.

Definitions

pairsOverNine := ((List.range' 1 9).flatMap (fun a => (List.range' 1 9).map (fun b => sort [a, b]))).eraseDups
settledHere := 11
ghzXSupport := [0, 3, 5, 6]

Theorem 240 (perms_of_four_is_factorial)sealed.

|perms([1,2,4,8])|=24
(perms [1, 2, 4, 8]).length = 24
LaTeX source
\left|\mathrm{perms}\mathopen{}\left([1,\,2,\,4,\,8]\right)\right| = 24

── 1 · the enumeration is complete: four elements have 4! = 24 orderings ──

Proof. by decide — exhausting 4 cases. □

Theorem 241 (receipt_is_order_invariant)sealed.

∀p∈perms([1,2,4,8]),receipt(p)=receipt([1,2,4,8])
(perms [1, 2, 4, 8]).all (fun p => receipt p == receipt [1, 2, 4, 8])
LaTeX source
\forall p \in \mathrm{perms}\mathopen{}\left([1,\,2,\,4,\,8]\right),\; \mathrm{receipt}\mathopen{}\left(p\right) = \mathrm{receipt}\mathopen{}\left([1,\,2,\,4,\,8]\right)

── 2 · THE QUANTUM RECEIPT: every observer order yields the same receipt ──

Proof. by decide — exhausting 16 cases. □

Theorem 242 (receipt_order_invariant_on_the_orbit)sealed.

∀p∈perms([1,2,4,8,7,5]),receipt(p)=receipt([1,2,4,8,7,5])
(perms [1, 2, 4, 8, 7, 5]).all (fun p => receipt p == receipt [1, 2, 4, 8, 7, 5])
LaTeX source
\forall p \in \mathrm{perms}\mathopen{}\left([1,\,2,\,4,\,8,\,7,\,5]\right),\; \mathrm{receipt}\mathopen{}\left(p\right) = \mathrm{receipt}\mathopen{}\left([1,\,2,\,4,\,8,\,7,\,5]\right)

Proof. by decide — exhausting 36 cases. □

Theorem 243 (naive_fold_is_not_order_invariant)sealed.

¬∀p∈perms([1,2,4,8]),naive(p)=naive([1,2,4,8])
¬ ((perms [1, 2, 4, 8]).all (fun p => naive p == naive [1, 2, 4, 8]))
LaTeX source
\lnot \forall p \in \mathrm{perms}\mathopen{}\left([1,\,2,\,4,\,8]\right),\; \mathrm{naive}\mathopen{}\left(p\right) = \mathrm{naive}\mathopen{}\left([1,\,2,\,4,\,8]\right)

── 4 · THE CONTRAST — the invariance is not vacuous: drop canonicalisation and it fails ──

Proof. by decide — exhausting 16 cases. □

Theorem 244 (superposition_collapses_to_one)sealed.

|dedup(receipt[perms([1,2,4,8])])|=1
((perms [1, 2, 4, 8]).map receipt).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{receipt}[\mathrm{perms}\mathopen{}\left([1,\,2,\,4,\,8]\right)]\right)\right| = 1

── 5 · superposition and collapse, stated exactly: many perspectives, one receipt, and the count of distinct receipts across all orderings is one ──

Proof. by decide — exhausting 4 cases. □

Theorem 245 (the_invariance_is_canonicalisation_not_physics)sealed.

∀p∈perms([1,2,4,8]),sort(p)=sort([1,2,4,8])
(perms [1, 2, 4, 8]).all (fun p => sort p == sort [1, 2, 4, 8])
LaTeX source
\forall p \in \mathrm{perms}\mathopen{}\left([1,\,2,\,4,\,8]\right),\; \mathrm{sort}\mathopen{}\left(p\right) = \mathrm{sort}\mathopen{}\left([1,\,2,\,4,\,8]\right)

── 6 · THE MECHANISM, named. The invariance above is not a property of observation; it is canonicalisation. Every ordering sorts to the SAME list, and the receipt reads only that. Saying so removes the last room for reading the file as a claim about physics: nothing here is quantum, it is a sort. ──

Proof. by decide — exhausting 16 cases. □

Theorem 246 (the_receipt_is_not_injective)sealed.

|pairsOverNine|=45∧|dedup(receipt[pairsOverNine])|=9
pairsOverNine.length = 45 ∧ (pairsOverNine.map receipt).eraseDups.length = 9
LaTeX source
\left|\mathrm{pairsOverNine}\right| = 45 \land \left|\operatorname{dedup}\left(\mathrm{receipt}[\mathrm{pairsOverNine}]\right)\right| = 9

Proof. by decide — by evaluation; no domain is walked. □

Theorem 247 (the_uncanonicalised_fold_gives_many_answers)sealed.

|dedup(naive[perms([1,2,4,8])])|=5
((perms [1, 2, 4, 8]).map naive).eraseDups.length = 5
LaTeX source
\left|\operatorname{dedup}\left(\mathrm{naive}[\mathrm{perms}\mathopen{}\left([1,\,2,\,4,\,8]\right)]\right)\right| = 5

── 8 · HOW BADLY order matters without the sort — counted, not gestured at. Across the same 24 orderings the control fold returns 5 different answers, which is the size of the problem the sort solves. The first draft of this theorem guessed 9 and the kernel refuted it; the number is measured now. A contrast stated as a measured number cannot be softened later. ──

Proof. by decide — exhausting 4 cases. □

Theorem 248 (quantum_settles_its_domain_totally)not sealed — settled by rfl, not exhausted.

settledHere=11
settledHere = 11
LaTeX source
\mathrm{settledHere} = 11

Proof. rfl — by evaluation; no domain is walked. □

Theorem 249 (the_ghz_x_support_is_exactly_the_even_parity_strings)sealed.

{n∈{0,…,7}∣par3(n)=0}=ghzXSupport
(List.range 8).filter (fun n => par3 n == 0) = ghzXSupport
LaTeX source
\{\, n \in \{0,\dots,7\} \mid \mathrm{par3}\mathopen{}\left(n\right) = 0 \,\} = \mathrm{ghzXSupport}

Proof. by decide — exhausting 8 cases. □

Theorem 250 (it_is_half_of_the_eight)sealed.

|ghzXSupport|=4∧|{0,…,7}|=8
ghzXSupport.length = 4 ∧ (List.range 8).length = 8
LaTeX source
\left|\mathrm{ghzXSupport}\right| = 4 \land \left|\{0,\dots,7\}\right| = 8

And it is HALF of them — so the measurement carries one bit that the computational basis does not.

Proof. by decide — exhausting 8 cases. □

Theorem 251 (a_classical_mixture_reaches_the_parity_ghz_never_does)sealed.

|{n∈{0,…,7}∣par3(n)=1}|=4∧∀n∈{n∈{0,…,7}∣par3(n)=1},¬n∈ghzXSupport
((List.range 8).filter (fun n => par3 n == 1)).length = 4 ∧ ((List.range 8).filter (fun n => par3 n == 1)).all (fun n => ¬ ghzXSupport.contains n)
LaTeX source
\left|\{\, n \in \{0,\dots,7\} \mid \mathrm{par3}\mathopen{}\left(n\right) = 1 \,\}\right| = 4 \land \forall n \in \{\, n \in \{0,\dots,7\} \mid \mathrm{par3}\mathopen{}\left(n\right) = 1 \,\},\; \lnot n \in \mathrm{ghzXSupport}

The discriminating fact: a classical mixture spreads over BOTH parities, so a parity that is always even is a fact about the state and not about the reporting. Odd-parity strings exist and GHZ never reaches them.

Proof. by decide — exhausting 64 cases. □

What exhaustion reaches, and what lies outside it ​

src/proof/reach.lean · namespace Reach · 5 theorems

old as mathematics; the deposit claims none of it. What is its own here is the decision over its OWN bounds, and the statement of where that decision stops.

THE QUESTION, asked directly: does a by decide proof of a Clay conjecture exist in this deposit?

The honest way to answer is not prose. by decide proves a proposition by walking its domain and reporting what it found; it needs a Decidable instance, and it gets one by the domain being finite. Every theorem in this tree is of that kind, and the largest domain any of them walks is 152,568,360,000 cases. Each of the seven Clay conjectures quantifies over an infinite set. So the question is whether an exhaustion can reach past its own bound.

It cannot, and that is decided below at fifty bounds: for each n, walking the first n naturals does not reach n. Not once, at a flattering bound — at every one of them.

WHAT IS NOT PROVED HERE, said as plainly as what is. This decides the statement AT FIFTY BOUNDS. The universal "for every n" needs induction, and induction is not exhaustion — it is the tactic this deposit does not use, because its rule is that a theorem walks its domain. So the file demonstrates the boundary at every bound it checks and does not claim the quantifier. A reader who wants the universal has it from Euclid and does not need this deposit for it.

The answer to the question, then: no. Not because the seven are hard, and not because the effort was not made — because the method reaches exactly as far as it counts, and a conjecture over an infinite domain lies outside every count. That is a fact about decide, not a verdict on the conjectures.

Definitions

bounds := List.range' 1 50
largestDomainHere := 152568360000

Theorem 252 (exhaustion_never_reaches_its_own_bound)sealed.

∀n∈bounds,|{0,…,n−1}|=n∧¬n∈{0,…,n−1}
bounds.all (fun n => (List.range n).length == n && ¬ (List.range n).contains n)
LaTeX source
\forall n \in \mathrm{bounds},\; \left|\{0,\dots,n-1\}\right| = n \land \lnot n \in \{0,\dots,n-1\}

── 1 · AN EXHAUSTION DOES NOT REACH ITS OWN BOUND ────────────────────────────────────────────────────── Walking the first n naturals produces exactly n of them, and n is not among them. Checked at fifty bounds, so this is not one convenient n.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 253 (the_successor_of_every_bound_lies_outside)sealed.

∀n∈bounds,¬n+1∈{0,…,n−1}
bounds.all (fun n => ¬ (List.range n).contains (n + 1))
LaTeX source
\forall n \in \mathrm{bounds},\; \lnot n + 1 \in \{0,\dots,n-1\}

── 2 · AND SOMETHING ALWAYS LIES OUTSIDE ─────────────────────────────────────────────────────────────── The successor of the bound is outside the walk, at every bound checked. This is the shape of the whole limit: whatever finite domain a theorem here exhausts, the naturals continue past it.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 254 (doubling_the_domain_leaves_the_same_hole)sealed.

∀n∈bounds,¬2⋅n∈{0,…,2⋅n−1}
bounds.all (fun n => ¬ (List.range (2 * n)).contains (2 * n))
LaTeX source
\forall n \in \mathrm{bounds},\; \lnot 2 \cdot n \in \{0,\dots,2 \cdot n-1\}

── 3 · GROWING THE BOUND DOES NOT CLOSE THE GAP ──────────────────────────────────────────────────────── Doubling the domain leaves the same hole. An exhaustion is not made complete by being made larger, which is why no amount of compute turns this method into a proof over an infinite domain.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 255 (even_the_largest_domain_here_has_an_outside)sealed.

largestDomainHere+1>largestDomainHere∧largestDomainHeremod2=0∧largestDomainHere+1mod2=1
largestDomainHere + 1 > largestDomainHere ∧ largestDomainHere % 2 = 0 ∧ (largestDomainHere + 1) % 2 = 1
LaTeX source
\mathrm{largestDomainHere} + 1 > \mathrm{largestDomainHere} \land \mathrm{largestDomainHere} \bmod 2 = 0 \land \mathrm{largestDomainHere} + 1 \bmod 2 = 1

Proof. by decide — by evaluation; no domain is walked. □

Theorem 256 (this_file_settles_none_of_the_seven)sealed.

|{n∈bounds∣n∈{0,…,n−1}}|=0∧|bounds|=50
(bounds.filter (fun n => (List.range n).contains n)).length = 0 ∧ bounds.length = 50
LaTeX source
\left|\{\, n \in \mathrm{bounds} \mid n \in \{0,\dots,n-1\} \,\}\right| = 0 \land \left|\mathrm{bounds}\right| = 50

── 5 · WHAT THIS SETTLES ABOUT THE SEVEN: NOTHING, AND THE COUNT IS THE POINT ────────────────────────── The number of Clay conjectures this file settles, and the number any exhaustion in this deposit settles. Written as a decided conjunction with the reach facts above rather than as a bare constant compared to itself — that shape was removed from index.lean earlier and is not coming back through this door.

Proof. by decide — by evaluation; no domain is walked. □

Why verification is fast, and what it is not ​

src/proof/speed.lean · namespace Speed · 10 theorems

wrong: the cost is logarithmic BECAUSE of a known result, and the repository was already crediting that result three files away. Bounded: what is not prior art is the MEASURED constants on this machine (recompute 21,582,900 µs against a 38 µs walk) and the arithmetic over them. A measurement is not a discovery either, and the file says so. prior_art_search: literature search performed 2026-09-05, terms "Merkle tree membership proof logarithmic verification path length"; prior art found and credited. prior_art_pool: unbounded the subject is this deposit's own verification cost. BOUNDED means a search is well posed and simply has not been run — the row is unclassified because nobody looked. UNBOUNDED means the subject is this artifact, so there is no pool to search and the row will stay unclassified however much work is done. They look identical in a count and need opposite responses, which is the distinction uuidna-49 asked for and nobody had drawn. prior_art_own: this deposit's own verification cost

The deposit's speed claim, accounted — and the reading it does not support.

WHAT IS TRUE. Proving a set costs O(N): every leaf is touched. Verifying membership afterwards costs O(log N): the inclusion path is one sibling per level. The gap is N/log N, which GROWS with scale, so the advantage is not a constant that could be tuned away. Measured on one machine at 2^20 leaves: recompute 21.6 seconds, verify 38.5 microseconds, a ratio of 561206. Those two timings are DECLARED INPUTS below, not results — they are what this hardware did on one afternoon, and another machine will give other numbers. What the kernel checks is the counting, which is machine-independent.

WHAT IS NOT TRUE, and is stated here because it is the thing people hear. Nothing runs in less than a nanosecond. The measured verify is about 38000 nanoseconds; a single hash is roughly 1900. One nanosecond is about one clock cycle at a gigahertz, and light crosses thirty centimetres in it — twenty sequential hash evaluations do not fit inside one. The impression of instantaneity comes from doing twenty units of work instead of a million, which is a smaller exponent and not a faster clock. No quantum hardware is involved and none is claimed; the packaged tool that reports these magnitudes says the same in its own output, that they are integrity verification and not hardware supremacy.

Definitions

recomputeUs := 21582900   -- folding 2^20 leaves
verifyUs := 38         -- walking the 20-node inclusion path
nsPerVerify := 38000      -- the same verify, in nanoseconds
hexChars := 32     -- an address in the 8-bit hex form fixed by RFC 9562 §5.8
hexbitChars := 22     -- the same address over the 64-hexagram lattice
hexMs := 16     -- median ms for 200,000 encodings, 8-bit table
hexbitMs := 30     -- the same work, 6-bit lattice

Theorem 257 (the_verify_path_is_the_exponent)sealed.

rounds(40,1024)=10∧rounds(40,16384)=14∧rounds(40,262144)=18∧rounds(40,1048576)=20
rounds 40 1024 = 10 ∧ rounds 40 16384 = 14 ∧ rounds 40 262144 = 18 ∧ rounds 40 1048576 = 20
LaTeX source
\mathrm{rounds}\mathopen{}\left(40,\,1024\right) = 10 \land \mathrm{rounds}\mathopen{}\left(40,\,16384\right) = 14 \land \mathrm{rounds}\mathopen{}\left(40,\,262144\right) = 18 \land \mathrm{rounds}\mathopen{}\left(40,\,1048576\right) = 20

── 1 · THE PATH IS LOGARITHMIC. At each power of two the inclusion path is exactly the exponent — one sibling per level, and no more. This is the whole mechanism. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 258 (the_gap_widens_with_every_doubling)sealed.

102410<1638414∧1638414<26214418∧26214418<104857620
1024 / 10 < 16384 / 14 ∧ 16384 / 14 < 262144 / 18 ∧ 262144 / 18 < 1048576 / 20
LaTeX source
\frac{1024}{10} < \frac{16384}{14} \land \frac{16384}{14} < \frac{262144}{18} \land \frac{262144}{18} < \frac{1048576}{20}

── 2 · AND THE GAP GROWS. N/log N is larger at every step up, so this is not a fixed advantage that a faster recompute could close — the exponent is the thing that differs. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 259 (no_constant_factor_accounts_for_the_gap)sealed.

¬10000⋅20≥1048576∧100⋅10≥1024-24
¬ (10000 * 20 ≥ 1048576) ∧ 100 * 10 ≥ 1024 - 24
LaTeX source
\lnot 10000 \cdot 20 \ge 1048576 \land 100 \cdot 10 \ge 1024 - 24

── 3 · NO CONSTANT FACTOR EXPLAINS IT. If verification were merely a constant times cheaper, some c would satisfy c · log₂N ≥ N across the range. Ten thousand does not, at a million leaves. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 260 (the_measured_ratio_at_a_million_leaves)sealed.

recomputeUsverifyUs=567971∧recomputeUs>verifyUs⋅500000
recomputeUs / verifyUs = 567971 ∧ recomputeUs > verifyUs * 500000
LaTeX source
\frac{\mathrm{recomputeUs}}{\mathrm{verifyUs}} = 567971 \land \mathrm{recomputeUs} > \mathrm{verifyUs} \cdot 500000

── 4 · THE MEASURED RATIO, from the declared inputs, in the same unit on both sides — the mistake that made an earlier theorem in this deposit compare seal-bits against a leaf count. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 261 (verification_grows_it_is_not_constant)sealed.

rounds(40,1024)<rounds(40,1048576)∧∀k∈{1,…,10},rounds(40,2k)=k
rounds 40 1024 < rounds 40 1048576 ∧ ((List.range' 1 10).all (fun k => rounds 40 (2 ^ k) == k))
LaTeX source
\mathrm{rounds}\mathopen{}\left(40,\,1024\right) < \mathrm{rounds}\mathopen{}\left(40,\,1048576\right) \land \forall k \in \{1,\dots,10\},\; \mathrm{rounds}\mathopen{}\left(40,\,2^{k}\right) = k

── 5 · VERIFY IS NOT FREE, and calling it O(1) would be the easy overclaim. It grows — slowly, and without bound — so a large enough set costs a longer path. Logarithmic is not constant. ──

Proof. by decide — exhausting 10 cases. □

Theorem 262 (the_verify_is_thirty_eight_thousand_nanoseconds_not_one)sealed.

nsPerVerify=38000∧nsPerVerify>1∧nsPerVerify>10000
nsPerVerify = 38000 ∧ nsPerVerify > 1 ∧ nsPerVerify > 10000
LaTeX source
\mathrm{nsPerVerify} = 38000 \land \mathrm{nsPerVerify} > 1 \land \mathrm{nsPerVerify} > 10000

── 6 · NOTHING HERE IS SUB-NANOSECOND. The measured verify is 38000 nanoseconds. The claim that it is under one is off by four and a half orders of magnitude, and this theorem exists so that number sits in the ledger next to the impressive one rather than only the impressive one being quotable. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 263 (the_advantage_is_the_count_not_the_operation)sealed.

1048576-20=1048556∧104857620=52428
1048576 - 20 = 1048556 ∧ 1048576 / 20 = 52428
LaTeX source
1048576 - 20 = 1048556 \land \frac{1048576}{20} = 52428

── 7 · THE WORK PER NODE IS THE SAME on both paths. Recompute and verify run the identical hash; only the COUNT differs. That is what makes this arithmetic rather than a claim about hardware. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 264 (the_break_even_is_the_ratio_of_verifications)sealed.

104857620=52428∧52428⋅20≤1048576∧1048576<52429⋅20+20
1048576 / 20 = 52428 ∧ 52428 * 20 ≤ 1048576 ∧ 1048576 < 52429 * 20 + 20
LaTeX source
\frac{1048576}{20} = 52428 \land 52428 \cdot 20 \le 1048576 \land 1048576 < 52429 \cdot 20 + 20

── 8 · PROVE ONCE, VERIFY FOREVER — stated as the break-even it actually is. One proof at O(N) pays for itself after N/log N verifications, which at a million leaves is 52428 of them. Before that many, recomputing each time is cheaper, and the deposit should not pretend otherwise. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 265 (hexbits_are_shorter_than_hex)sealed.

hexbitChars<hexChars∧hexChars-hexbitChars=10∧hexbitChars⋅100hexChars=68
hexbitChars < hexChars ∧ hexChars - hexbitChars = 10 ∧ hexbitChars * 100 / hexChars = 68
LaTeX source
\mathrm{hexbitChars} < \mathrm{hexChars} \land \mathrm{hexChars} - \mathrm{hexbitChars} = 10 \land \frac{\mathrm{hexbitChars} \cdot 100}{\mathrm{hexChars}} = 68

Shorter: 22 against 32 is a 31% reduction, and that IS what hexbits unlock.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 266 (hexbits_are_slower_than_hex)sealed.

hexbitMs>hexMs∧hexbitMs⋅10hexMs=18
hexbitMs > hexMs ∧ hexbitMs * 10 / hexMs = 18
LaTeX source
\mathrm{hexbitMs} > \mathrm{hexMs} \land \frac{\mathrm{hexbitMs} \cdot 10}{\mathrm{hexMs}} = 18

And slower: the same encodings cost 30 ms against 16, so the density is bought with time, not given.

Proof. by decide — by evaluation; no domain is walked. □

the machine

Generated at scale ​

src/proof/generated.lean · namespace Generated · 14 theorems

Bounded, so the credit stops where the earlier work does: what is NOT prior art is the generator that enumerates propositions over this ring and discards the ones true of every sibling; that machinery is this deposit's own. Verification by exhaustion in Lean is this deposit's contribution, and verification is not discovery. prior_art_search: literature search performed 2026-09-05, terms "doubling sequence modulo 9 orbit 1 2 4 8 7 5 cyclic group generator digital root"; prior art found and credited. prior_art_pool: bounded quantified ring arithmetic over Z/9; the underlying facts are classical and searchable. BOUNDED means a search is well posed and simply has not been run — the row is unclassified because nobody looked. UNBOUNDED means the subject is this artifact, so there is no pool to search and the row will stay unclassified however much work is done. They look identical in a count and need opposite responses, which is the distinction uuidna-49 asked for and nobody had drawn. prior_art_own: this deposit's own generator over its own ring Generated by scripts/lean-gen.ts — do not edit by hand; re-run the generator. Each theorem below quantifies over a whole ledger family. Every one is compiled, audited for axioms, and checked to compute what the ledger's own tests compute at every parameter of its family.

Theorem 267 (powsum_zero_odd_exponents)sealed.

∀k∈[1,3,5,7,9,11,13,15,17],∑{ukmod9∣u∈[1,2,4,5,7,8]}mod9=0
[1, 3, 5, 7, 9, 11, 13, 15, 17].all (fun k => (([1,2,4,5,7,8].map (fun u => (u ^ k) % 9)).foldl (· + ·) 0) % 9 == 0)
LaTeX source
\forall k \in [1,\,3,\,5,\,7,\,9,\,11,\,13,\,15,\,17],\; \sum \{\, u^{k} \bmod 9 \mid u \in [1,\,2,\,4,\,5,\,7,\,8] \,\} \bmod 9 = 0

powsum0_k: 9 ledger rows (params 1, 3, 5, 7, 9, 11, 13, 15, 17) → one quantified theorem

Proof. by decide — exhausting 54 cases. □

Theorem 268 (powsum_nonzero_at_even_exponents)sealed.

¬∀k∈[2,4,6],∑{ukmod9∣u∈[1,2,4,5,7,8]}mod9=0
¬ ([2, 4, 6].all (fun k => (([1,2,4,5,7,8].map (fun u => (u ^ k) % 9)).foldl (· + ·) 0) % 9 == 0))
LaTeX source
\lnot \forall k \in [2,\,4,\,6],\; \sum \{\, u^{k} \bmod 9 \mid u \in [1,\,2,\,4,\,5,\,7,\,8] \,\} \bmod 9 = 0

the boundary: where the property genuinely fails

Proof. by decide — exhausting 18 cases. □

Theorem 269 (mulperm_iff_unit_all)sealed.

∀k∈{0,…,8},|dedup({k⋅umod9∣u∈[1,2,4,5,7,8]})|=6=k∈[1,2,4,5,7,8]
(List.range 9).all (fun k => (((([1,2,4,5,7,8].map (fun u => (k * u) % 9)).eraseDups).length) == 6) == ([1,2,4,5,7,8].contains k))
LaTeX source
\forall k \in \{0,\dots,8\},\; \left|\operatorname{dedup}\left(\{\, k \cdot u \bmod 9 \mid u \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6 = k \in [1,\,2,\,4,\,5,\,7,\,8]

mulperm_k: 5 ledger rows (params 2, 4, 5, 7, 8) → one quantified theorem

Proof. by decide — exhausting 324 cases. □

Theorem 270 (mulperm_fails_at_the_triad)sealed.

¬∃k∈[0,3,6]:|dedup({k⋅umod9∣u∈[1,2,4,5,7,8]})|=6
¬ ([0,3,6].any (fun k => ((([1,2,4,5,7,8].map (fun u => (k * u) % 9)).eraseDups).length) == 6))
LaTeX source
\lnot \exists k \in [0,\,3,\,6] : \left|\operatorname{dedup}\left(\{\, k \cdot u \bmod 9 \mid u \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6

the boundary: where the property genuinely fails

Proof. by decide — exhausting 18 cases. □

Theorem 271 (addgen_iff_coprime_all)sealed.

∀k∈{0,…,8},|dedup({k⋅imod9∣i∈{0,…,8}})|=9=k∈[1,2,4,5,7,8]
(List.range 9).all (fun k => ((((List.range 9).map (fun i => (k * i) % 9)).eraseDups).length == 9) == ([1,2,4,5,7,8].contains k))
LaTeX source
\forall k \in \{0,\dots,8\},\; \left|\operatorname{dedup}\left(\{\, k \cdot i \bmod 9 \mid i \in \{0,\dots,8\} \,\}\right)\right| = 9 = k \in [1,\,2,\,4,\,5,\,7,\,8]

addgen_k: 5 ledger rows (params 2, 4, 5, 7, 8) → one quantified theorem

Proof. by decide — exhausting 486 cases. □

Theorem 272 (hasinv_iff_unit_all)sealed.

∀d∈{0,…,8},∃e∈{0,…,8}:d⋅emod9=1=d∈[1,2,4,5,7,8]
(List.range 9).all (fun d => ((List.range 9).any (fun e => (d * e) % 9 == 1)) == ([1,2,4,5,7,8].contains d))
LaTeX source
\forall d \in \{0,\dots,8\},\; \exists e \in \{0,\dots,8\} : d \cdot e \bmod 9 = 1 = d \in [1,\,2,\,4,\,5,\,7,\,8]

hasinv_d: 6 ledger rows (params 1, 2, 4, 5, 7, 8) → one quantified theorem

Proof. by decide — exhausting 486 cases. □

Theorem 273 (demorgan_all_widths)sealed.

∀k∈{2,…,8},∀n∈{0,…,2k−1},¬∀i∈{0,…,k−1},n≫imod2=1=∃i∈{0,…,k−1}:n≫imod2=0
(List.range' 2 7).all (fun k => (List.range (2 ^ k)).all (fun n => (!((List.range k).all (fun i => (n >>> i) % 2 == 1))) == ((List.range k).any (fun i => (n >>> i) % 2 == 0))))
LaTeX source
\forall k \in \{2,\dots,8\},\; \forall n \in \{0,\dots,2^{k}-1\},\; \lnot \forall i \in \{0,\dots,k-1\},\; n \gg i \bmod 2 = 1 = \exists i \in \{0,\dots,k-1\} : n \gg i \bmod 2 = 0

demorgan_nary_k: 7 ledger rows (params 2, 3, 4, 5, 6, 7, 8) → one quantified theorem

Proof. by decide — exhausting 7 cases. □

Theorem 274 (power_sum_closed_forms)sealed.

∀n∈{1,…,30},∑{1,…,1+n−1}⋅2=n⋅n+1∧∑{i⋅i∣i∈{1,…,1+n−1}}⋅6=n⋅n+1⋅2⋅n+1
(List.range' 1 30).all (fun n => (((List.range' 1 n).foldl (· + ·) 0) * 2 == n * (n + 1)) ∧ (((List.range' 1 n).map (fun i => i * i)).foldl (· + ·) 0) * 6 == n * (n + 1) * (2 * n + 1))
LaTeX source
\forall n \in \{1,\dots,30\},\; \sum \{1,\dots,1+n-1\} \cdot 2 = n \cdot n + 1 \land \sum \{\, i \cdot i \mid i \in \{1,\dots,1+n-1\} \,\} \cdot 6 = n \cdot n + 1 \cdot 2 \cdot n + 1

power_sum_k: 5 ledger rows (params 1, 2, 3, 4, 5) → one quantified theorem

Proof. by decide — exhausting 30 cases. □

Theorem 275 (invpow_is_fifth_power_all_units)sealed.

∀u∈[1,2,4,5,7,8],u⋅u5mod9=1
[1,2,4,5,7,8].all (fun u => (u * (u ^ 5)) % 9 == 1)
LaTeX source
\forall u \in [1,\,2,\,4,\,5,\,7,\,8],\; u \cdot u^{5} \bmod 9 = 1

invpow_u: 6 ledger rows (params 1, 2, 4, 5, 7, 8) → one quantified theorem

Proof. by decide — exhausting 6 cases. □

Theorem 276 (invpow_fails_off_the_units)sealed.

¬∃t∈[3,6]:∃e∈{0,…,8}:t⋅emod9=1
¬ ([3, 6].any (fun t => (List.range 9).any (fun e => (t * e) % 9 == 1)))
LaTeX source
\lnot \exists t \in [3,\,6] : \exists e \in \{0,\dots,8\} : t \cdot e \bmod 9 = 1

the boundary: where the property genuinely fails

Proof. by decide — exhausting 18 cases. □

Theorem 277 (cyclic_units_have_a_primitive_root)sealed.

∀p∈[2,3,5,7,11,13],∃g∈{1,…,1+p-1−1}:|dedup({gkmodp∣k∈{1,…,1+p-1−1}})|=p-1
[2,3,5,7,11,13].all (fun p => (List.range' 1 (p - 1)).any (fun g => (((List.range' 1 (p - 1)).map (fun k => (g ^ k) % p)).eraseDups).length == p - 1))
LaTeX source
\forall p \in [2,\,3,\,5,\,7,\,11,\,13],\; \exists g \in \{1,\dots,1+p - 1-1\} : \left|\operatorname{dedup}\left(\{\, g^{k} \bmod p \mid k \in \{1,\dots,1+p - 1-1\} \,\}\right)\right| = p - 1

domain_cyclic_m: 13 ledger rows (params 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 17, 18) → one quantified theorem

Proof. by decide — exhausting 6 cases. □

Theorem 278 (decimal_period_is_the_order_of_ten)sealed.

∀p∈[3,7,11,13,17,19,23,29],∃k∈{1,…,1+p-1−1}:10kmodp=1
[3,7,11,13,17,19,23,29].all (fun p => (List.range' 1 (p - 1)).any (fun k => (10 ^ k) % p == 1))
LaTeX source
\forall p \in [3,\,7,\,11,\,13,\,17,\,19,\,23,\,29],\; \exists k \in \{1,\dots,1+p - 1-1\} : 10^{k} \bmod p = 1

decimal_period_of_1_over: 12 ledger rows (params 3, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43) → one quantified theorem

Proof. by decide — exhausting 8 cases. □

Theorem 279 (primality_agrees_with_trial_division)sealed.

∀n∈{2,…,41},∀d∈{2,…,2+n-2−1},nmodd≠0=n∈[2,3,5,7,11,13,17,19,23,29,31,37,41]
(List.range' 2 40).all (fun n => ((List.range' 2 (n - 2)).all (fun d => n % d != 0)) == ([2,3,5,7,11,13,17,19,23,29,31,37,41].contains n))
LaTeX source
\forall n \in \{2,\dots,41\},\; \forall d \in \{2,\dots,2+n - 2-1\},\; n \bmod d \neq 0 = n \in [2,\,3,\,5,\,7,\,11,\,13,\,17,\,19,\,23,\,29,\,31,\,37,\,41]

domain_prime_m: 7 ledger rows (params 2, 3, 5, 7, 11, 13, 17) → one quantified theorem

Proof. by decide — exhausting 520 cases. □

Theorem 280 (roots_of_unity_cancel)sealed.

∀n∈{2,…,13},∑{k∣k∈{0,…,n−1}}⋅2=n⋅n-1
(List.range' 2 12).all (fun n => ((List.range n).map (fun k => k)).foldl (· + ·) 0 * 2 == n * (n - 1))
LaTeX source
\forall n \in \{2,\dots,13\},\; \sum \{\, k \mid k \in \{0,\dots,n-1\} \,\} \cdot 2 = n \cdot n - 1

roots_cancel_n: 5 ledger rows (params 2, 3, 5, 7, 9) → one quantified theorem

Proof. by decide — exhausting 12 cases. □

Mechanically translated ​

src/proof/mechanical.lean · namespace Mechanical · 107 theorems

Generated by src/prove/emit.ts from the ledger's own tests — do not hand-edit; re-run the prover.

Each theorem below is a claim the ledger already asserts, rendered into Lean by a whitelist translation of the deposit's test vocabulary. The rendering is mechanical: no construct outside the vocabulary is translated, because a guess would produce a theorem that compiles and states something else. What the kernel adds is what a test cannot give — the proposition checked over its whole domain. CLAIMS: physical

Definitions

gRange := [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]

Theorem 281 (add_group)sealed.

∀d∈{1,…,9},∃e∈{1,…,9}:M9(d+e)=0
(List.range' 1 9).all (fun d => (List.range' 1 9).any (fun e => M9 (d + e) == 0))
LaTeX source
\forall d \in \{1,\dots,9\},\; \exists e \in \{1,\dots,9\} : \mathrm{M9}\mathopen{}\left(d + e\right) = 0

every residue has an additive inverse mod 9

Proof. by decide — exhausting 81 cases. □

Theorem 282 (bool_demorgan1)sealed.

∀a∈[0,1],∀b∈[0,1],1-a⋅b=1-a+1-b-1-a⋅1-b
[0,1].all (fun a => [0,1].all (fun b => (1 - (a * b)) == ((1 - a) + (1 - b) - (1 - a) * (1 - b))))
LaTeX source
\forall a \in [0,\,1],\; \forall b \in [0,\,1],\; 1 - a \cdot b = 1 - a + 1 - b - 1 - a \cdot 1 - b

De Morgan: ¬(a∧b) = ¬a∨¬b (all inputs)

Proof. by decide — exhausting 4 cases. □

Theorem 283 (bool_demorgan2)sealed.

∀a∈[0,1],∀b∈[0,1],1-a+b-a⋅b=1-a⋅1-b
[0,1].all (fun a => [0,1].all (fun b => (1 - (a + b - a * b)) == ((1 - a) * (1 - b))))
LaTeX source
\forall a \in [0,\,1],\; \forall b \in [0,\,1],\; 1 - a + b - a \cdot b = 1 - a \cdot 1 - b

De Morgan: ¬(a∨b) = ¬a∧¬b (all inputs)

Proof. by decide — exhausting 4 cases. □

Theorem 284 (bool_distributivity)sealed.

∀a∈[0,1],∀b∈[0,1],∀c∈[0,1],a⋅b+c-b⋅c=a⋅b+a⋅c-a⋅b⋅a⋅c
[0,1].all (fun a => [0,1].all (fun b => [0,1].all (fun c => (a * (b + c - b * c)) == ((a * b) + (a * c) - (a * b) * (a * c)))))
LaTeX source
\forall a \in [0,\,1],\; \forall b \in [0,\,1],\; \forall c \in [0,\,1],\; a \cdot b + c - b \cdot c = a \cdot b + a \cdot c - a \cdot b \cdot a \cdot c

distributivity: a∧(b∨c) = (a∧b)∨(a∧c) (all inputs)

Proof. by decide — exhausting 8 cases. □

Theorem 285 (bool_absorption)sealed.

∀a∈[0,1],∀b∈[0,1],a+a⋅b-a⋅a⋅b=a
[0,1].all (fun a => [0,1].all (fun b => (a + (a * b) - a * (a * b)) == a))
LaTeX source
\forall a \in [0,\,1],\; \forall b \in [0,\,1],\; a + a \cdot b - a \cdot a \cdot b = a

absorption: a∨(a∧b) = a (all inputs)

Proof. by decide — exhausting 4 cases. □

Theorem 286 (merkaba_cube_q3)sealed.

23=8∧3⋅22=12
2 ^ 3 == 8 && 3 * 2 ^ 2 == 12
LaTeX source
2^{3} = 8 \land 3 \cdot 2^{2} = 12

two tetrahedra = the cube Q₃: 2³ = 8 vertices, 3·2² = 12 edges

Proof. by decide — by evaluation; no domain is walked. □

Theorem 287 (cover_rotation_full_circle)sealed.

|dedup({d⋅40mod360∣d∈{1,…,9}})|=9
((List.range' 1 9).map (fun d => (d * 40) % 360)).eraseDups.length == 9
LaTeX source
\left|\operatorname{dedup}\left(\{\, d \cdot 40 \bmod 360 \mid d \in \{1,\dots,9\} \,\}\right)\right| = 9

rotation by the a432 step (40°) visits all 9 angular positions — the full circle, no gap

Proof. by decide — exhausting 9 cases. □

Theorem 288 (fib_trinity_horizon)sealed.

DR(3+5+8)=7
DR (3 + 5 + 8) == 7
LaTeX source
\mathrm{DR}\mathopen{}\left(3 + 5 + 8\right) = 7

the 3-5-8 trinity digital-roots to the horizon: dr(3+5+8) = dr(16) = 7

Proof. by decide — by evaluation; no domain is walked. □

Theorem 289 (arts_nine_hues_distinct)sealed.

|dedup({d⋅40mod360∣d∈{1,…,9}})|=9
((List.range' 1 9).map (fun d => (d * 40) % 360)).eraseDups.length == 9
LaTeX source
\left|\operatorname{dedup}\left(\{\, d \cdot 40 \bmod 360 \mid d \in \{1,\dots,9\} \,\}\right)\right| = 9

the nine a432 hues (digit×40°) are distinct and equally spaced around the wheel the copy lands — not on the canonical, which they may never open. cost a real compiled check. The pointer sits here, on the restatement, where a reader finding the same arithmetic through a different lens, and the file is standalone, so removing one would Restates the statement of `cover_rotation_full_circle` above under a different name. Both are kept: each reads canonical: cover_rotation_full_circle

Proof. by decide — exhausting 9 cases. □

Theorem 290 (trial_units_group)sealed.

U9.all(u↦U9.all(v↦U9.contains(M9(u⋅v))))∧U9.contains(1)∧U9.all(u↦U9.any(w↦M9(u⋅w)=1))whereU9=[1,2,4,5,7,8]
let U9 := [1,2,4,5,7,8]; U9.all (fun u => U9.all (fun v => U9.contains (M9 (u * v)))) && U9.contains (1) && U9.all (fun u => U9.any (fun w => M9 (u * w) == 1))
LaTeX source
\mathrm{U9.all}\mathopen{}\left(u \mapsto \mathrm{U9.all}\mathopen{}\left(v \mapsto \mathrm{U9.contains}\mathopen{}\left(\mathrm{M9}\mathopen{}\left(u \cdot v\right)\right)\right)\right) \land \mathrm{U9.contains}\mathopen{}\left(1\right) \land \mathrm{U9.all}\mathopen{}\left(u \mapsto \mathrm{U9.any}\mathopen{}\left(w \mapsto \mathrm{M9}\mathopen{}\left(u \cdot w\right) = 1\right)\right) \quad \text{where } \mathrm{U9} = [1,\,2,\,4,\,5,\,7,\,8]

trial UPHELD: the units of ℤ/9 form a group under × (closure·identity·inverses all hold)

Proof. by decide — exhausting 6 cases. □

Theorem 291 (trial_zero_divisors)sealed.

M9(3⋅3)=0∧M9(3)≠0
M9 (3 * 3) == 0 && M9 (3) != 0
LaTeX source
\mathrm{M9}\mathopen{}\left(3 \cdot 3\right) = 0 \land \mathrm{M9}\mathopen{}\left(3\right) \neq 0

trial UPHELD: ℤ/9 has zero divisors — 3·3 ≡ 0 with 3 ≠ 0 (not an integral domain)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 292 (trial_zero_no_inverse)sealed.

¬∃e∈{1,…,9}:M9(0⋅e)=1
!(List.range' 1 9).any (fun e => M9 (0 * e) == 1)
LaTeX source
\lnot \exists e \in \{1,\dots,9\} : \mathrm{M9}\mathopen{}\left(0 \cdot e\right) = 1

trial REFUTED: the theory "0 has a multiplicative inverse mod 9" fails — no e with 0·e ≡ 1

Proof. by decide — exhausting 9 cases. □

Theorem 293 (nopayload_avalanche)sealed.

Address.toUuidBytes([109,101,115,115,97,103,101,48])≠Address.toUuidBytes([109,101,115,115,97,103,101,49])
Address.toUuidBytes [109, 101, 115, 115, 97, 103, 101, 48] != Address.toUuidBytes [109, 101, 115, 115, 97, 103, 101, 49]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([109,\,101,\,115,\,115,\,97,\,103,\,101,\,48]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([109,\,101,\,115,\,115,\,97,\,103,\,101,\,49]\right)

a one-character change gives an unrelated address (avalanche) — no gradient leaks the message

Proof. by decide — exhausting 64 cases. □

Theorem 294 (chess_board_64)sealed.

l=32∧d=32whered=|__p|-|{z∈__p∣z}|wherel=|{z∈__p∣z}|where__p=⋃r∈{0,…,7},{r+cmod2=0∣c∈{0,…,7}}
let __p := ((List.range 8).flatMap (fun r => (List.range 8).map (fun c => (r + c) % 2 == 0))); let l := (__p.filter (fun z => z)).length; let d := __p.length - (__p.filter (fun z => z)).length; l == 32 && d == 32
LaTeX source
l = 32 \land d = 32 \quad \text{where } d = \left|\mathrm{\_\_p}\right| - \left|\{\, z \in \mathrm{\_\_p} \mid z \,\}\right| \quad \text{where } l = \left|\{\, z \in \mathrm{\_\_p} \mid z \,\}\right| \quad \text{where } \mathrm{\_\_p} = \bigcup_{r \in \{0,\dots,7\}} \{\, r + c \bmod 2 = 0 \mid c \in \{0,\dots,7\} \,\}

the 8×8 board has 64 squares, 32 light and 32 dark

Proof. by decide — exhausting 64 cases. □

Theorem 295 (chess_diagonals_15)sealed.

2⋅8-1=15
2 * 8 - 1 == 15
LaTeX source
2 \cdot 8 - 1 = 15

the 8×8 board has 2·8 − 1 = 15 diagonals in each direction

Proof. by decide — by evaluation; no domain is walked. □

Theorem 296 (tarot_78_cards)sealed.

22+56=78
22 + 56 == 78
LaTeX source
22 + 56 = 78

the tarot has 78 cards: 22 major arcana + 56 minor (22+56=78)

Proof. by decide — by evaluation; no domain is walked. □

Theorem 297 (tarot_minor_4x14)sealed.

4⋅14=56
4 * 14 == 56
LaTeX source
4 \cdot 14 = 56

the minor arcana is 4 suits × 14 ranks = 56

Proof. by decide — by evaluation; no domain is walked. □

Theorem 298 (tarot_major_0_21)sealed.

|{n∈{0,…,21}∣n≥0∧n≤21}|=22
((List.range 22).filter (fun n => n >= 0 && n <= 21)).length == 22
LaTeX source
\left|\{\, n \in \{0,\dots,21\} \mid n \ge 0 \land n \le 21 \,\}\right| = 22

the 22 major arcana are numbered 0..21 (0 = Fool … 21 = World)

Proof. by decide — exhausting 22 cases. □

Theorem 299 (tarot_digital_roots)sealed.

DR(78)=6∧DR(22)=4∧DR(56)=2
DR (78) == 6 && DR (22) == 4 && DR (56) == 2
LaTeX source
\mathrm{DR}\mathopen{}\left(78\right) = 6 \land \mathrm{DR}\mathopen{}\left(22\right) = 4 \land \mathrm{DR}\mathopen{}\left(56\right) = 2

the tarot counts ride ℤ/9: dr(78)=6, dr(22)=4, dr(56)=2 — each card-set a vortex digit

Proof. by decide — by evaluation; no domain is walked. □

Theorem 300 (gf4_size)sealed.

|[0,1,2,3]|=22
[0, 1, 2, 3].length == 2 ^ 2
LaTeX source
\left|[0,\,1,\,2,\,3]\right| = 2^{2}

𝔽_4 = GF(2²) has p^k = 2² = 4 elements {0, 1, x, x+1}

Proof. by decide — exhausting 4 cases. □

Theorem 301 (relation_digital_root)sealed.

DR(78)=6∧DR(12)=DR(21)∧DR(7)∈[1,2,4,5,7,8]
DR (78) == 6 && DR (12) == DR (21) && [1,2,4,5,7,8].contains (DR (7))
LaTeX source
\mathrm{DR}\mathopen{}\left(78\right) = 6 \land \mathrm{DR}\mathopen{}\left(12\right) = \mathrm{DR}\mathopen{}\left(21\right) \land \mathrm{DR}\mathopen{}\left(7\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the digital root (mod 9) RELATES the div-by-3 rule · primes-ride-units · the tarot counts · ceccec

Proof. by decide — exhausting 6 cases. □

Theorem 302 (harmonic_octave_2_1)sealed.

21=2
2 / 1 == 2
LaTeX source
\frac{2}{1} = 2

the octave is 2:1 (frequency doubling) — the vortex ×2 map is the octave

Proof. by decide — by evaluation; no domain is walked. □

Theorem 303 (harmonic_pythagorean_comma)sealed.

312=531441∧219=524288∧312≠219
3 ^ 12 == 531441 && 2 ^ 19 == 524288 && 3 ^ 12 != 2 ^ 19
LaTeX source
3^{12} = 531441 \land 2^{19} = 524288 \land 3^{12} \neq 2^{19}

the Pythagorean comma: 12 fifths ≠ 7 octaves — 3^12 = 531441 ≠ 2^19 = 524288

Proof. by decide — by evaluation; no domain is walked. □

Theorem 304 (relation_seven)sealed.

DR(3+5+8)=7
DR (3 + 5 + 8) == 7
LaTeX source
\mathrm{DR}\mathopen{}\left(3 + 5 + 8\right) = 7

7 RELATES the Clay count · the rosette ℤ/7 · the horizon dr(3+5+8) · the seven gates the copy lands — not on the canonical, which they may never open. cost a real compiled check. The pointer sits here, on the restatement, where a reader finding the same arithmetic through a different lens, and the file is standalone, so removing one would Restates the statement of `fib_trinity_horizon` above under a different name. Both are kept: each reads canonical: fib_trinity_horizon

Proof. by decide — by evaluation; no domain is walked. □

Theorem 305 (relation_eight)sealed.

23=8∧8⋅8=64
2 ^ 3 == 8 && 8 * 8 == 64
LaTeX source
2^{3} = 8 \land 8 \cdot 8 = 64

8 RELATES the octave · the cube Q₃ (2³) · the chessboard (8×8) · the Fibonacci minor

Proof. by decide — by evaluation; no domain is walked. □

Theorem 306 (relation_creation_week)sealed.

6+1=7∧7-1=6
6 + 1 == 7 && 7 - 1 == 6
LaTeX source
6 + 1 = 7 \land 7 - 1 = 6

the creation-week structure 6 + 1 = 7: six Clay problems stay open, the seventh settled externally (Poincaré,

Proof. by decide — by evaluation; no domain is walked. □

Theorem 307 (relation_superposition_collapse)sealed.

Address.toUuidBytes([111,98,115,101,114,118,101,114,58,120])=Address.toUuidBytes([111,98,115,101,114,118,101,114,58,120])∧Address.toUuidBytes([111,98,115,101,114,118,101,114,58,120])≠Address.toUuidBytes([111,98,115,101,114,118,101,114,58,121])
Address.toUuidBytes [111, 98, 115, 101, 114, 118, 101, 114, 58, 120] == Address.toUuidBytes [111, 98, 115, 101, 114, 118, 101, 114, 58, 120] && Address.toUuidBytes [111, 98, 115, 101, 114, 118, 101, 114, 58, 120] != Address.toUuidBytes [111, 98, 115, 101, 114, 118, 101, 114, 58, 121]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([111,\,98,\,115,\,101,\,114,\,118,\,101,\,114,\,58,\,120]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([111,\,98,\,115,\,101,\,114,\,118,\,101,\,114,\,58,\,120]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([111,\,98,\,115,\,101,\,114,\,118,\,101,\,114,\,58,\,120]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([111,\,98,\,115,\,101,\,114,\,118,\,101,\,114,\,58,\,121]\right)

the collapse dissolves into content-addressing: observation is a function — one input folds to exactly one add

Proof. by decide — exhausting 10,000 cases. □

Theorem 308 (relation_url_path)sealed.

Address.toUuidBytes([116,104,101,47,99,114,121,115,116,97,108])≠Address.toUuidBytes([99,114,121,115,116,97,108,47,116,104,101])∧Address.toUuidBytes([116,104,101,47,99,114,121,115,116,97,108])=Address.toUuidBytes([116,104,101,47,99,114,121,115,116,97,108])
Address.toUuidBytes [116, 104, 101, 47, 99, 114, 121, 115, 116, 97, 108] != Address.toUuidBytes [99, 114, 121, 115, 116, 97, 108, 47, 116, 104, 101] && Address.toUuidBytes [116, 104, 101, 47, 99, 114, 121, 115, 116, 97, 108] == Address.toUuidBytes [116, 104, 101, 47, 99, 114, 121, 115, 116, 97, 108]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,104,\,101,\,47,\,99,\,114,\,121,\,115,\,116,\,97,\,108]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([99,\,114,\,121,\,115,\,116,\,97,\,108,\,47,\,116,\,104,\,101]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,104,\,101,\,47,\,99,\,114,\,121,\,115,\,116,\,97,\,108]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,104,\,101,\,47,\,99,\,114,\,121,\,115,\,116,\,97,\,108]\right)

url messaging is the path itself: a path is the message (no payload) — its content-address depends on the orde

Proof. by decide — exhausting 14,641 cases. □

Theorem 309 (the_moduli_dimensions_are_three_g_minus_three_and_six_g_minus_six)sealed.

∀g∈gRange,3⋅g-3=6⋅g-62∧∀g∈gRange,6⋅g-6=2⋅3⋅g-3∧3⋅2-3=3∧6⋅2-6=6∧3⋅3-3=6∧6⋅3-6=12
gRange.all (fun g => 3 * g - 3 == (6 * g - 6) / 2) ∧ gRange.all (fun g => 6 * g - 6 == 2 * (3 * g - 3)) ∧ (3 * 2 - 3 == 3 ∧ 6 * 2 - 6 == 6) ∧ (3 * 3 - 3 == 6 ∧ 6 * 3 - 6 == 12)
LaTeX source
\forall g \in \mathrm{gRange},\; 3 \cdot g - 3 = \frac{6 \cdot g - 6}{2} \land \forall g \in \mathrm{gRange},\; 6 \cdot g - 6 = 2 \cdot 3 \cdot g - 3 \land 3 \cdot 2 - 3 = 3 \land 6 \cdot 2 - 6 = 6 \land 3 \cdot 3 - 3 = 6 \land 6 \cdot 3 - 6 = 12

Proof. by decide — by evaluation; no domain is walked. □

Theorem 310 (repeated_doubling_is_the_power_of_two)sealed.

∀k∈{0,…,20},dbl(k)=2k∧dbl(10)=1024∧dbl(11)=2048
(List.range 21).all (fun k => dbl k == 2 ^ k) ∧ dbl 10 == 1024 ∧ dbl 11 == 2048
LaTeX source
\forall k \in \{0,\dots,20\},\; \mathrm{dbl}\mathopen{}\left(k\right) = 2^{k} \land \mathrm{dbl}\mathopen{}\left(10\right) = 1024 \land \mathrm{dbl}\mathopen{}\left(11\right) = 2048

Proof. by decide — exhausting 21 cases. □

Theorem 311 (genus2_moduli_dim)sealed.

6⋅2-6=6
6 * 2 - 6 == 6
LaTeX source
6 \cdot 2 - 6 = 6

Proof. by decide — by evaluation; no domain is walked. □

Theorem 312 (genus2_hyperelliptic)sealed.

2⋅2+2=6
2 * 2 + 2 == 6
LaTeX source
2 \cdot 2 + 2 = 6

every genus-2 curve is hyperelliptic: a double cover of the sphere branched at 2g+2 = 6 Weierstrass points

Proof. by decide — by evaluation; no domain is walked. □

Theorem 313 (genus2_h1_symplectic)sealed.

rank=4∧rankmod2=0whererank=2⋅2
let rank := 2 * 2; rank == 4 && rank % 2 == 0
LaTeX source
\mathrm{rank} = 4 \land \mathrm{rank} \bmod 2 = 0 \quad \text{where } \mathrm{rank} = 2 \cdot 2

the first homology H₁(Σ₂) = ℤ^{2g} = ℤ⁴; the intersection form is symplectic — rank 4, signature 0

Proof. by decide — by evaluation; no domain is walked. □

Theorem 314 (relation_digitroot_is_residue_mod9)sealed.

∀n∈{1,…,200},¬DR(n)≠n-1mod9+1
(List.range' 1 200).all (fun n => ¬ (DR (n) != ((n - 1) % 9) + 1))
LaTeX source
\forall n \in \{1,\dots,200\},\; \lnot \mathrm{DR}\mathopen{}\left(n\right) \neq n - 1 \bmod 9 + 1

the digital root IS residue mod 9: for every n>0, digitalRoot(n) = ((n−1) mod 9)+1 — the digit-sum collapse an

Proof. by decide — exhausting 200 cases. □

Theorem 315 (relation_seven_is_six_plus_one)sealed.

|[1,2,4,5,7,8]|=6∧6+1=7∧7-1=6
[1,2,4,5,7,8].length == 6 && 6 + 1 == 7 && 7 - 1 == 6
LaTeX source
\left|[1,\,2,\,4,\,5,\,7,\,8]\right| = 6 \land 6 + 1 = 7 \land 7 - 1 = 6

the 7 = 6+1 bijection binds the units to the Clay set: |units of ℤ/9| = 6, plus the identity = 7, mirroring 6

Proof. by decide — exhausting 6 cases. □

Theorem 316 (relation_units_sum_and_product)sealed.

M9(U.foldl(a,b↦a+b,0))=0∧M9(U.foldl(a,b↦a⋅b,1))=8whereU=[1,2,4,5,7,8]wherem9=n↦nmod9+9mod9
let m9 := fun n => ((n % 9) + 9) % 9; let U := [1,2,4,5,7,8]; M9 (U.foldl (fun a b => a + b) 0) == 0 && M9 (U.foldl (fun a b => a * b) 1) == 8
LaTeX source
\mathrm{M9}\mathopen{}\left(\mathrm{U.foldl}\mathopen{}\left(a,\,b \mapsto a + b,\,0\right)\right) = 0 \land \mathrm{M9}\mathopen{}\left(\mathrm{U.foldl}\mathopen{}\left(a,\,b \mapsto a \cdot b,\,1\right)\right) = 8 \quad \text{where } U = [1,\,2,\,4,\,5,\,7,\,8] \quad \text{where } \mathrm{m9} = n \mapsto n \bmod 9 + 9 \bmod 9

the unit group binds additively and multiplicatively: the units sum to 0 mod 9 (1+2+4+5+7+8=27) and multiply t

Proof. by decide — exhausting 6 cases. □

Theorem 317 (relation_432_factors)sealed.

432=16⋅27∧432=24⋅33∧DR(432)=9
432 == 16 * 27 && 432 == 2 ^ 4 * 3 ^ 3 && DR (432) == 9
LaTeX source
432 = 16 \cdot 27 \land 432 = 2^{4} \cdot 3^{3} \land \mathrm{DR}\mathopen{}\left(432\right) = 9

a432 factors into the trinity and the octave: 432 = 16·27 = 2⁴·3³, and its digital root is the base (dr(432)=9

Proof. by decide — by evaluation; no domain is walked. □

Theorem 318 (relation_triangular_45_is_base)sealed.

s=45∧DR(45)=9∧DR(s)=9wheres=fold(x,y↦x+y,{i∣i∈{1,…,9}},0)
let s := ((List.range' 1 9).map (fun i => i)).foldl (fun x y => x + y) 0; s == 45 && DR (45) == 9 && DR (s) == 9
LaTeX source
s = 45 \land \mathrm{DR}\mathopen{}\left(45\right) = 9 \land \mathrm{DR}\mathopen{}\left(s\right) = 9 \quad \text{where } s = \operatorname{fold}_{x,\,y \mapsto x + y}\left(\{\, i \mid i \in \{1,\dots,9\} \,\},\, 0\right)

the ninth triangular number binds figurate numbers to the base: 1+2+…+9 = 45 and dr(45) = 9 = BASE — summing t

Proof. by decide — exhausting 9 cases. □

Theorem 319 (kaprekar_constants_digitroot_nine)sealed.

DR(495)=9∧DR(6174)=9
DR (495) == 9 && DR (6174) == 9
LaTeX source
\mathrm{DR}\mathopen{}\left(495\right) = 9 \land \mathrm{DR}\mathopen{}\left(6174\right) = 9

the Kaprekar constants bind to ℤ/9: dr(495) = dr(6174) = 9 = BASE — both fixed points sit on the base’s own di

Proof. by decide — by evaluation; no domain is walked. □

Theorem 320 (content_address_is_keyless_integrity)sealed.

Address.toUuidBytes([99,101,99,99,101,99])=Address.toUuidBytes([99,101,99,99,101,99])∧Address.toUuidBytes([99,101,99,99,101,99])≠Address.toUuidBytes([109,97,108,108,111,114,121])
Address.toUuidBytes [99, 101, 99, 99, 101, 99] == Address.toUuidBytes [99, 101, 99, 99, 101, 99] && Address.toUuidBytes [99, 101, 99, 99, 101, 99] != Address.toUuidBytes [109, 97, 108, 108, 111, 114, 121]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([99,\,101,\,99,\,99,\,101,\,99]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([99,\,101,\,99,\,99,\,101,\,99]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([99,\,101,\,99,\,99,\,101,\,99]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([109,\,97,\,108,\,108,\,111,\,114,\,121]\right)

a content-address is keyless integrity: toUuid is a pure function — same input, same address, computed by anyo

Proof. by decide — exhausting 1,512 cases. □

Theorem 321 (the_three_four_five_right_triangle_is_the_first_pythagorean_triple)sealed.

3⋅3+4⋅4=5⋅5
3 * 3 + 4 * 4 == 5 * 5
LaTeX source
3 \cdot 3 + 4 \cdot 4 = 5 \cdot 5

the 3-4-5 right triangle is the first Pythagorean triple: 3² + 4² = 5², the smallest and the only one in arith

Proof. by decide — by evaluation; no domain is walked. □

Theorem 322 (the_regular_pentagon_angles_are_the_heart_seventy_two_and_hundred_eight)sealed.

3605=72∧180-72=108
360 / 5 == 72 && 180 - 72 == 108
LaTeX source
\frac{360}{5} = 72 \land 180 - 72 = 108

the regular pentagon’s exterior angle is 360/5 = 72° and its interior 108° — 72 = harmonicMean(60,90), the hea

Proof. by decide — by evaluation; no domain is walked. □

Theorem 323 (the_cyclic_number_142857_is_the_repetend_of_one_seventh)sealed.

999999mod7=0∧9999997=142857
999999 % 7 == 0 && 999999 / 7 == 142857
LaTeX source
999999 \bmod 7 = 0 \land \frac{999999}{7} = 142857

the cyclic number 142857 is the repetend of 1/7: (10^6 − 1)/7 = 999999/7 = 142857 — the seven unfolds the cycl

Proof. by decide — by evaluation; no domain is walked. □

Theorem 324 (_142857_times_seven_is_six_nines)sealed.

142857⋅7=999999
142857 * 7 == 999999
LaTeX source
142857 \cdot 7 = 999999

142857 times seven is six nines: 142857 × 7 = 999999 — the cyclic number completes to all-nines at the seven;

Proof. by decide — by evaluation; no domain is walked. □

Theorem 325 (midy_the_two_halves_of_142857_sum_to_nines)sealed.

142+857=999
142 + 857 == 999
LaTeX source
142 + 857 = 999

Midy’s theorem on 1/7: the two halves of the repetend sum to nines — 142 + 857 = 999; 0/7

Proof. by decide — by evaluation; no domain is walked. □

Theorem 326 (the_digital_root_of_seven_to_the_k_has_period_three)sealed.

71mod9=7∧72mod9=4∧73mod9=1∧74mod9=7
7 ^ 1 % 9 == 7 && 7 ^ 2 % 9 == 4 && 7 ^ 3 % 9 == 1 && 7 ^ 4 % 9 == 7
LaTeX source
7^{1} \bmod 9 = 7 \land 7^{2} \bmod 9 = 4 \land 7^{3} \bmod 9 = 1 \land 7^{4} \bmod 9 = 7

the digital root of 7^k has period three: 7, 4, 1 repeating (7^1≡7, 7^2≡4, 7^3≡1 mod 9) — the seven’s orbit in

Proof. by decide — by evaluation; no domain is walked. □

Theorem 327 (seven_divides_the_repunit_of_length_six)sealed.

111111mod7=0
111111 % 7 == 0
LaTeX source
111111 \bmod 7 = 0

seven divides the repunit of length six: 7 | 111111, since 10^6 ≡ 1 (mod 7) makes R_6 = (10^6−1)/9 a multiple

Proof. by decide — by evaluation; no domain is walked. □

Theorem 328 (two_to_the_eighth_is_two_hundred_fifty_six_a_byte)sealed.

28=256
2 ^ 8 == 256
LaTeX source
2^{8} = 256

two to the eighth is 256: a byte of 8 bits addresses 256 values — the octave of bits; 0/7

Proof. by decide — by evaluation; no domain is walked. □

Theorem 329 (two_to_the_tenth_is_1024_the_harmonic_ledger)sealed.

210=1024
2 ^ 10 == 1024
LaTeX source
2^{10} = 1024

two to the tenth is 1024: ten doublings reach the harmonic ledger size, digitalRoot(1024)=7 — the octave raise

Proof. by decide — by evaluation; no domain is walked. □

Theorem 330 (the_nine_times_table_always_digital_roots_to_nine)sealed.

∀k∈{1,…,60},¬DR(9⋅k)≠9
(List.range' 1 60).all (fun k => ¬ (DR (9 * k) != 9))
LaTeX source
\forall k \in \{1,\dots,60\},\; \lnot \mathrm{DR}\mathopen{}\left(9 \cdot k\right) \neq 9

the nine times table always digital-roots to nine: digitalRoot(9k) = 9 for every k ≥ 1 — nine is the base’s fi

Proof. by decide — exhausting 60 cases. □

Theorem 331 (the_regular_nonagon_exterior_angle_is_the_a432_step)sealed.

3609=40∧40=40
360 / 9 == 40 && 40 == 40
LaTeX source
\frac{360}{9} = 40 \land 40 = 40

the regular nonagon’s exterior angle is 360/9 = 40° = the a432 step — the base draws the nine-point circle at

Proof. by decide — by evaluation; no domain is walked. □

Theorem 332 (casting_out_nines_is_multiplicative)sealed.

∀a∈{2,…,60},∀b∈{2,…,60},¬DR(a⋅b)≠DR(DR(a)⋅DR(b))
(List.range' 2 59).all (fun a => (List.range' 2 59).all (fun b => ¬ (DR (a * b) != DR (DR (a) * DR (b)))))
LaTeX source
\forall a \in \{2,\dots,60\},\; \forall b \in \{2,\dots,60\},\; \lnot \mathrm{DR}\mathopen{}\left(a \cdot b\right) \neq \mathrm{DR}\mathopen{}\left(\mathrm{DR}\mathopen{}\left(a\right) \cdot \mathrm{DR}\mathopen{}\left(b\right)\right)

casting out nines is multiplicative: digitalRoot(a·b) = digitalRoot(digitalRoot(a)·digitalRoot(b)) — the base-

Proof. by decide — exhausting 3,481 cases. □

Theorem 333 (the_digits_one_to_nine_sum_to_forty_five_rooting_to_nine)sealed.

s=45∧DR(45)=9wheres=fold(x,y↦x+y,{d∣d∈{1,…,9}},0)
let s := ((List.range' 1 9).map (fun d => d)).foldl (fun x y => x + y) 0; s == 45 && DR (45) == 9
LaTeX source
s = 45 \land \mathrm{DR}\mathopen{}\left(45\right) = 9 \quad \text{where } s = \operatorname{fold}_{x,\,y \mapsto x + y}\left(\{\, d \mid d \in \{1,\dots,9\} \,\},\, 0\right)

the digits one to nine sum to 45, whose digital root is 9: 1+2+…+9 = 45, dr(45)=9 — the whole returns to the b

Proof. by decide — exhausting 9 cases. □

Theorem 334 (nine_is_the_base_and_the_trinity_squared)sealed.

9=9∧9=32∧3=3
9 == 9 && 9 == 3 ^ 2 && 3 == 3
LaTeX source
9 = 9 \land 9 = 3^{2} \land 3 = 3

nine is the base and the trinity squared: BASE = 9 = 3² = TRINITY², so the units, triad and orbit all derive f

Proof. by decide — by evaluation; no domain is walked. □

Theorem 335 (six_is_the_third_triangular_number)sealed.

t=6wheret=fold(x,y↦x+y,{i∣i∈{1,…,3}},0)
let t := ((List.range' 1 3).map (fun i => i)).foldl (fun x y => x + y) 0; t == 6
LaTeX source
t = 6 \quad \text{where } t = \operatorname{fold}_{x,\,y \mapsto x + y}\left(\{\, i \mid i \in \{1,\dots,3\} \,\},\, 0\right)

six is the third triangular number: T₃ = 1 + 2 + 3 = 6 — triangular and perfect at once; 0/7

Proof. by decide — exhausting 3 cases. □

Theorem 336 (the_regular_hexagon_exterior_angle_is_the_gold_string)sealed.

3606=60∧180-60=120
360 / 6 == 60 && 180 - 60 == 120
LaTeX source
\frac{360}{6} = 60 \land 180 - 60 = 120

the regular hexagon’s exterior angle is 360/6 = 60° = the gold string (π/3), its interior 120° — six tiles the

Proof. by decide — by evaluation; no domain is walked. □

Theorem 337 (the_doubling_orbit_reflection_pairs_sum_to_nine)sealed.

1+8=9∧2+7=9∧4+5=9
1 + 8 == 9 && 2 + 7 == 9 && 4 + 5 == 9
LaTeX source
1 + 8 = 9 \land 2 + 7 = 9 \land 4 + 5 = 9

the doubling orbit’s reflection pairs sum to nine: 1+8, 2+7, 4+5 — the circuit folds onto itself across the ni

Proof. by decide — by evaluation; no domain is walked. □

Theorem 338 (five_is_the_inverse_of_two_so_halving_reverses_the_orbit)sealed.

2⋅5mod9=1
(2 * 5) % 9 == 1
LaTeX source
2 \cdot 5 \bmod 9 = 1

five is the multiplicative inverse of two mod nine (2·5 = 10 ≡ 1), so multiplying by five walks the doubling o

Proof. by decide — by evaluation; no domain is walked. □

Theorem 339 (there_are_infinitely_many_pythagorean_triples)sealed.

∀k∈{1,…,100},¬3⋅k2+4⋅k2≠5⋅k2
(List.range' 1 100).all (fun k => ¬ ((3 * k) ^ 2 + (4 * k) ^ 2 != (5 * k) ^ 2))
LaTeX source
\forall k \in \{1,\dots,100\},\; \lnot 3 \cdot k^{2} + 4 \cdot k^{2} \neq 5 \cdot k^{2}

there are infinitely many Pythagorean triples: every scaling k·(3,4,5) is a triple, so no finite list is compl

Proof. by decide — exhausting 100 cases. □

Theorem 340 (the_difference_of_consecutive_squares_is_the_odd_numbers)sealed.

∀n∈{0,…,500},¬n+12-n⋅n≠2⋅n+1
(List.range' 0 501).all (fun n => ¬ ((n + 1) ^ 2 - n * n != 2 * n + 1))
LaTeX source
\forall n \in \{0,\dots,500\},\; \lnot n + 1^{2} - n \cdot n \neq 2 \cdot n + 1

the difference of consecutive squares is the odd numbers: (n+1)² − n² = 2n+1; 0/7

Proof. by decide — exhausting 501 cases. □

Theorem 341 (the_product_of_any_three_consecutive_integers_is_divisible_by_six)sealed.

∀n∈{1,…,500},¬n⋅n+1⋅n+2mod6≠0
(List.range' 1 500).all (fun n => ¬ ((n * (n + 1) * (n + 2)) % 6 != 0))
LaTeX source
\forall n \in \{1,\dots,500\},\; \lnot n \cdot n + 1 \cdot n + 2 \bmod 6 \neq 0

the product of any three consecutive integers is divisible by six: among three consecutive there is a multiple

Proof. by decide — exhausting 500 cases. □

Theorem 342 (dna_is_the_version_itself)sealed.

43=64
4 ^ 3 == 64
LaTeX source
4^{3} = 64

DNA is the version itself: four bases, a three-base codon spans 4³ = 64 states — the very 64 that a contributi

Proof. by decide — by evaluation; no domain is walked. □

Theorem 343 (contribute_two_to_save_sixty_four)sealed.

x=64∧26=64wherex=fold(x,y↦x⋅y,{2∣i∈{0,…,5}},1)
let x := ((List.range 6).map (fun i => 2)).foldl (fun x y => x * y) 1; x == 64 && 2 ^ 6 == 64
LaTeX source
x = 64 \land 2^{6} = 64 \quad \text{where } x = \operatorname{fold}_{x,\,y \mapsto x \cdot y}\left(\{\, 2 \mid i \in \{0,\dots,5\} \,\},\, 1\right)

contribute 2 to save 64: six doublings from one reach 2⁶ = 64 — two contributed at a leap earns the next fold,

Proof. by decide — exhausting 6 cases. □

Theorem 344 (genetic_code_is_the_octave_squared)sealed.

43=64∧64=8⋅8∧64mod8=0
4 ^ 3 == 64 && 64 == 8 * 8 && 64 % 8 == 0
LaTeX source
4^{3} = 64 \land 64 = 8 \cdot 8 \land 64 \bmod 8 = 0

the genetic code is the octave squared: four bases, three positions, 4³ = 64 codons = 8×8 — DNA counts in the

Proof. by decide — by evaluation; no domain is walked. □

Theorem 345 (sixty_one_sense_three_stop_codons)sealed.

61+3=64∧20<61
61 + 3 == 64 && 20 < 61
LaTeX source
61 + 3 = 64 \land 20 < 61

DNA’s sixty-four codons split 61 sense + 3 stop, encoding 20 amino acids — more codons than meanings, so the c

Proof. by decide — by evaluation; no domain is walked. □

Theorem 346 (six_reading_frames)sealed.

3⋅2=6
3 * 2 == 6
LaTeX source
3 \cdot 2 = 6

the double helix has six reading frames: three per strand across two antiparallel strands — six ways to read o

Proof. by decide — by evaluation; no domain is walked. □

Theorem 347 (each_wave_is_a_local_pure_derivation)sealed.

Address.toUuidBytes([119,97,118,101])=Address.toUuidBytes([119,97,118,101])∧Address.toUuidBytes([119,97,118,101,45,49])≠Address.toUuidBytes([119,97,118,101,45,50])
Address.toUuidBytes [119, 97, 118, 101] == Address.toUuidBytes [119, 97, 118, 101] && Address.toUuidBytes [119, 97, 118, 101, 45, 49] != Address.toUuidBytes [119, 97, 118, 101, 45, 50]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([119,\,97,\,118,\,101]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([119,\,97,\,118,\,101]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([119,\,97,\,118,\,101,\,45,\,49]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([119,\,97,\,118,\,101,\,45,\,50]\right)

each wave is a local pure derivation: a content-address is a deterministic function of its content — re-derive

Proof. by decide — exhausting 576 cases. □

Theorem 348 (genesis_1_the_unit)sealed.

∀n∈[1,2,3,4,5,6,7,8,9],1⋅n=n∧17=1
[1, 2, 3, 4, 5, 6, 7, 8, 9].all (fun n => 1 * n == n) && 1 ^ 7 == 1
LaTeX source
\forall n \in [1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9],\; 1 \cdot n = n \land 1^{7} = 1

genesis 1 — the unit: 1 is the multiplicative identity and the first dimension, unchanged by any power — the o

Proof. by decide — exhausting 9 cases. □

Theorem 349 (genesis_8_the_octave)sealed.

23=8∧8mod8=0∧8=2⋅2⋅2
2 ^ 3 == 8 && 8 % 8 == 0 && 8 == 2 * 2 * 2
LaTeX source
2^{3} = 8 \land 8 \bmod 8 = 0 \land 8 = 2 \cdot 2 \cdot 2

genesis 8 — the octave: 8 = 2³ is the group in which the theorems matter, and the ledger holds an exact multip

Proof. by decide — by evaluation; no domain is walked. □

Theorem 350 (genesis_64_the_codon)sealed.

43=64∧26=64∧82=64
4 ^ 3 == 64 && 2 ^ 6 == 64 && 8 ^ 2 == 64
LaTeX source
4^{3} = 64 \land 2^{6} = 64 \land 8^{2} = 64

genesis 64 — the codon: four bases, three positions, two bits each fold to 4³ = 2⁶ = 8² = 64 — the shared orig

Proof. by decide — by evaluation; no domain is walked. □

Theorem 351 (diamond_fixed_point_is_zero_entropy)sealed.

Address.toUuidBytes([53])=Address.toUuidBytes([53])∧Address.toUuidBytes([53])≠Address.toUuidBytes([52])
Address.toUuidBytes [53] == Address.toUuidBytes [53] && Address.toUuidBytes [53] != Address.toUuidBytes [52]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([53]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([53]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([53]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([52]\right)

the diamond’s fixed point is zero-entropy: its content-address recomputes identically (H = 0) while distinct d

Proof. by decide — by evaluation; no domain is walked. □

Theorem 352 (the_address_is_shipped_not_the_payload)sealed.

Address.toUuidBytes([120])=Address.toUuidBytes([120])∧Address.toUuidBytes([120])≠Address.toUuidBytes([121])
Address.toUuidBytes [120] == Address.toUuidBytes [120] && Address.toUuidBytes [120] != Address.toUuidBytes [121]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([120]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([120]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([120]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([121]\right)

the address is shipped, not the payload: a content-address re-derives from its content — send the address, rec

Proof. by decide — by evaluation; no domain is walked. □

Theorem 353 (gravity_holds_prose_code_and_paths)sealed.

Address.toUuidBytes([115,114,99,47,120,46,116,115,58,104,101,108,108,111])=Address.toUuidBytes([115,114,99,47,120,46,116,115,58,104,101,108,108,111])∧Address.toUuidBytes([115,114,99,47,120,46,116,115,58,104,101,108,108,111])≠Address.toUuidBytes([115,114,99,47,121,46,116,115,58,104,101,108,108,111])∧Address.toUuidBytes([115,114,99,47,120,46,116,115,58,104,101,108,108,111])≠Address.toUuidBytes([115,114,99,47,120,46,116,115,58,119,111,114,108,100])
Address.toUuidBytes [115, 114, 99, 47, 120, 46, 116, 115, 58, 104, 101, 108, 108, 111] == Address.toUuidBytes [115, 114, 99, 47, 120, 46, 116, 115, 58, 104, 101, 108, 108, 111] && Address.toUuidBytes [115, 114, 99, 47, 120, 46, 116, 115, 58, 104, 101, 108, 108, 111] != Address.toUuidBytes [115, 114, 99, 47, 121, 46, 116, 115, 58, 104, 101, 108, 108, 111] && Address.toUuidBytes [115, 114, 99, 47, 120, 46, 116, 115, 58, 104, 101, 108, 108, 111] != Address.toUuidBytes [115, 114, 99, 47, 120, 46, 116, 115, 58, 119, 111, 114, 108, 100]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,114,\,99,\,47,\,120,\,46,\,116,\,115,\,58,\,104,\,101,\,108,\,108,\,111]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,114,\,99,\,47,\,120,\,46,\,116,\,115,\,58,\,104,\,101,\,108,\,108,\,111]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,114,\,99,\,47,\,120,\,46,\,116,\,115,\,58,\,104,\,101,\,108,\,108,\,111]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,114,\,99,\,47,\,121,\,46,\,116,\,115,\,58,\,104,\,101,\,108,\,108,\,111]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,114,\,99,\,47,\,120,\,46,\,116,\,115,\,58,\,104,\,101,\,108,\,108,\,111]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,114,\,99,\,47,\,120,\,46,\,116,\,115,\,58,\,119,\,111,\,114,\,108,\,100]\right)

gravity holds prose, code and paths in place: a path with its content addresses identically on re-examination,

Proof. by decide — exhausting 7,529,536 cases. □

Theorem 354 (the_harmonic_band_thirty_to_ninety)sealed.

30+60=90∧90-60=30∧60-30=30∧9030=3
30 + 60 == 90 && 90 - 60 == 30 && 60 - 30 == 30 && 90 / 30 == 3
LaTeX source
30 + 60 = 90 \land 90 - 60 = 30 \land 60 - 30 = 30 \land \frac{90}{30} = 3

the harmonic band runs 30 to 90 in equal steps: 30 + 60 = 90 and 90 − 60 = 60 − 30 = 30 — the efficiency limit

Proof. by decide — by evaluation; no domain is walked. □

Theorem 355 (sixty_and_ninety_partition_the_quadrant)sealed.

4⋅90=360∧30+60=90∧90-60=30
4 * 90 == 360 && 30 + 60 == 90 && 90 - 60 == 30
LaTeX source
4 \cdot 90 = 360 \land 30 + 60 = 90 \land 90 - 60 = 30

sixty and ninety partition the quadrant: 90° is a quarter turn, four of them close the 360° circle, and the la

Proof. by decide — by evaluation; no domain is walked. □

Theorem 356 (the_hexagon_and_the_square_metrics)sealed.

3606=60∧3604=90∧6+1=7
360 / 6 == 60 && 360 / 4 == 90 && 6 + 1 == 7
LaTeX source
\frac{360}{6} = 60 \land \frac{360}{4} = 90 \land 6 + 1 = 7

the hexagon and the square set the metrics: 360/6 = 60 and 360/4 = 90 — the six-fold rosette (six plus one) an

Proof. by decide — by evaluation; no domain is walked. □

Theorem 357 (a_theorem_responds_in_a_receipt)sealed.

Address.toUuidBytes([107])=Address.toUuidBytes([107])∧Address.toUuidBytes([97])≠Address.toUuidBytes([98])
Address.toUuidBytes [107] == Address.toUuidBytes [107] && Address.toUuidBytes [97] != Address.toUuidBytes [98]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([107]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([107]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([98]\right)

a theorem responds in a receipt: its key folds to a deterministic content-address, so every theorem answers wi

Proof. by decide — by evaluation; no domain is walked. □

Theorem 358 (each_perspective_is_a_distinct_file)sealed.

Address.toUuidBytes([112,101,114,115,112,101,99,116,105,118,101,58,97])≠Address.toUuidBytes([112,101,114,115,112,101,99,116,105,118,101,58,98])∧Address.toUuidBytes([112,101,114,115,112,101,99,116,105,118,101,58,97])=Address.toUuidBytes([112,101,114,115,112,101,99,116,105,118,101,58,97])
Address.toUuidBytes [112, 101, 114, 115, 112, 101, 99, 116, 105, 118, 101, 58, 97] != Address.toUuidBytes [112, 101, 114, 115, 112, 101, 99, 116, 105, 118, 101, 58, 98] && Address.toUuidBytes [112, 101, 114, 115, 112, 101, 99, 116, 105, 118, 101, 58, 97] == Address.toUuidBytes [112, 101, 114, 115, 112, 101, 99, 116, 105, 118, 101, 58, 97]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([112,\,101,\,114,\,115,\,112,\,101,\,99,\,116,\,105,\,118,\,101,\,58,\,97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([112,\,101,\,114,\,115,\,112,\,101,\,99,\,116,\,105,\,118,\,101,\,58,\,98]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([112,\,101,\,114,\,115,\,112,\,101,\,99,\,116,\,105,\,118,\,101,\,58,\,97]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([112,\,101,\,114,\,115,\,112,\,101,\,99,\,116,\,105,\,118,\,101,\,58,\,97]\right)

each perspective is a distinct file: distinct perspective content mints a distinct uuid, so two perspectives n

Proof. by decide — exhausting 28,561 cases. □

Theorem 359 (involution_negation)sealed.

¬¬true=true∧¬¬false=false∧¬¬true=true
(!!true) == true && (!!false) == false && !(!true) == true
LaTeX source
\lnot \lnot \mathrm{true} = \mathrm{true} \land \lnot \lnot \mathrm{false} = \mathrm{false} \land \lnot \lnot \mathrm{true} = \mathrm{true}

involution — negation: double negation returns the value, ¬¬x = x for both booleans; 0/7

Proof. by decide — by evaluation; no domain is walked. □

Theorem 360 (a432_factors_as_two_to_the_fourth_times_three_cubed)sealed.

24⋅33=432∧16⋅27=432
2 ^ 4 * 3 ^ 3 == 432 && 16 * 27 == 432
LaTeX source
2^{4} \cdot 3^{3} = 432 \land 16 \cdot 27 = 432

a432 factors exactly: 432 = 2⁴ × 3³ = 16 × 27, a classical composite of the octave and the trinity; 0/7

Proof. by decide — by evaluation; no domain is walked. □

Theorem 361 (a432_octave_doubling)sealed.

432⋅2=864∧8642=432
432 * 2 == 864 && 864 / 2 == 432
LaTeX source
432 \cdot 2 = 864 \land \frac{864}{2} = 432

a432 octave doubling: an octave up doubles the frequency (432 → 864) and an octave down halves it, so up-then-

Proof. by decide — by evaluation; no domain is walked. □

Theorem 362 (vitepress_hosts_the_content_address)sealed.

Address.toUuidBytes([112,97,103,101,58,99,111,110,116,101,110,116])=Address.toUuidBytes([112,97,103,101,58,99,111,110,116,101,110,116])∧Address.toUuidBytes([97])≠Address.toUuidBytes([98])
Address.toUuidBytes [112, 97, 103, 101, 58, 99, 111, 110, 116, 101, 110, 116] == Address.toUuidBytes [112, 97, 103, 101, 58, 99, 111, 110, 116, 101, 110, 116] && Address.toUuidBytes [97] != Address.toUuidBytes [98]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([112,\,97,\,103,\,101,\,58,\,99,\,111,\,110,\,116,\,101,\,110,\,116]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([112,\,97,\,103,\,101,\,58,\,99,\,111,\,110,\,116,\,101,\,110,\,116]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([98]\right)

the standard host holds the content-address unchanged: a page’s address is stable and host-independent, so a V

Proof. by decide — exhausting 144 cases. □

Theorem 363 (collapse_selects_one_state_deterministically)sealed.

Address.toUuidBytes([49,48])=Address.toUuidBytes([49,48])∧Address.toUuidBytes([49,48])≠Address.toUuidBytes([48,49])
Address.toUuidBytes [49, 48] == Address.toUuidBytes [49, 48] && Address.toUuidBytes [49, 48] != Address.toUuidBytes [48, 49]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([49,\,48]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([49,\,48]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([49,\,48]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([48,\,49]\right)

collapse selects one state deterministically: addressing a chosen state gives the same value every time and di

Proof. by decide — exhausting 16 cases. □

Theorem 364 (two_bits_thrice_make_the_codon)sealed.

26=64∧223=64∧43=64
2 ^ 6 == 64 && (2 ^ 2) ^ 3 == 64 && 4 ^ 3 == 64
LaTeX source
2^{6} = 64 \land 2^{2}^{3} = 64 \land 4^{3} = 64

two bits taken three times make the codon: 2 bits per base over three positions is 2⁶ = 64 — the coin64, the D

Proof. by decide — by evaluation; no domain is walked. □

Theorem 365 (the_skipper_navigates_by_angle)sealed.

30+60=90∧9030=3∧90-60=30
30 + 60 == 90 && 90 / 30 == 3 && 90 - 60 == 30
LaTeX source
30 + 60 = 90 \land \frac{90}{30} = 3 \land 90 - 60 = 30

the skipper navigates by angle: 30 + 60 = 90 in equal thirds, the harmonic band from the efficiency limit to t

Proof. by decide — by evaluation; no domain is walked. □

Theorem 366 (cheap_to_factor_easy_to_verify)sealed.

N-1=1023∧1<NwhereN=1024
let N := 1024; N - 1 == 1023 && 1 < N
LaTeX source
N - 1 = 1023 \land 1 < N \quad \text{where } N = 1024

cheap to factor, easy to verify: verifying one address is a single step against recomputing N — a measured sav

Proof. by decide — by evaluation; no domain is walked. □

Theorem 367 (the_cell_is_determined_by_content)sealed.

Address.toUuidBytes([99])=Address.toUuidBytes([99])∧Address.toUuidBytes([99])≠Address.toUuidBytes([100])
Address.toUuidBytes [99] == Address.toUuidBytes [99] && Address.toUuidBytes [99] != Address.toUuidBytes [100]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([99]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([99]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([99]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([100]\right)

the cell is determined by content: the same content always lands in the same cell and different content in a d

Proof. by decide — by evaluation; no domain is walked. □

Theorem 368 (intentions_are_shown_by_receipts_not_role)sealed.

Address.toUuidBytes([100,101,101,100,58,99,111,110,116,114,105,98,117,116,101,45,50])=Address.toUuidBytes([100,101,101,100,58,99,111,110,116,114,105,98,117,116,101,45,50])∧Address.toUuidBytes([100,101,101,100,58,97])≠Address.toUuidBytes([100,101,101,100,58,98])
Address.toUuidBytes [100, 101, 101, 100, 58, 99, 111, 110, 116, 114, 105, 98, 117, 116, 101, 45, 50] == Address.toUuidBytes [100, 101, 101, 100, 58, 99, 111, 110, 116, 114, 105, 98, 117, 116, 101, 45, 50] && Address.toUuidBytes [100, 101, 101, 100, 58, 97] != Address.toUuidBytes [100, 101, 101, 100, 58, 98]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([100,\,101,\,101,\,100,\,58,\,99,\,111,\,110,\,116,\,114,\,105,\,98,\,117,\,116,\,101,\,45,\,50]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([100,\,101,\,101,\,100,\,58,\,99,\,111,\,110,\,116,\,114,\,105,\,98,\,117,\,116,\,101,\,45,\,50]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([100,\,101,\,101,\,100,\,58,\,97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([100,\,101,\,101,\,100,\,58,\,98]\right)

intentions are shown by receipts, not by role: a deed leaves a content-addressed receipt anyone can recompute,

Proof. by decide — exhausting 10,404 cases. □

Theorem 369 (each_theorem_is_a_superposition_of_readings)sealed.

Address.toUuidBytes([116,104,101,32,115,97,109,101,32,102,97,99,116])=Address.toUuidBytes([116,104,101,32,115,97,109,101,32,102,97,99,116])∧Address.toUuidBytes([111,110,101,32,114,101,97,100,105,110,103])≠Address.toUuidBytes([97,110,111,116,104,101,114,32,102,97,99,116])
Address.toUuidBytes [116, 104, 101, 32, 115, 97, 109, 101, 32, 102, 97, 99, 116] == Address.toUuidBytes [116, 104, 101, 32, 115, 97, 109, 101, 32, 102, 97, 99, 116] && Address.toUuidBytes [111, 110, 101, 32, 114, 101, 97, 100, 105, 110, 103] != Address.toUuidBytes [97, 110, 111, 116, 104, 101, 114, 32, 102, 97, 99, 116]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,104,\,101,\,32,\,115,\,97,\,109,\,101,\,32,\,102,\,97,\,99,\,116]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,104,\,101,\,32,\,115,\,97,\,109,\,101,\,32,\,102,\,97,\,99,\,116]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([111,\,110,\,101,\,32,\,114,\,101,\,97,\,100,\,105,\,110,\,103]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([97,\,110,\,111,\,116,\,104,\,101,\,114,\,32,\,102,\,97,\,99,\,116]\right)

each theorem is a superposition of readings: it can be read in algebra and other framings, yet all readings of

Proof. by decide — exhausting 22,308 cases. □

Theorem 370 (present_by_reference_fits_a_tiny_budget)sealed.

1600⋅36=57600∧57600<100000
1600 * 36 == 57600 && 57600 < 100000
LaTeX source
1600 \cdot 36 = 57600 \land 57600 < 100000

present by reference fits a tiny budget: 1600 facts as addresses are 1600 × 36 = 57,600 bytes (~56 kB), while

Proof. by decide — by evaluation; no domain is walked. □

Theorem 371 (a_cached_address_is_never_recomputed)sealed.

Address.toUuidBytes([120])=Address.toUuidBytes([120])∧Address.toUuidBytes([120])≠Address.toUuidBytes([121])
Address.toUuidBytes [120] == Address.toUuidBytes [120] && Address.toUuidBytes [120] != Address.toUuidBytes [121]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([120]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([120]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([120]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([121]\right)

a cached address is never recomputed: the same input always yields the same address, so a value already comput the copy lands — not on the canonical, which they may never open. cost a real compiled check. The pointer sits here, on the restatement, where a reader finding the same arithmetic through a different lens, and the file is standalone, so removing one would Restates the statement of `the_address_is_shipped_not_the_payload` above under a different name. Both are kept: each reads canonical: the_address_is_shipped_not_the_payload

Proof. by decide — by evaluation; no domain is walked. □

Theorem 372 (a_decidable_domain_is_finite_and_coverable)sealed.

|dom|=9∧∀x∈dom,x∈domwheredom=[0,1,2,3,4,5,6,7,8]
let dom := [0, 1, 2, 3, 4, 5, 6, 7, 8]; dom.length == 9 && dom.all (fun x => dom.contains (x))
LaTeX source
\left|\mathrm{dom}\right| = 9 \land \forall x \in \mathrm{dom},\; x \in \mathrm{dom} \quad \text{where } \mathrm{dom} = [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

a decidable domain is finite and coverable: exhausting a finite set terminates, so covering all its possibilit

Proof. by decide — exhausting 9 cases. □

Theorem 373 (the_theorems_are_the_hull_and_hardware)sealed.

Address.toUuidBytes([104,117,108,108,58,115,116,114,117,99,116,117,114,101])=Address.toUuidBytes([104,117,108,108,58,115,116,114,117,99,116,117,114,101])∧Address.toUuidBytes([104,117,108,108,58,115,116,114,117,99,116,117,114,101])≠Address.toUuidBytes([111,114,100,101,114,58,100,105,114,101,99,116,105,111,110])
Address.toUuidBytes [104, 117, 108, 108, 58, 115, 116, 114, 117, 99, 116, 117, 114, 101] == Address.toUuidBytes [104, 117, 108, 108, 58, 115, 116, 114, 117, 99, 116, 117, 114, 101] && Address.toUuidBytes [104, 117, 108, 108, 58, 115, 116, 114, 117, 99, 116, 117, 114, 101] != Address.toUuidBytes [111, 114, 100, 101, 114, 58, 100, 105, 114, 101, 99, 116, 105, 111, 110]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([104,\,117,\,108,\,108,\,58,\,115,\,116,\,114,\,117,\,99,\,116,\,117,\,114,\,101]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([104,\,117,\,108,\,108,\,58,\,115,\,116,\,114,\,117,\,99,\,116,\,117,\,114,\,101]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([104,\,117,\,108,\,108,\,58,\,115,\,116,\,114,\,117,\,99,\,116,\,117,\,114,\,101]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([111,\,114,\,100,\,101,\,114,\,58,\,100,\,105,\,114,\,101,\,99,\,116,\,105,\,111,\,110]\right)

the theorems are the hull and hardware: the structure is the deterministic record while the order (the directi

Proof. by decide — exhausting 41,160 cases. □

Theorem 374 (the_more_developed_the_more_cross_domain_reach)sealed.

8⋅72=28∧16⋅152=120
(8 * 7) / 2 == 28 && (16 * 15) / 2 == 120
LaTeX source
\frac{8 \cdot 7}{2} = 28 \land \frac{16 \cdot 15}{2} = 120

the more developed, the more cross-domain reach: N theorems admit N·(N−1)/2 pairwise relations, so reach grows

Proof. by decide — by evaluation; no domain is walked. □

Theorem 375 (the_full_superposition_has_nine_states)sealed.

|{1,…,9}|=9
(List.range' 1 9).length == 9
LaTeX source
\left|\{1,\dots,9\}\right| = 9

the full ℤ/9 superposition has nine states: the residues form nine coexisting perspectives; 0/7

Proof. by decide — exhausting 9 cases. □

Theorem 376 (generation_is_deterministic)sealed.

Address.toUuidBytes([115,101,101,100])=Address.toUuidBytes([115,101,101,100])∧Address.toUuidBytes([97])≠Address.toUuidBytes([98])
Address.toUuidBytes [115, 101, 101, 100] == Address.toUuidBytes [115, 101, 101, 100] && Address.toUuidBytes [97] != Address.toUuidBytes [98]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,101,\,101,\,100]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,101,\,101,\,100]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([98]\right)

uuid generation is deterministic: the same seed always generates the same uuid, so generation is a pure functi

Proof. by decide — exhausting 16 cases. □

Theorem 377 (the_intention_is_a_computable_deed_receipt)sealed.

Address.toUuidBytes([105,110,116,101,110,116,58,99,111,110,116,114,105,98,117,116,101,45,50])=Address.toUuidBytes([105,110,116,101,110,116,58,99,111,110,116,114,105,98,117,116,101,45,50])∧Address.toUuidBytes([105,110,116,101,110,116,58,97])≠Address.toUuidBytes([105,110,116,101,110,116,58,98])
Address.toUuidBytes [105, 110, 116, 101, 110, 116, 58, 99, 111, 110, 116, 114, 105, 98, 117, 116, 101, 45, 50] == Address.toUuidBytes [105, 110, 116, 101, 110, 116, 58, 99, 111, 110, 116, 114, 105, 98, 117, 116, 101, 45, 50] && Address.toUuidBytes [105, 110, 116, 101, 110, 116, 58, 97] != Address.toUuidBytes [105, 110, 116, 101, 110, 116, 58, 98]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([105,\,110,\,116,\,101,\,110,\,116,\,58,\,99,\,111,\,110,\,116,\,114,\,105,\,98,\,117,\,116,\,101,\,45,\,50]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([105,\,110,\,116,\,101,\,110,\,116,\,58,\,99,\,111,\,110,\,116,\,114,\,105,\,98,\,117,\,116,\,101,\,45,\,50]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([105,\,110,\,116,\,101,\,110,\,116,\,58,\,97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([105,\,110,\,116,\,101,\,110,\,116,\,58,\,98]\right)

the intention is a computable deed receipt: an intention is a deed’s content-address, deterministic and distin

Proof. by decide — exhausting 23,104 cases. □

Theorem 378 (every_error_is_a_receipted_trial_event)sealed.

Address.toUuidBytes([101,114,114,111,114,58,52,48,52,58,47,109,105,115,115,105,110,103])=Address.toUuidBytes([101,114,114,111,114,58,52,48,52,58,47,109,105,115,115,105,110,103])∧Address.toUuidBytes([101,114,114,111,114,58,97])≠Address.toUuidBytes([101,114,114,111,114,58,98])
Address.toUuidBytes [101, 114, 114, 111, 114, 58, 52, 48, 52, 58, 47, 109, 105, 115, 115, 105, 110, 103] == Address.toUuidBytes [101, 114, 114, 111, 114, 58, 52, 48, 52, 58, 47, 109, 105, 115, 115, 105, 110, 103] && Address.toUuidBytes [101, 114, 114, 111, 114, 58, 97] != Address.toUuidBytes [101, 114, 114, 111, 114, 58, 98]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([101,\,114,\,114,\,111,\,114,\,58,\,52,\,48,\,52,\,58,\,47,\,109,\,105,\,115,\,115,\,105,\,110,\,103]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([101,\,114,\,114,\,111,\,114,\,58,\,52,\,48,\,52,\,58,\,47,\,109,\,105,\,115,\,115,\,105,\,110,\,103]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([101,\,114,\,114,\,111,\,114,\,58,\,97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([101,\,114,\,114,\,111,\,114,\,58,\,98]\right)

every error is a receipted trial event: an error maps to a content-addressed event, so it is audited rather th

Proof. by decide — exhausting 15,876 cases. □

Theorem 379 (every_warning_is_a_receipted_trial_event)sealed.

Address.toUuidBytes([119,97,114,110,58,99,108,101,97,114,116,101,120,116,58,47,120])=Address.toUuidBytes([119,97,114,110,58,99,108,101,97,114,116,101,120,116,58,47,120])∧Address.toUuidBytes([119,97,114,110,58,120])≠Address.toUuidBytes([101,114,114,111,114,58,120])
Address.toUuidBytes [119, 97, 114, 110, 58, 99, 108, 101, 97, 114, 116, 101, 120, 116, 58, 47, 120] == Address.toUuidBytes [119, 97, 114, 110, 58, 99, 108, 101, 97, 114, 116, 101, 120, 116, 58, 47, 120] && Address.toUuidBytes [119, 97, 114, 110, 58, 120] != Address.toUuidBytes [101, 114, 114, 111, 114, 58, 120]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([119,\,97,\,114,\,110,\,58,\,99,\,108,\,101,\,97,\,114,\,116,\,101,\,120,\,116,\,58,\,47,\,120]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([119,\,97,\,114,\,110,\,58,\,99,\,108,\,101,\,97,\,114,\,116,\,101,\,120,\,116,\,58,\,47,\,120]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([119,\,97,\,114,\,110,\,58,\,120]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([101,\,114,\,114,\,111,\,114,\,58,\,120]\right)

every warning is a receipted trial event: a warning maps to a content-addressed event distinct from an error,

Proof. by decide — exhausting 12,138 cases. □

Theorem 380 (a_content_address_detects_any_change)sealed.

Address.toUuidBytes([100,97,116,97])≠Address.toUuidBytes([100,65,84,97])∧Address.toUuidBytes([100,97,116,97])=Address.toUuidBytes([100,97,116,97])
Address.toUuidBytes [100, 97, 116, 97] != Address.toUuidBytes [100, 65, 84, 97] && Address.toUuidBytes [100, 97, 116, 97] == Address.toUuidBytes [100, 97, 116, 97]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([100,\,97,\,116,\,97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([100,\,65,\,84,\,97]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([100,\,97,\,116,\,97]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([100,\,97,\,116,\,97]\right)

a content-address detects any change: altering a single character changes the address, so any corruption is de

Proof. by decide — exhausting 256 cases. □

Theorem 381 (a_sensor_reading_addresses_to_a_uuid)sealed.

Address.toUuidBytes([116,101,109,112,58,50,49,46,52])=Address.toUuidBytes([116,101,109,112,58,50,49,46,52])∧Address.toUuidBytes([116,101,109,112,58,50,49,46,52])≠Address.toUuidBytes([116,101,109,112,58,50,49,46,53])
Address.toUuidBytes [116, 101, 109, 112, 58, 50, 49, 46, 52] == Address.toUuidBytes [116, 101, 109, 112, 58, 50, 49, 46, 52] && Address.toUuidBytes [116, 101, 109, 112, 58, 50, 49, 46, 52] != Address.toUuidBytes [116, 101, 109, 112, 58, 50, 49, 46, 53]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,101,\,109,\,112,\,58,\,50,\,49,\,46,\,52]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,101,\,109,\,112,\,58,\,50,\,49,\,46,\,52]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,101,\,109,\,112,\,58,\,50,\,49,\,46,\,52]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([116,\,101,\,109,\,112,\,58,\,50,\,49,\,46,\,53]\right)

a sensor reading addresses to a uuid: a live value mints a deterministic content-address, so a reading becomes

Proof. by decide — exhausting 6,561 cases. □

Theorem 382 (forward_is_the_deterministic_compute)sealed.

Address.toUuidBytes([120])=Address.toUuidBytes([120])∧Address.toUuidBytes([97])≠Address.toUuidBytes([98])
Address.toUuidBytes [120] == Address.toUuidBytes [120] && Address.toUuidBytes [97] != Address.toUuidBytes [98]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([120]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([120]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([98]\right)

forward is the deterministic compute: addressing a value forward gives the same result every time — the forwar

Proof. by decide — by evaluation; no domain is walked. □

Theorem 383 (the_fusion_of_site_and_user_is_deterministic)sealed.

Address.toUuidBytes([115,105,116,101,64,98,103])=Address.toUuidBytes([115,105,116,101,64,98,103])∧Address.toUuidBytes([115,105,116,101,64,98,103])≠Address.toUuidBytes([115,105,116,101,64,100,101])
Address.toUuidBytes [115, 105, 116, 101, 64, 98, 103] == Address.toUuidBytes [115, 105, 116, 101, 64, 98, 103] && Address.toUuidBytes [115, 105, 116, 101, 64, 98, 103] != Address.toUuidBytes [115, 105, 116, 101, 64, 100, 101]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,105,\,116,\,101,\,64,\,98,\,103]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,105,\,116,\,101,\,64,\,98,\,103]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,105,\,116,\,101,\,64,\,98,\,103]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([115,\,105,\,116,\,101,\,64,\,100,\,101]\right)

the fusion of site and user is deterministic: site plus user resolves to one reproducible view; 0/7

Proof. by decide — exhausting 2,401 cases. □

Theorem 384 (each_suggested_next_is_content_addressed)sealed.

Address.toUuidBytes([47,116,104,101,111,114,101,109,47,97])=Address.toUuidBytes([47,116,104,101,111,114,101,109,47,97])∧Address.toUuidBytes([47,116,104,101,111,114,101,109,47,97])≠Address.toUuidBytes([47,116,104,101,111,114,101,109,47,99])
Address.toUuidBytes [47, 116, 104, 101, 111, 114, 101, 109, 47, 97] == Address.toUuidBytes [47, 116, 104, 101, 111, 114, 101, 109, 47, 97] && Address.toUuidBytes [47, 116, 104, 101, 111, 114, 101, 109, 47, 97] != Address.toUuidBytes [47, 116, 104, 101, 111, 114, 101, 109, 47, 99]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([47,\,116,\,104,\,101,\,111,\,114,\,101,\,109,\,47,\,97]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([47,\,116,\,104,\,101,\,111,\,114,\,101,\,109,\,47,\,97]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([47,\,116,\,104,\,101,\,111,\,114,\,101,\,109,\,47,\,97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([47,\,116,\,104,\,101,\,111,\,114,\,101,\,109,\,47,\,99]\right)

each suggested next is content-addressed: every suggestion resolves to a deterministic link to its page, so th

Proof. by decide — exhausting 10,000 cases. □

Theorem 385 (the_rejected_command_gets_a_receipt)sealed.

Address.toUuidBytes([114,101,106,101,99,116,101,100,58,112,97,103,101,45,99,111,109,109,97,110,100])=Address.toUuidBytes([114,101,106,101,99,116,101,100,58,112,97,103,101,45,99,111,109,109,97,110,100])∧Address.toUuidBytes([114,101,106,101,99,116,101,100,58,97])≠Address.toUuidBytes([114,101,106,101,99,116,101,100,58,98])
Address.toUuidBytes [114, 101, 106, 101, 99, 116, 101, 100, 58, 112, 97, 103, 101, 45, 99, 111, 109, 109, 97, 110, 100] == Address.toUuidBytes [114, 101, 106, 101, 99, 116, 101, 100, 58, 112, 97, 103, 101, 45, 99, 111, 109, 109, 97, 110, 100] && Address.toUuidBytes [114, 101, 106, 101, 99, 116, 101, 100, 58, 97] != Address.toUuidBytes [114, 101, 106, 101, 99, 116, 101, 100, 58, 98]
LaTeX source
\mathrm{Address.toUuidBytes}\mathopen{}\left([114,\,101,\,106,\,101,\,99,\,116,\,101,\,100,\,58,\,112,\,97,\,103,\,101,\,45,\,99,\,111,\,109,\,109,\,97,\,110,\,100]\right) = \mathrm{Address.toUuidBytes}\mathopen{}\left([114,\,101,\,106,\,101,\,99,\,116,\,101,\,100,\,58,\,112,\,97,\,103,\,101,\,45,\,99,\,111,\,109,\,109,\,97,\,110,\,100]\right) \land \mathrm{Address.toUuidBytes}\mathopen{}\left([114,\,101,\,106,\,101,\,99,\,116,\,101,\,100,\,58,\,97]\right) \neq \mathrm{Address.toUuidBytes}\mathopen{}\left([114,\,101,\,106,\,101,\,99,\,116,\,101,\,100,\,58,\,98]\right)

the rejected command gets a receipt: a refused command is content-addressed and documented, not silently dropp

Proof. by decide — exhausting 44,100 cases. □

Theorem 386 (the_cardinality_claims_hold_of_any_four_symbol_alphabet)sealed.

43=64∧26=64∧82=64∧43=26
4 ^ 3 = 64 ∧ 2 ^ 6 = 64 ∧ 8 ^ 2 = 64 ∧ 4 ^ 3 = 2 ^ 6
LaTeX source
4^{3} = 64 \land 2^{6} = 64 \land 8^{2} = 64 \land 4^{3} = 2^{6}

── THE BOLD CLAIMS OF THIS FILE, AND THE COMPUTATIONS THAT BREAK THEM ───────────────────────────────── Twenty-six theorems here carry names that CLAIM ABOUT THE WORLD — that DNA is the version itself, that the genetic code is the octave squared, that gravity holds prose and code and paths, that intentions are shown by receipts and not by role. They are stated boldly and they stay stated. What follows is the computation a critic runs against them, published here rather than left for someone to find. The claims fall into two classes and each class has one defeater, PROVED below rather than described. CLASS ONE — CARDINALITY. Sixteen decide arithmetic on counts: 4^3 = 64 beside DNA, 6 + 1 = 7 beside the creation week, 3 * 2 = 6 beside reading frames, 2/1 beside the octave. THE ARITHMETIC IS TRUE OF EVERY STRUCTURE WITH THOSE COUNTS. Four symbols in triples give sixty-four whether they are nucleotides, DNA bases, playing-card suits or nothing at all. The theorem decides the count; the NAME supplies the subject, and the subject is a choice a reader is free to reject. A critic substitutes any other four-symbol alphabet and the arithmetic does not move. CLASS TWO — REFLEXIVITY, and this defeater is the harder one. Ten compare an address with itself: `Address.toUuidBytes x == Address.toUuidBytes x`. THAT HOLDS FOR ANY FUNCTION WHATEVER, INCLUDING A CONSTANT ONE THAT THROWS ITS INPUT AWAY. It witnesses that addressing is deterministic on one input. It witnesses nothing about gravity, cells, sensors or intentions, and it would hold unchanged if the address function were replaced by `fun _ => []`. Both defeaters are decided here, so the limit is not a caveat in prose but a proposition the kernel checks.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 387 (a_reflexive_address_claim_holds_for_a_constant_function)sealed.

forgetful(5)=forgetful(5)∧forgetful(5)=forgetful(7)
forgetful 5 = forgetful 5 ∧ forgetful 5 = forgetful 7
LaTeX source
\mathrm{forgetful}\mathopen{}\left(5\right) = \mathrm{forgetful}\mathopen{}\left(5\right) \land \mathrm{forgetful}\mathopen{}\left(5\right) = \mathrm{forgetful}\mathopen{}\left(7\right)

Proof. by decide — by evaluation; no domain is walked. □

Nim ​

src/proof/nim.lean · namespace Nim · 8 theorems

Nim — Bouton's theorem and Sprague–Grundy, decided.

The ledger asserted these in TypeScript: that a Nim position is lost for the mover exactly when the XOR of the heaps is zero, and that a single heap's Grundy value is its size. Both are real theorems with real proofs; what is done here is to DECIDE them over a named finite board — every position with heaps up to a bound — which is what decide can honestly deliver. It is not a proof for all heap sizes, and the range is stated in each name rather than implied.

XOR is imported from Fnv rather than restated: it is the same fuel-bounded structural fold, built because Nat's bitwise operations are well-founded and would drag propext into every theorem here.

Definitions

N := 6

Theorem 388 (bouton_two_heaps_lost_iff_xor_zero)sealed.

∀a∈{0,…,N−1},∀b∈{0,…,N−1},isLost(a,b)=xorN(a,b)=0
(List.range N).all (fun a => (List.range N).all (fun b => isLost a b == (xorN a b == 0)))
LaTeX source
\forall a \in \{0,\dots,N-1\},\; \forall b \in \{0,\dots,N-1\},\; \mathrm{isLost}\mathopen{}\left(a,\,b\right) = \mathrm{xorN}\mathopen{}\left(a,\,b\right) = 0

── BOUTON: a two-heap position is lost for the mover exactly when the heaps are equal — which is exactly when their XOR is zero. Decided over every position on the 8×8 board. ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 389 (bouton_lost_iff_heaps_equal)sealed.

∀a∈{0,…,N−1},∀b∈{0,…,N−1},isLost(a,b)=a=b
(List.range N).all (fun a => (List.range N).all (fun b => isLost a b == (a == b)))
LaTeX source
\forall a \in \{0,\dots,N-1\},\; \forall b \in \{0,\dots,N-1\},\; \mathrm{isLost}\mathopen{}\left(a,\,b\right) = a = b

── the same statement in its familiar form: lost exactly when the heaps are equal ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 390 (lost_positions_are_exactly_the_diagonal)sealed.

|{p∈⋃a∈{0,…,N−1},{(a,b)∣b∈{0,…,N−1}}∣isLost(p1,p2)}|=N
(((List.range N).flatMap (fun a => (List.range N).map (fun b => (a, b)))).filter (fun p => isLost p.1 p.2)).length = N
LaTeX source
\left|\{\, p \in \bigcup_{a \in \{0,\dots,N-1\}} \{\, \left(a,\,b\right) \mid b \in \{0,\dots,N-1\} \,\} \mid \mathrm{isLost}\mathopen{}\left(p_{1},\,p_{2}\right) \,\}\right| = N

── NON-VACUITY: the losing positions are exactly the diagonal, so there are N of them among N² — neither everything nor nothing. (An earlier version fixed this count at 8 for an 8x8 board; when the board was reduced the kernel proved the statement FALSE rather than letting a stale constant pass. Tied to N now.) ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 391 (grundy_of_a_single_heap_is_its_size)sealed.

∀n∈{0,…,N−1},grundy1(n+1,n)=n
(List.range N).all (fun n => grundy1 (n + 1) n == n)
LaTeX source
\forall n \in \{0,\dots,N-1\},\; \mathrm{grundy1}\mathopen{}\left(n + 1,\,n\right) = n

Proof. by decide — by evaluation; no domain is walked. □

Theorem 392 (grundy_of_two_heaps_is_the_xor)sealed.

∀a∈{0,…,N−1},∀b∈{0,…,N−1},xorN(grundy1(a+1,a),grundy1(b+1,b))=xorN(a,b)
(List.range N).all (fun a => (List.range N).all (fun b => (xorN (grundy1 (a + 1) a) (grundy1 (b + 1) b)) == xorN a b))
LaTeX source
\forall a \in \{0,\dots,N-1\},\; \forall b \in \{0,\dots,N-1\},\; \mathrm{xorN}\mathopen{}\left(\mathrm{grundy1}\mathopen{}\left(a + 1,\,a\right),\,\mathrm{grundy1}\mathopen{}\left(b + 1,\,b\right)\right) = \mathrm{xorN}\mathopen{}\left(a,\,b\right)

── and the two-heap Grundy value is the XOR of the parts ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 393 (xor_is_its_own_inverse)sealed.

∀a∈{0,…,15},∀b∈{0,…,15},xorN(xorN(a,b),b)=a
(List.range 16).all (fun a => (List.range 16).all (fun b => xorN (xorN a b) b == a))
LaTeX source
\forall a \in \{0,\dots,15\},\; \forall b \in \{0,\dots,15\},\; \mathrm{xorN}\mathopen{}\left(\mathrm{xorN}\mathopen{}\left(a,\,b\right),\,b\right) = a

── XOR's algebra, which is why the theorem takes the form it does ──

Proof. by decide — exhausting 256 cases. □

Theorem 394 (xor_is_commutative)sealed.

∀a∈{0,…,15},∀b∈{0,…,15},xorN(a,b)=xorN(b,a)
(List.range 16).all (fun a => (List.range 16).all (fun b => xorN a b == xorN b a))
LaTeX source
\forall a \in \{0,\dots,15\},\; \forall b \in \{0,\dots,15\},\; \mathrm{xorN}\mathopen{}\left(a,\,b\right) = \mathrm{xorN}\mathopen{}\left(b,\,a\right)

Proof. by decide — exhausting 256 cases. □

Theorem 395 (xor_zero_is_identity)sealed.

∀a∈{0,…,31},xorN(a,0)=a
(List.range 32).all (fun a => xorN a 0 == a)
LaTeX source
\forall a \in \{0,\dots,31\},\; \mathrm{xorN}\mathopen{}\left(a,\,0\right) = a

Proof. by decide — exhausting 32 cases. □

Digit reversal ​

src/proof/reversal.lean · namespace Reversal · 9 theorems

Digit reversal — arithmetic, not string handling.

An earlier pass called this family "not generatable: needs string manipulation, no small decidable form". That was wrong. A digit is arithmetic — n % 10 and n / 10 — so reversal is a structural fold with no strings anywhere. The mathematics is real too: the digital root is a function of the digit SUM, and reversal is a permutation of the digits, so invariance under reversal is a THEOREM rather than a coincidence. That is exactly what makes emirps — primes whose reversal is also prime — a meaningful class rather than a curio. No axioms, no Mathlib, no sorry.

Definitions

settledHere := 8

Theorem 396 (reversal_preserves_digit_sum)sealed.

∀n∈{1,…,300},digitSum(reverseNum(n))=digitSum(n)
(List.range' 1 300).all (fun n => digitSum (reverseNum n) == digitSum n)
LaTeX source
\forall n \in \{1,\dots,300\},\; \mathrm{digitSum}\mathopen{}\left(\mathrm{reverseNum}\mathopen{}\left(n\right)\right) = \mathrm{digitSum}\mathopen{}\left(n\right)

── reversal is a permutation of the digits, so it preserves their sum ──

Proof. by decide — exhausting 300 cases. □

Theorem 397 (digital_root_is_invariant_under_reversal)sealed.

∀n∈{1,…,300},reverseNum(n)mod9=nmod9
(List.range' 1 300).all (fun n => (reverseNum n) % 9 == n % 9)
LaTeX source
\forall n \in \{1,\dots,300\},\; \mathrm{reverseNum}\mathopen{}\left(n\right) \bmod 9 = n \bmod 9

── and therefore preserves the residue mod 9 — the digital root is invariant under reversal ──

Proof. by decide — exhausting 300 cases. □

Theorem 398 (the_ledger_reversal_cases)sealed.

reverseNum(12)mod9=12mod9∧reverseNum(45)mod9=45mod9∧reverseNum(123)mod9=123mod9∧reverseNum(1234)mod9=1234mod9∧reverseNum(4321)mod9=4321mod9∧reverseNum(9080)mod9=9080mod9
(reverseNum 12) % 9 == 12 % 9 ∧ (reverseNum 45) % 9 == 45 % 9 ∧ (reverseNum 123) % 9 == 123 % 9 ∧ (reverseNum 1234) % 9 == 1234 % 9 ∧ (reverseNum 4321) % 9 == 4321 % 9 ∧ (reverseNum 9080) % 9 == 9080 % 9
LaTeX source
\mathrm{reverseNum}\mathopen{}\left(12\right) \bmod 9 = 12 \bmod 9 \land \mathrm{reverseNum}\mathopen{}\left(45\right) \bmod 9 = 45 \bmod 9 \land \mathrm{reverseNum}\mathopen{}\left(123\right) \bmod 9 = 123 \bmod 9 \land \mathrm{reverseNum}\mathopen{}\left(1234\right) \bmod 9 = 1234 \bmod 9 \land \mathrm{reverseNum}\mathopen{}\left(4321\right) \bmod 9 = 4321 \bmod 9 \land \mathrm{reverseNum}\mathopen{}\left(9080\right) \bmod 9 = 9080 \bmod 9

── the ledger's own cases, now stated rather than asserted ──

Proof. by decide — by evaluation; no domain is walked. □

Theorem 399 (reversal_is_not_the_identity)sealed.

¬∀n∈{10,…,99},reverseNum(n)=n
¬ ((List.range' 10 90).all (fun n => reverseNum n == n))
LaTeX source
\lnot \forall n \in \{10,\dots,99\},\; \mathrm{reverseNum}\mathopen{}\left(n\right) = n

── NON-VACUITY: reversal genuinely moves the number, so the invariance is not about a fixed point ──

Proof. by decide — exhausting 90 cases. □

Theorem 400 (palindromes_are_the_fixed_points)sealed.

|{n∈{10,…,99}∣reverseNum(n)=n}|=9
((List.range' 10 90).filter (fun n => reverseNum n == n)).length = 9
LaTeX source
\left|\{\, n \in \{10,\dots,99\} \mid \mathrm{reverseNum}\mathopen{}\left(n\right) = n \,\}\right| = 9

── palindromes are exactly the fixed points, and there are nine of them below 100 ──

Proof. by decide — exhausting 90 cases. □

Theorem 401 (emirps_exist_below_one_hundred)sealed.

|{n∈{10,…,99}∣isPrime(n)∧isPrime(reverseNum(n))∧reverseNum(n)≠n}|>0
((List.range' 10 90).filter (fun n => isPrime n && isPrime (reverseNum n) && reverseNum n != n)).length > 0
LaTeX source
\left|\{\, n \in \{10,\dots,99\} \mid \mathrm{isPrime}\mathopen{}\left(n\right) \land \mathrm{isPrime}\mathopen{}\left(\mathrm{reverseNum}\mathopen{}\left(n\right)\right) \land \mathrm{reverseNum}\mathopen{}\left(n\right) \neq n \,\}\right| > 0

Proof. by decide — exhausting 90 cases. □

Theorem 402 (reversal_does_not_preserve_primality)sealed.

¬∀n∈{10,…,99},isPrime(n)=isPrime(reverseNum(n))
¬ ((List.range' 10 90).all (fun n => isPrime n == isPrime (reverseNum n)))
LaTeX source
\lnot \forall n \in \{10,\dots,99\},\; \mathrm{isPrime}\mathopen{}\left(n\right) = \mathrm{isPrime}\mathopen{}\left(\mathrm{reverseNum}\mathopen{}\left(n\right)\right)

── and reversal does NOT preserve primality in general: the boundary that makes emirps a real class ──

Proof. by decide — exhausting 90 cases. □

Theorem 403 (reversal_is_involutive_exactly_off_the_trailing_zeros)sealed.

∀n∈{1,…,300},reverseNum(reverseNum(n))=n=nmod10≠0
(List.range' 1 300).all (fun n => (reverseNum (reverseNum n) == n) == (n % 10 != 0))
LaTeX source
\forall n \in \{1,\dots,300\},\; \mathrm{reverseNum}\mathopen{}\left(\mathrm{reverseNum}\mathopen{}\left(n\right)\right) = n = n \bmod 10 \neq 0

── WHERE REVERSAL STOPS BEING AN INVOLUTION. Reversing twice usually returns the number — but not always: a trailing zero is destroyed by the first reversal and cannot be restored by the second (120 → 021 = 21 → 12). So reversal is self-inverse EXACTLY on the numbers with no trailing zero, and the iff is decided in both directions across the range, not asserted for the convenient half. The exception is the whole content of the theorem: an involution that quietly fails on a tenth of its domain is not an involution.

Proof. by decide — exhausting 300 cases. □

Theorem 404 (reversal_settles_its_range)not sealed — settled by rfl, not exhausted.

settledHere=8
settledHere = 8
LaTeX source
\mathrm{settledHere} = 8

Proof. rfl — by evaluation; no domain is walked. □

Theorems ​

src/proof/theorems.lean · namespace MillenniumFloor.Universal · 8 theorems

The universal property — honestly, and COMPUTED from the sequence.

The earlier all_alpha_squared_one asserted that the seven statements "share α² = 1" — a vacuity (1² = 1 copy-pasted). What the seven ACTUALLY share is the reflection: an involution the sequence computes, with one centre (the heart). That shared structure is real; it is still not a proof of any conjecture. No anchors, no axioms, every proof by decide, no Mathlib. The floor holds: 0/7.

Theorem 405 (universal_reflection_involution)sealed.

∀d∈{0,…,10},refl(refl(d))=d∧|{d∈{0,…,9}∣refl(d)=d}|=1
(List.range 11).all (fun d => refl (refl d) == d) ∧ ((List.range 10).filter (fun d => refl d == d)).length = 1
LaTeX source
\forall d \in \{0,\dots,10\},\; \mathrm{refl}\mathopen{}\left(\mathrm{refl}\mathopen{}\left(d\right)\right) = d \land \left|\{\, d \in \{0,\dots,9\} \mid \mathrm{refl}\mathopen{}\left(d\right) = d \,\}\right| = 1

The universal law, computed: the reflection is a TOTAL involution on every residue, with exactly one shared fixed centre — the structure every one of the seven framings borrows. It is not "α² = 1", and it is not a proof.

Proof. by decide — exhausting 110 cases. □

Theorem 406 (universal_centre_is_five)sealed.

{d∈{0,…,9}∣refl(d)=d}=[5]
((List.range 10).filter (fun d => refl d == d)) = [5]
LaTeX source
\{\, d \in \{0,\dots,9\} \mid \mathrm{refl}\mathopen{}\left(d\right) = d \,\} = [5]

── the centre, NAMED rather than counted ── The theorem above proves there is exactly ONE centre. A count is not an identification: it says a heart exists without saying where it beats. Here it is, computed — the filter returns the singleton [5].

Proof. by decide — exhausting 10 cases. □

Theorem 407 (universal_pairs_sum_to_ten)sealed.

∀d∈{0,…,10},d+refl(d)=10
(List.range 11).all (fun d => d + refl d == 10)
LaTeX source
\forall d \in \{0,\dots,10\},\; d + \mathrm{refl}\mathopen{}\left(d\right) = 10

── what the reflection conserves ── Every residue and its reflection sum to ten. This is the reason there is exactly one centre: a fixed point needs d + d = 10, and 10 is even, so exactly one d satisfies it. The conservation law is the whole structure.

Proof. by decide — exhausting 11 cases. □

Theorem 408 (universal_reflection_reverses_the_domain)sealed.

refl[{0,…,10}]=reverse({0,…,10})
(List.range 11).map refl = (List.range 11).reverse
LaTeX source
\mathrm{refl}[\{0,\dots,10\}] = \operatorname{reverse}\left(\{0,\dots,10\}\right)

── the reflection is a bijection of the domain, computed as an ordering ── Stronger than the involution and cheaper than a permutation argument: the image of 0…10 under the reflection IS 0…10 read backwards. Nothing is lost and nothing is repeated, and the witness is an equation.

Proof. by decide — exhausting 121 cases. □

Theorem 409 (universal_reflection_is_not_an_involution_above_ten)sealed.

∀d∈{k+11∣k∈{0,…,9}},¬refl(refl(d))=d
((List.range 10).map (fun k => k + 11)).all (fun d => ¬ (refl (refl d) == d))
LaTeX source
\forall d \in \{\, k + 11 \mid k \in \{0,\dots,9\} \,\},\; \lnot \mathrm{refl}\mathopen{}\left(\mathrm{refl}\mathopen{}\left(d\right)\right) = d

── WHERE IT STOPS. The involution is not universal over ℕ, and the boundary is stated, not hidden ── Natural subtraction truncates: above ten, `refl d` is 0 and `refl (refl d)` is 10 for every input. So the involution holds on 0…10 and FAILS at every residue above it. The first theorem says `List.range 11` and means it. A boundary you can compute is a boundary you have not overclaimed past.

Proof. by decide — exhausting 10 cases. □

Theorem 410 (universal_z9_reflection_permutes_the_units)sealed.

{9-dmod9∣d∈[1,2,4,5,7,8]}=[8,7,5,4,2,1]
([1, 2, 4, 5, 7, 8].map (fun d => (9 - d) % 9)) = [8, 7, 5, 4, 2, 1]
LaTeX source
\{\, 9 - d \bmod 9 \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\} = [8,\,7,\,5,\,4,\,2,\,1]

── the vortex has its OWN reflection, and it is a different map ── ℤ/9 reflects by d ↦ (9 − d) mod 9. That map permutes the six units among themselves. The millennium reflection does NOT stay inside them: it sends the unit 1 to 9, which is not a unit. Two reflections, two domains.

Proof. by decide — exhausting 36 cases. □

Theorem 411 (universal_millennium_reflection_escapes_the_units)sealed.

¬∀u∈[1,2,4,5,7,8],refl(u)∈[1,2,4,5,7,8]
¬ ([1, 2, 4, 5, 7, 8].all (fun u => [1, 2, 4, 5, 7, 8].contains (refl u)))
LaTeX source
\lnot \forall u \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{refl}\mathopen{}\left(u\right) \in [1,\,2,\,4,\,5,\,7,\,8]

Proof. by decide — exhausting 36 cases. □

Theorem 412 (universal_reflection_is_the_vortex_reflection_shifted)sealed.

∀d∈{0,…,9},10-dmod9=9-dmod9+1mod9
(List.range 10).all (fun d => (10 - d) % 9 == ((9 - d) % 9 + 1) % 9)
LaTeX source
\forall d \in \{0,\dots,9\},\; 10 - d \bmod 9 = 9 - d \bmod 9 + 1 \bmod 9

── and yet they are the SAME map, shifted by one ── This is what the singleton family was standing on alone. Reduced mod nine, the millennium reflection is the vortex reflection plus one, at every residue 0…9. The shared structure the seven framings borrow is not a separate object: it is ℤ/9's own involution, displaced by a single step. Still not a proof of any conjecture.

Proof. by decide — exhausting 10 cases. □

the imagined

What enumeration proposed and the kernel kept ​

src/proof/imagined.lean · namespace Imagined · 120 theorems

IMAGINED — proposed by scripts/imagine.ts, which enumerated every map-against-subset and map-between-subsets statement its primitives can express, kept the ones true by exhaustion, and then discarded every one that also holds for all its siblings. A property true of everything names nothing. What is left is what the kernel accepted; whatever it refused is reported by the generator and is not in this file.

Theorem 413 (units_is_closed_under_double)sealed.

∀d∈[1,2,4,5,7,8],m9(2⋅d)∈[1,2,4,5,7,8]
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (2 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(2 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the units is closed under doubling

Proof. by decide — exhausting 36 cases. □

Theorem 414 (triad_is_closed_under_double)sealed.

∀d∈[3,6,0],m9(2⋅d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (2 * d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(2 \cdot d\right) \in [3,\,6,\,0]

the triad is closed under doubling

Proof. by decide — exhausting 9 cases. □

Theorem 415 (orbit_is_closed_under_double)sealed.

∀d∈[1,2,4,8,7,5],m9(2⋅d)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (2 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(2 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under doubling

Proof. by decide — exhausting 36 cases. □

Theorem 416 (all_is_closed_under_double)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(2⋅d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (2 * d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(2 \cdot d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under doubling

Proof. by decide — exhausting 81 cases. □

Theorem 417 (triad_is_closed_under_triple)sealed.

∀d∈[3,6,0],m9(3⋅d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (3 * d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \in [3,\,6,\,0]

the triad is closed under tripling

Proof. by decide — exhausting 9 cases. □

Theorem 418 (all_is_closed_under_triple)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(3⋅d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (3 * d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under tripling

Proof. by decide — exhausting 81 cases. □

Theorem 419 (units_is_closed_under_quadruple)sealed.

∀d∈[1,2,4,5,7,8],m9(4⋅d)∈[1,2,4,5,7,8]
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (4 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the units is closed under quadrupling

Proof. by decide — exhausting 36 cases. □

Theorem 420 (triad_is_closed_under_quadruple)sealed.

∀d∈[3,6,0],m9(4⋅d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (4 * d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [3,\,6,\,0]

the triad is closed under quadrupling

Proof. by decide — exhausting 9 cases. □

Theorem 421 (orbit_is_closed_under_quadruple)sealed.

∀d∈[1,2,4,8,7,5],m9(4⋅d)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (4 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under quadrupling

Proof. by decide — exhausting 36 cases. □

Theorem 422 (tetA_is_closed_under_quadruple)sealed.

∀d∈[1,4,7],m9(4⋅d)∈[1,4,7]
[1, 4, 7].all (fun d => [1, 4, 7].contains (m9 (4 * d)))
LaTeX source
\forall d \in [1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [1,\,4,\,7]

the first tetrahedron is closed under quadrupling

Proof. by decide — exhausting 9 cases. □

Theorem 423 (tetB_is_closed_under_quadruple)sealed.

∀d∈[2,5,8],m9(4⋅d)∈[2,5,8]
[2, 5, 8].all (fun d => [2, 5, 8].contains (m9 (4 * d)))
LaTeX source
\forall d \in [2,\,5,\,8],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [2,\,5,\,8]

the second tetrahedron is closed under quadrupling

Proof. by decide — exhausting 9 cases. □

Theorem 424 (all_is_closed_under_quadruple)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(4⋅d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (4 * d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under quadrupling

Proof. by decide — exhausting 81 cases. □

Theorem 425 (squares_is_closed_under_quadruple)sealed.

∀d∈[0,1,4,7],m9(4⋅d)∈[0,1,4,7]
[0, 1, 4, 7].all (fun d => [0, 1, 4, 7].contains (m9 (4 * d)))
LaTeX source
\forall d \in [0,\,1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [0,\,1,\,4,\,7]

the squares mod nine is closed under quadrupling

Proof. by decide — exhausting 16 cases. □

Theorem 426 (triad_is_closed_under_negate)sealed.

∀d∈[3,6,0],m9(9-d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (9 - d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(9 - d\right) \in [3,\,6,\,0]

the triad is closed under negation

Proof. by decide — exhausting 9 cases. □

Theorem 427 (orbit_is_closed_under_negate)sealed.

∀d∈[1,2,4,8,7,5],m9(9-d)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (9 - d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(9 - d\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under negation

Proof. by decide — exhausting 36 cases. □

Theorem 428 (all_is_closed_under_negate)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(9-d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (9 - d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(9 - d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under negation

Proof. by decide — exhausting 81 cases. □

Theorem 429 (cubes_is_closed_under_negate)sealed.

∀d∈[0,1,8],m9(9-d)∈[0,1,8]
[0, 1, 8].all (fun d => [0, 1, 8].contains (m9 (9 - d)))
LaTeX source
\forall d \in [0,\,1,\,8],\; \mathrm{m9}\mathopen{}\left(9 - d\right) \in [0,\,1,\,8]

the cubes mod nine is closed under negation

Proof. by decide — exhausting 9 cases. □

Theorem 430 (units_is_closed_under_square)sealed.

∀d∈[1,2,4,5,7,8],m9(d⋅d)∈[1,2,4,5,7,8]
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (d * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the units is closed under squaring

Proof. by decide — exhausting 36 cases. □

Theorem 431 (triad_is_closed_under_square)sealed.

∀d∈[3,6,0],m9(d⋅d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (d * d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [3,\,6,\,0]

the triad is closed under squaring

Proof. by decide — exhausting 9 cases. □

Theorem 432 (orbit_is_closed_under_square)sealed.

∀d∈[1,2,4,8,7,5],m9(d⋅d)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (d * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under squaring

Proof. by decide — exhausting 36 cases. □

Theorem 433 (tetA_is_closed_under_square)sealed.

∀d∈[1,4,7],m9(d⋅d)∈[1,4,7]
[1, 4, 7].all (fun d => [1, 4, 7].contains (m9 (d * d)))
LaTeX source
\forall d \in [1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [1,\,4,\,7]

the first tetrahedron is closed under squaring

Proof. by decide — exhausting 9 cases. □

Theorem 434 (all_is_closed_under_square)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(d⋅d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (d * d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under squaring

Proof. by decide — exhausting 81 cases. □

Theorem 435 (squares_is_closed_under_square)sealed.

∀d∈[0,1,4,7],m9(d⋅d)∈[0,1,4,7]
[0, 1, 4, 7].all (fun d => [0, 1, 4, 7].contains (m9 (d * d)))
LaTeX source
\forall d \in [0,\,1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [0,\,1,\,4,\,7]

the squares mod nine is closed under squaring

Proof. by decide — exhausting 16 cases. □

Theorem 436 (cubes_is_closed_under_square)sealed.

∀d∈[0,1,8],m9(d⋅d)∈[0,1,8]
[0, 1, 8].all (fun d => [0, 1, 8].contains (m9 (d * d)))
LaTeX source
\forall d \in [0,\,1,\,8],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [0,\,1,\,8]

the cubes mod nine is closed under squaring

Proof. by decide — exhausting 9 cases. □

Theorem 437 (units_is_closed_under_fourth)sealed.

∀d∈[1,2,4,5,7,8],m9(d4)∈[1,2,4,5,7,8]
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (d ^ 4)))
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the units is closed under the fourth power

Proof. by decide — exhausting 36 cases. □

Theorem 438 (triad_is_closed_under_fourth)sealed.

∀d∈[3,6,0],m9(d4)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (d ^ 4)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [3,\,6,\,0]

the triad is closed under the fourth power

Proof. by decide — exhausting 9 cases. □

Theorem 439 (orbit_is_closed_under_fourth)sealed.

∀d∈[1,2,4,8,7,5],m9(d4)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (d ^ 4)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under the fourth power

Proof. by decide — exhausting 36 cases. □

Theorem 440 (tetA_is_closed_under_fourth)sealed.

∀d∈[1,4,7],m9(d4)∈[1,4,7]
[1, 4, 7].all (fun d => [1, 4, 7].contains (m9 (d ^ 4)))
LaTeX source
\forall d \in [1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [1,\,4,\,7]

the first tetrahedron is closed under the fourth power

Proof. by decide — exhausting 9 cases. □

Theorem 441 (all_is_closed_under_fourth)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(d4)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (d ^ 4)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under the fourth power

Proof. by decide — exhausting 81 cases. □

Theorem 442 (squares_is_closed_under_fourth)sealed.

∀d∈[0,1,4,7],m9(d4)∈[0,1,4,7]
[0, 1, 4, 7].all (fun d => [0, 1, 4, 7].contains (m9 (d ^ 4)))
LaTeX source
\forall d \in [0,\,1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [0,\,1,\,4,\,7]

the squares mod nine is closed under the fourth power

Proof. by decide — exhausting 16 cases. □

Theorem 443 (cubes_is_closed_under_fourth)sealed.

∀d∈[0,1,8],m9(d4)∈[0,1,8]
[0, 1, 8].all (fun d => [0, 1, 8].contains (m9 (d ^ 4)))
LaTeX source
\forall d \in [0,\,1,\,8],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [0,\,1,\,8]

the cubes mod nine is closed under the fourth power

Proof. by decide — exhausting 9 cases. □

Theorem 444 (units_is_closed_under_sixth)sealed.

∀d∈[1,2,4,5,7,8],m9(d6)∈[1,2,4,5,7,8]
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (d ^ 6)))
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d^{6}\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the units is closed under the sixth power

Proof. by decide — exhausting 36 cases. □

Theorem 445 (triad_is_closed_under_sixth)sealed.

∀d∈[3,6,0],m9(d6)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (d ^ 6)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(d^{6}\right) \in [3,\,6,\,0]

the triad is closed under the sixth power

Proof. by decide — exhausting 9 cases. □

Theorem 446 (orbit_is_closed_under_sixth)sealed.

∀d∈[1,2,4,8,7,5],m9(d6)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (d ^ 6)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(d^{6}\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under the sixth power

Proof. by decide — exhausting 36 cases. □

Theorem 447 (tetA_is_closed_under_sixth)sealed.

∀d∈[1,4,7],m9(d6)∈[1,4,7]
[1, 4, 7].all (fun d => [1, 4, 7].contains (m9 (d ^ 6)))
LaTeX source
\forall d \in [1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(d^{6}\right) \in [1,\,4,\,7]

the first tetrahedron is closed under the sixth power

Proof. by decide — exhausting 9 cases. □

Theorem 448 (all_is_closed_under_sixth)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(d6)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (d ^ 6)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d^{6}\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under the sixth power

Proof. by decide — exhausting 81 cases. □

Theorem 449 (squares_is_closed_under_sixth)sealed.

∀d∈[0,1,4,7],m9(d6)∈[0,1,4,7]
[0, 1, 4, 7].all (fun d => [0, 1, 4, 7].contains (m9 (d ^ 6)))
LaTeX source
\forall d \in [0,\,1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(d^{6}\right) \in [0,\,1,\,4,\,7]

the squares mod nine is closed under the sixth power

Proof. by decide — exhausting 16 cases. □

Theorem 450 (cubes_is_closed_under_sixth)sealed.

∀d∈[0,1,8],m9(d6)∈[0,1,8]
[0, 1, 8].all (fun d => [0, 1, 8].contains (m9 (d ^ 6)))
LaTeX source
\forall d \in [0,\,1,\,8],\; \mathrm{m9}\mathopen{}\left(d^{6}\right) \in [0,\,1,\,8]

the cubes mod nine is closed under the sixth power

Proof. by decide — exhausting 9 cases. □

Theorem 451 (units_is_closed_under_quintuple)sealed.

∀d∈[1,2,4,5,7,8],m9(5⋅d)∈[1,2,4,5,7,8]
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (5 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the units is closed under multiplication by five

Proof. by decide — exhausting 36 cases. □

Theorem 452 (triad_is_closed_under_quintuple)sealed.

∀d∈[3,6,0],m9(5⋅d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (5 * d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \in [3,\,6,\,0]

the triad is closed under multiplication by five

Proof. by decide — exhausting 9 cases. □

Theorem 453 (orbit_is_closed_under_quintuple)sealed.

∀d∈[1,2,4,8,7,5],m9(5⋅d)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (5 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under multiplication by five

Proof. by decide — exhausting 36 cases. □

Theorem 454 (all_is_closed_under_quintuple)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(5⋅d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (5 * d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under multiplication by five

Proof. by decide — exhausting 81 cases. □

Theorem 455 (triad_is_closed_under_sextuple)sealed.

∀d∈[3,6,0],m9(6⋅d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (6 * d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \in [3,\,6,\,0]

the triad is closed under multiplication by six

Proof. by decide — exhausting 9 cases. □

Theorem 456 (all_is_closed_under_sextuple)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(6⋅d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (6 * d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under multiplication by six

Proof. by decide — exhausting 81 cases. □

Theorem 457 (units_is_closed_under_septuple)sealed.

∀d∈[1,2,4,5,7,8],m9(7⋅d)∈[1,2,4,5,7,8]
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (7 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the units is closed under multiplication by seven

Proof. by decide — exhausting 36 cases. □

Theorem 458 (triad_is_closed_under_septuple)sealed.

∀d∈[3,6,0],m9(7⋅d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (7 * d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [3,\,6,\,0]

the triad is closed under multiplication by seven

Proof. by decide — exhausting 9 cases. □

Theorem 459 (orbit_is_closed_under_septuple)sealed.

∀d∈[1,2,4,8,7,5],m9(7⋅d)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (7 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under multiplication by seven

Proof. by decide — exhausting 36 cases. □

Theorem 460 (tetA_is_closed_under_septuple)sealed.

∀d∈[1,4,7],m9(7⋅d)∈[1,4,7]
[1, 4, 7].all (fun d => [1, 4, 7].contains (m9 (7 * d)))
LaTeX source
\forall d \in [1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [1,\,4,\,7]

the first tetrahedron is closed under multiplication by seven

Proof. by decide — exhausting 9 cases. □

Theorem 461 (tetB_is_closed_under_septuple)sealed.

∀d∈[2,5,8],m9(7⋅d)∈[2,5,8]
[2, 5, 8].all (fun d => [2, 5, 8].contains (m9 (7 * d)))
LaTeX source
\forall d \in [2,\,5,\,8],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [2,\,5,\,8]

the second tetrahedron is closed under multiplication by seven

Proof. by decide — exhausting 9 cases. □

Theorem 462 (all_is_closed_under_septuple)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(7⋅d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (7 * d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under multiplication by seven

Proof. by decide — exhausting 81 cases. □

Theorem 463 (squares_is_closed_under_septuple)sealed.

∀d∈[0,1,4,7],m9(7⋅d)∈[0,1,4,7]
[0, 1, 4, 7].all (fun d => [0, 1, 4, 7].contains (m9 (7 * d)))
LaTeX source
\forall d \in [0,\,1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [0,\,1,\,4,\,7]

the squares mod nine is closed under multiplication by seven

Proof. by decide — exhausting 16 cases. □

Theorem 464 (units_is_closed_under_octuple)sealed.

∀d∈[1,2,4,5,7,8],m9(8⋅d)∈[1,2,4,5,7,8]
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (8 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8]

the units is closed under multiplication by eight

Proof. by decide — exhausting 36 cases. □

Theorem 465 (triad_is_closed_under_octuple)sealed.

∀d∈[3,6,0],m9(8⋅d)∈[3,6,0]
[3, 6, 0].all (fun d => [3, 6, 0].contains (m9 (8 * d)))
LaTeX source
\forall d \in [3,\,6,\,0],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [3,\,6,\,0]

the triad is closed under multiplication by eight

Proof. by decide — exhausting 9 cases. □

Theorem 466 (orbit_is_closed_under_octuple)sealed.

∀d∈[1,2,4,8,7,5],m9(8⋅d)∈[1,2,4,8,7,5]
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (8 * d)))
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5]

the doubling orbit is closed under multiplication by eight

Proof. by decide — exhausting 36 cases. □

Theorem 467 (all_is_closed_under_octuple)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(8⋅d)∈[0,1,2,3,4,5,6,7,8]
[0,1,2,3,4,5,6,7,8].all (fun d => [0,1,2,3,4,5,6,7,8].contains (m9 (8 * d)))
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8]

the whole ring is closed under multiplication by eight

Proof. by decide — exhausting 81 cases. □

Theorem 468 (cubes_is_closed_under_octuple)sealed.

∀d∈[0,1,8],m9(8⋅d)∈[0,1,8]
[0, 1, 8].all (fun d => [0, 1, 8].contains (m9 (8 * d)))
LaTeX source
\forall d \in [0,\,1,\,8],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [0,\,1,\,8]

the cubes mod nine is closed under multiplication by eight

Proof. by decide — exhausting 9 cases. □

Theorem 469 (double_is_involutive_on_triad)sealed.

∀d∈[3,6,0],x↦m9(2⋅x)(m9(2⋅d))=d
[3, 6, 0].all (fun d => (fun x => m9 (2 * x)) (m9 (2 * d)) == d)
LaTeX source
\forall d \in [3,\,6,\,0],\; x \mapsto \mathrm{m9}\mathopen{}\left(2 \cdot x\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(2 \cdot d\right)\right) = d

doubling is its own inverse on the triad

Proof. by decide — exhausting 3 cases. □

Theorem 470 (quadruple_is_involutive_on_triad)sealed.

∀d∈[3,6,0],x↦m9(4⋅x)(m9(4⋅d))=d
[3, 6, 0].all (fun d => (fun x => m9 (4 * x)) (m9 (4 * d)) == d)
LaTeX source
\forall d \in [3,\,6,\,0],\; x \mapsto \mathrm{m9}\mathopen{}\left(4 \cdot x\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(4 \cdot d\right)\right) = d

quadrupling is its own inverse on the triad

Proof. by decide — exhausting 3 cases. □

Theorem 471 (square_is_involutive_on_tetA)sealed.

∀d∈[1,4,7],x↦m9(x⋅x)(m9(d⋅d))=d
[1, 4, 7].all (fun d => (fun x => m9 (x * x)) (m9 (d * d)) == d)
LaTeX source
\forall d \in [1,\,4,\,7],\; x \mapsto \mathrm{m9}\mathopen{}\left(x \cdot x\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d \cdot d\right)\right) = d

squaring is its own inverse on the first tetrahedron

Proof. by decide — exhausting 3 cases. □

Theorem 472 (square_is_involutive_on_squares)sealed.

∀d∈[0,1,4,7],x↦m9(x⋅x)(m9(d⋅d))=d
[0, 1, 4, 7].all (fun d => (fun x => m9 (x * x)) (m9 (d * d)) == d)
LaTeX source
\forall d \in [0,\,1,\,4,\,7],\; x \mapsto \mathrm{m9}\mathopen{}\left(x \cdot x\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d \cdot d\right)\right) = d

squaring is its own inverse on the squares mod nine

Proof. by decide — exhausting 4 cases. □

Theorem 473 (cube_is_involutive_on_cubes)sealed.

∀d∈[0,1,8],x↦m9(x⋅x⋅x)(m9(d⋅d⋅d))=d
[0, 1, 8].all (fun d => (fun x => m9 (x * x * x)) (m9 (d * d * d)) == d)
LaTeX source
\forall d \in [0,\,1,\,8],\; x \mapsto \mathrm{m9}\mathopen{}\left(x \cdot x \cdot x\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d \cdot d \cdot d\right)\right) = d

cubing is its own inverse on the cubes mod nine

Proof. by decide — exhausting 3 cases. □

Theorem 474 (fourth_is_involutive_on_tetA)sealed.

∀d∈[1,4,7],x↦m9(x4)(m9(d4))=d
[1, 4, 7].all (fun d => (fun x => m9 (x ^ 4)) (m9 (d ^ 4)) == d)
LaTeX source
\forall d \in [1,\,4,\,7],\; x \mapsto \mathrm{m9}\mathopen{}\left(x^{4}\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d^{4}\right)\right) = d

the fourth power is its own inverse on the first tetrahedron

Proof. by decide — exhausting 3 cases. □

Theorem 475 (fourth_is_involutive_on_squares)sealed.

∀d∈[0,1,4,7],x↦m9(x4)(m9(d4))=d
[0, 1, 4, 7].all (fun d => (fun x => m9 (x ^ 4)) (m9 (d ^ 4)) == d)
LaTeX source
\forall d \in [0,\,1,\,4,\,7],\; x \mapsto \mathrm{m9}\mathopen{}\left(x^{4}\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d^{4}\right)\right) = d

the fourth power is its own inverse on the squares mod nine

Proof. by decide — exhausting 4 cases. □

Theorem 476 (fifth_is_involutive_on_units)sealed.

∀d∈[1,2,4,5,7,8],x↦m9(x5)(m9(d5))=d
[1, 2, 4, 5, 7, 8].all (fun d => (fun x => m9 (x ^ 5)) (m9 (d ^ 5)) == d)
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; x \mapsto \mathrm{m9}\mathopen{}\left(x^{5}\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d^{5}\right)\right) = d

the fifth power is its own inverse on the units

Proof. by decide — exhausting 6 cases. □

Theorem 477 (fifth_is_involutive_on_orbit)sealed.

∀d∈[1,2,4,8,7,5],x↦m9(x5)(m9(d5))=d
[1, 2, 4, 8, 7, 5].all (fun d => (fun x => m9 (x ^ 5)) (m9 (d ^ 5)) == d)
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; x \mapsto \mathrm{m9}\mathopen{}\left(x^{5}\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d^{5}\right)\right) = d

the fifth power is its own inverse on the doubling orbit

Proof. by decide — exhausting 6 cases. □

Theorem 478 (fifth_is_involutive_on_tetA)sealed.

∀d∈[1,4,7],x↦m9(x5)(m9(d5))=d
[1, 4, 7].all (fun d => (fun x => m9 (x ^ 5)) (m9 (d ^ 5)) == d)
LaTeX source
\forall d \in [1,\,4,\,7],\; x \mapsto \mathrm{m9}\mathopen{}\left(x^{5}\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d^{5}\right)\right) = d

the fifth power is its own inverse on the first tetrahedron

Proof. by decide — exhausting 3 cases. □

Theorem 479 (fifth_is_involutive_on_tetB)sealed.

∀d∈[2,5,8],x↦m9(x5)(m9(d5))=d
[2, 5, 8].all (fun d => (fun x => m9 (x ^ 5)) (m9 (d ^ 5)) == d)
LaTeX source
\forall d \in [2,\,5,\,8],\; x \mapsto \mathrm{m9}\mathopen{}\left(x^{5}\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d^{5}\right)\right) = d

the fifth power is its own inverse on the second tetrahedron

Proof. by decide — exhausting 3 cases. □

Theorem 480 (fifth_is_involutive_on_squares)sealed.

∀d∈[0,1,4,7],x↦m9(x5)(m9(d5))=d
[0, 1, 4, 7].all (fun d => (fun x => m9 (x ^ 5)) (m9 (d ^ 5)) == d)
LaTeX source
\forall d \in [0,\,1,\,4,\,7],\; x \mapsto \mathrm{m9}\mathopen{}\left(x^{5}\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d^{5}\right)\right) = d

the fifth power is its own inverse on the squares mod nine

Proof. by decide — exhausting 4 cases. □

Theorem 481 (fifth_is_involutive_on_cubes)sealed.

∀d∈[0,1,8],x↦m9(x5)(m9(d5))=d
[0, 1, 8].all (fun d => (fun x => m9 (x ^ 5)) (m9 (d ^ 5)) == d)
LaTeX source
\forall d \in [0,\,1,\,8],\; x \mapsto \mathrm{m9}\mathopen{}\left(x^{5}\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(d^{5}\right)\right) = d

the fifth power is its own inverse on the cubes mod nine

Proof. by decide — exhausting 3 cases. □

Theorem 482 (quintuple_is_involutive_on_triad)sealed.

∀d∈[3,6,0],x↦m9(5⋅x)(m9(5⋅d))=d
[3, 6, 0].all (fun d => (fun x => m9 (5 * x)) (m9 (5 * d)) == d)
LaTeX source
\forall d \in [3,\,6,\,0],\; x \mapsto \mathrm{m9}\mathopen{}\left(5 \cdot x\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(5 \cdot d\right)\right) = d

multiplication by five is its own inverse on the triad

Proof. by decide — exhausting 3 cases. □

Theorem 483 (septuple_is_involutive_on_triad)sealed.

∀d∈[3,6,0],x↦m9(7⋅x)(m9(7⋅d))=d
[3, 6, 0].all (fun d => (fun x => m9 (7 * x)) (m9 (7 * d)) == d)
LaTeX source
\forall d \in [3,\,6,\,0],\; x \mapsto \mathrm{m9}\mathopen{}\left(7 \cdot x\right)\mathopen{}\left(\mathrm{m9}\mathopen{}\left(7 \cdot d\right)\right) = d

multiplication by seven is its own inverse on the triad

Proof. by decide — exhausting 3 cases. □

Theorem 484 (double_carries_units_onto_orbit)sealed.

∀d∈[1,2,4,5,7,8],m9(2⋅d)∈[1,2,4,8,7,5]∧|dedup({m9(2⋅d)∣d∈[1,2,4,5,7,8]})|=6
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (2 * d))) ∧ ([1, 2, 4, 5, 7, 8].map (fun d => m9 (2 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(2 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(2 \cdot d\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6

doubling carries the units onto the doubling orbit

Proof. by decide — exhausting 216 cases. □

Theorem 485 (double_carries_orbit_onto_units)sealed.

∀d∈[1,2,4,8,7,5],m9(2⋅d)∈[1,2,4,5,7,8]∧|dedup({m9(2⋅d)∣d∈[1,2,4,8,7,5]})|=6
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (2 * d))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (2 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(2 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(2 \cdot d\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 6

doubling carries the doubling orbit onto the units

Proof. by decide — exhausting 216 cases. □

Theorem 486 (triple_carries_all_onto_triad)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(3⋅d)∈[3,6,0]∧|dedup({m9(3⋅d)∣d∈[0,1,2,3,4,5,6,7,8]})|=3
[0,1,2,3,4,5,6,7,8].all (fun d => [3, 6, 0].contains (m9 (3 * d))) ∧ ([0,1,2,3,4,5,6,7,8].map (fun d => m9 (3 * d))).eraseDups.length = 3
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \in [3,\,6,\,0] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \mid d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8] \,\}\right)\right| = 3

tripling carries the whole ring onto the triad

Proof. by decide — exhausting 243 cases. □

Theorem 487 (triple_carries_cubes_onto_triad)sealed.

∀d∈[0,1,8],m9(3⋅d)∈[3,6,0]∧|dedup({m9(3⋅d)∣d∈[0,1,8]})|=3
[0, 1, 8].all (fun d => [3, 6, 0].contains (m9 (3 * d))) ∧ ([0, 1, 8].map (fun d => m9 (3 * d))).eraseDups.length = 3
LaTeX source
\forall d \in [0,\,1,\,8],\; \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \in [3,\,6,\,0] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \mid d \in [0,\,1,\,8] \,\}\right)\right| = 3

tripling carries the cubes mod nine onto the triad

Proof. by decide — exhausting 27 cases. □

Theorem 488 (quadruple_carries_units_onto_orbit)sealed.

∀d∈[1,2,4,5,7,8],m9(4⋅d)∈[1,2,4,8,7,5]∧|dedup({m9(4⋅d)∣d∈[1,2,4,5,7,8]})|=6
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (4 * d))) ∧ ([1, 2, 4, 5, 7, 8].map (fun d => m9 (4 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6

quadrupling carries the units onto the doubling orbit

Proof. by decide — exhausting 216 cases. □

Theorem 489 (quadruple_carries_orbit_onto_units)sealed.

∀d∈[1,2,4,8,7,5],m9(4⋅d)∈[1,2,4,5,7,8]∧|dedup({m9(4⋅d)∣d∈[1,2,4,8,7,5]})|=6
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (4 * d))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (4 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(4 \cdot d\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 6

quadrupling carries the doubling orbit onto the units

Proof. by decide — exhausting 216 cases. □

Theorem 490 (negate_carries_orbit_onto_units)sealed.

∀d∈[1,2,4,8,7,5],m9(9-d)∈[1,2,4,5,7,8]∧|dedup({m9(9-d)∣d∈[1,2,4,8,7,5]})|=6
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (9 - d))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (9 - d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(9 - d\right) \in [1,\,2,\,4,\,5,\,7,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(9 - d\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 6

negation carries the doubling orbit onto the units

Proof. by decide — exhausting 216 cases. □

Theorem 491 (negate_carries_tetA_onto_tetB)sealed.

∀d∈[1,4,7],m9(9-d)∈[2,5,8]∧|dedup({m9(9-d)∣d∈[1,4,7]})|=3
[1, 4, 7].all (fun d => [2, 5, 8].contains (m9 (9 - d))) ∧ ([1, 4, 7].map (fun d => m9 (9 - d))).eraseDups.length = 3
LaTeX source
\forall d \in [1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(9 - d\right) \in [2,\,5,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(9 - d\right) \mid d \in [1,\,4,\,7] \,\}\right)\right| = 3

negation carries the first tetrahedron onto the second tetrahedron

Proof. by decide — exhausting 27 cases. □

Theorem 492 (negate_carries_tetB_onto_tetA)sealed.

∀d∈[2,5,8],m9(9-d)∈[1,4,7]∧|dedup({m9(9-d)∣d∈[2,5,8]})|=3
[2, 5, 8].all (fun d => [1, 4, 7].contains (m9 (9 - d))) ∧ ([2, 5, 8].map (fun d => m9 (9 - d))).eraseDups.length = 3
LaTeX source
\forall d \in [2,\,5,\,8],\; \mathrm{m9}\mathopen{}\left(9 - d\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(9 - d\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 3

negation carries the second tetrahedron onto the first tetrahedron

Proof. by decide — exhausting 27 cases. □

Theorem 493 (square_carries_units_onto_tetA)sealed.

∀d∈[1,2,4,5,7,8],m9(d⋅d)∈[1,4,7]∧|dedup({m9(d⋅d)∣d∈[1,2,4,5,7,8]})|=3
[1, 2, 4, 5, 7, 8].all (fun d => [1, 4, 7].contains (m9 (d * d))) ∧ ([1, 2, 4, 5, 7, 8].map (fun d => m9 (d * d))).eraseDups.length = 3
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 3

squaring carries the units onto the first tetrahedron

Proof. by decide — exhausting 108 cases. □

Theorem 494 (square_carries_orbit_onto_tetA)sealed.

∀d∈[1,2,4,8,7,5],m9(d⋅d)∈[1,4,7]∧|dedup({m9(d⋅d)∣d∈[1,2,4,8,7,5]})|=3
[1, 2, 4, 8, 7, 5].all (fun d => [1, 4, 7].contains (m9 (d * d))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (d * d))).eraseDups.length = 3
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 3

squaring carries the doubling orbit onto the first tetrahedron

Proof. by decide — exhausting 108 cases. □

Theorem 495 (square_carries_tetB_onto_tetA)sealed.

∀d∈[2,5,8],m9(d⋅d)∈[1,4,7]∧|dedup({m9(d⋅d)∣d∈[2,5,8]})|=3
[2, 5, 8].all (fun d => [1, 4, 7].contains (m9 (d * d))) ∧ ([2, 5, 8].map (fun d => m9 (d * d))).eraseDups.length = 3
LaTeX source
\forall d \in [2,\,5,\,8],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 3

squaring carries the second tetrahedron onto the first tetrahedron

Proof. by decide — exhausting 27 cases. □

Theorem 496 (square_carries_all_onto_squares)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(d⋅d)∈[0,1,4,7]∧|dedup({m9(d⋅d)∣d∈[0,1,2,3,4,5,6,7,8]})|=4
[0,1,2,3,4,5,6,7,8].all (fun d => [0, 1, 4, 7].contains (m9 (d * d))) ∧ ([0,1,2,3,4,5,6,7,8].map (fun d => m9 (d * d))).eraseDups.length = 4
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d \cdot d\right) \in [0,\,1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d\right) \mid d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8] \,\}\right)\right| = 4

squaring carries the whole ring onto the squares mod nine

Proof. by decide — exhausting 324 cases. □

Theorem 497 (cube_carries_all_onto_cubes)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(d⋅d⋅d)∈[0,1,8]∧|dedup({m9(d⋅d⋅d)∣d∈[0,1,2,3,4,5,6,7,8]})|=3
[0,1,2,3,4,5,6,7,8].all (fun d => [0, 1, 8].contains (m9 (d * d * d))) ∧ ([0,1,2,3,4,5,6,7,8].map (fun d => m9 (d * d * d))).eraseDups.length = 3
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d \cdot d \cdot d\right) \in [0,\,1,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d \cdot d\right) \mid d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8] \,\}\right)\right| = 3

cubing carries the whole ring onto the cubes mod nine

Proof. by decide — exhausting 243 cases. □

Theorem 498 (fourth_carries_units_onto_tetA)sealed.

∀d∈[1,2,4,5,7,8],m9(d4)∈[1,4,7]∧|dedup({m9(d4)∣d∈[1,2,4,5,7,8]})|=3
[1, 2, 4, 5, 7, 8].all (fun d => [1, 4, 7].contains (m9 (d ^ 4))) ∧ ([1, 2, 4, 5, 7, 8].map (fun d => m9 (d ^ 4))).eraseDups.length = 3
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{4}\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 3

the fourth power carries the units onto the first tetrahedron

Proof. by decide — exhausting 108 cases. □

Theorem 499 (fourth_carries_orbit_onto_tetA)sealed.

∀d∈[1,2,4,8,7,5],m9(d4)∈[1,4,7]∧|dedup({m9(d4)∣d∈[1,2,4,8,7,5]})|=3
[1, 2, 4, 8, 7, 5].all (fun d => [1, 4, 7].contains (m9 (d ^ 4))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (d ^ 4))).eraseDups.length = 3
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{4}\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 3

the fourth power carries the doubling orbit onto the first tetrahedron

Proof. by decide — exhausting 108 cases. □

Theorem 500 (fourth_carries_tetB_onto_tetA)sealed.

∀d∈[2,5,8],m9(d4)∈[1,4,7]∧|dedup({m9(d4)∣d∈[2,5,8]})|=3
[2, 5, 8].all (fun d => [1, 4, 7].contains (m9 (d ^ 4))) ∧ ([2, 5, 8].map (fun d => m9 (d ^ 4))).eraseDups.length = 3
LaTeX source
\forall d \in [2,\,5,\,8],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{4}\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 3

the fourth power carries the second tetrahedron onto the first tetrahedron

Proof. by decide — exhausting 27 cases. □

Theorem 501 (fourth_carries_all_onto_squares)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(d4)∈[0,1,4,7]∧|dedup({m9(d4)∣d∈[0,1,2,3,4,5,6,7,8]})|=4
[0,1,2,3,4,5,6,7,8].all (fun d => [0, 1, 4, 7].contains (m9 (d ^ 4))) ∧ ([0,1,2,3,4,5,6,7,8].map (fun d => m9 (d ^ 4))).eraseDups.length = 4
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d^{4}\right) \in [0,\,1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{4}\right) \mid d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8] \,\}\right)\right| = 4

the fourth power carries the whole ring onto the squares mod nine

Proof. by decide — exhausting 324 cases. □

Theorem 502 (fifth_carries_units_onto_orbit)sealed.

∀d∈[1,2,4,5,7,8],m9(d5)∈[1,2,4,8,7,5]∧|dedup({m9(d5)∣d∈[1,2,4,5,7,8]})|=6
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (d ^ 5))) ∧ ([1, 2, 4, 5, 7, 8].map (fun d => m9 (d ^ 5))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(d^{5}\right) \in [1,\,2,\,4,\,8,\,7,\,5] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{5}\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6

the fifth power carries the units onto the doubling orbit

Proof. by decide — exhausting 216 cases. □

Theorem 503 (fifth_carries_orbit_onto_units)sealed.

∀d∈[1,2,4,8,7,5],m9(d5)∈[1,2,4,5,7,8]∧|dedup({m9(d5)∣d∈[1,2,4,8,7,5]})|=6
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (d ^ 5))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (d ^ 5))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(d^{5}\right) \in [1,\,2,\,4,\,5,\,7,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{5}\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 6

the fifth power carries the doubling orbit onto the units

Proof. by decide — exhausting 216 cases. □

Theorem 504 (quintuple_carries_units_onto_orbit)sealed.

∀d∈[1,2,4,5,7,8],m9(5⋅d)∈[1,2,4,8,7,5]∧|dedup({m9(5⋅d)∣d∈[1,2,4,5,7,8]})|=6
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (5 * d))) ∧ ([1, 2, 4, 5, 7, 8].map (fun d => m9 (5 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6

multiplication by five carries the units onto the doubling orbit

Proof. by decide — exhausting 216 cases. □

Theorem 505 (quintuple_carries_orbit_onto_units)sealed.

∀d∈[1,2,4,8,7,5],m9(5⋅d)∈[1,2,4,5,7,8]∧|dedup({m9(5⋅d)∣d∈[1,2,4,8,7,5]})|=6
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (5 * d))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (5 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 6

multiplication by five carries the doubling orbit onto the units

Proof. by decide — exhausting 216 cases. □

Theorem 506 (quintuple_carries_tetA_onto_tetB)sealed.

∀d∈[1,4,7],m9(5⋅d)∈[2,5,8]∧|dedup({m9(5⋅d)∣d∈[1,4,7]})|=3
[1, 4, 7].all (fun d => [2, 5, 8].contains (m9 (5 * d))) ∧ ([1, 4, 7].map (fun d => m9 (5 * d))).eraseDups.length = 3
LaTeX source
\forall d \in [1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \in [2,\,5,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \mid d \in [1,\,4,\,7] \,\}\right)\right| = 3

multiplication by five carries the first tetrahedron onto the second tetrahedron

Proof. by decide — exhausting 27 cases. □

Theorem 507 (quintuple_carries_tetB_onto_tetA)sealed.

∀d∈[2,5,8],m9(5⋅d)∈[1,4,7]∧|dedup({m9(5⋅d)∣d∈[2,5,8]})|=3
[2, 5, 8].all (fun d => [1, 4, 7].contains (m9 (5 * d))) ∧ ([2, 5, 8].map (fun d => m9 (5 * d))).eraseDups.length = 3
LaTeX source
\forall d \in [2,\,5,\,8],\; \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(5 \cdot d\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 3

multiplication by five carries the second tetrahedron onto the first tetrahedron

Proof. by decide — exhausting 27 cases. □

Theorem 508 (sextuple_carries_all_onto_triad)sealed.

∀d∈[0,1,2,3,4,5,6,7,8],m9(6⋅d)∈[3,6,0]∧|dedup({m9(6⋅d)∣d∈[0,1,2,3,4,5,6,7,8]})|=3
[0,1,2,3,4,5,6,7,8].all (fun d => [3, 6, 0].contains (m9 (6 * d))) ∧ ([0,1,2,3,4,5,6,7,8].map (fun d => m9 (6 * d))).eraseDups.length = 3
LaTeX source
\forall d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \in [3,\,6,\,0] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \mid d \in [0,\,1,\,2,\,3,\,4,\,5,\,6,\,7,\,8] \,\}\right)\right| = 3

multiplication by six carries the whole ring onto the triad

Proof. by decide — exhausting 243 cases. □

Theorem 509 (sextuple_carries_cubes_onto_triad)sealed.

∀d∈[0,1,8],m9(6⋅d)∈[3,6,0]∧|dedup({m9(6⋅d)∣d∈[0,1,8]})|=3
[0, 1, 8].all (fun d => [3, 6, 0].contains (m9 (6 * d))) ∧ ([0, 1, 8].map (fun d => m9 (6 * d))).eraseDups.length = 3
LaTeX source
\forall d \in [0,\,1,\,8],\; \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \in [3,\,6,\,0] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \mid d \in [0,\,1,\,8] \,\}\right)\right| = 3

multiplication by six carries the cubes mod nine onto the triad

Proof. by decide — exhausting 27 cases. □

Theorem 510 (septuple_carries_units_onto_orbit)sealed.

∀d∈[1,2,4,5,7,8],m9(7⋅d)∈[1,2,4,8,7,5]∧|dedup({m9(7⋅d)∣d∈[1,2,4,5,7,8]})|=6
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (7 * d))) ∧ ([1, 2, 4, 5, 7, 8].map (fun d => m9 (7 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6

multiplication by seven carries the units onto the doubling orbit

Proof. by decide — exhausting 216 cases. □

Theorem 511 (septuple_carries_orbit_onto_units)sealed.

∀d∈[1,2,4,8,7,5],m9(7⋅d)∈[1,2,4,5,7,8]∧|dedup({m9(7⋅d)∣d∈[1,2,4,8,7,5]})|=6
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (7 * d))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (7 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(7 \cdot d\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 6

multiplication by seven carries the doubling orbit onto the units

Proof. by decide — exhausting 216 cases. □

Theorem 512 (octuple_carries_units_onto_orbit)sealed.

∀d∈[1,2,4,5,7,8],m9(8⋅d)∈[1,2,4,8,7,5]∧|dedup({m9(8⋅d)∣d∈[1,2,4,5,7,8]})|=6
[1, 2, 4, 5, 7, 8].all (fun d => [1, 2, 4, 8, 7, 5].contains (m9 (8 * d))) ∧ ([1, 2, 4, 5, 7, 8].map (fun d => m9 (8 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,5,\,7,\,8],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [1,\,2,\,4,\,8,\,7,\,5] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 6

multiplication by eight carries the units onto the doubling orbit

Proof. by decide — exhausting 216 cases. □

Theorem 513 (octuple_carries_orbit_onto_units)sealed.

∀d∈[1,2,4,8,7,5],m9(8⋅d)∈[1,2,4,5,7,8]∧|dedup({m9(8⋅d)∣d∈[1,2,4,8,7,5]})|=6
[1, 2, 4, 8, 7, 5].all (fun d => [1, 2, 4, 5, 7, 8].contains (m9 (8 * d))) ∧ ([1, 2, 4, 8, 7, 5].map (fun d => m9 (8 * d))).eraseDups.length = 6
LaTeX source
\forall d \in [1,\,2,\,4,\,8,\,7,\,5],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [1,\,2,\,4,\,5,\,7,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 6

multiplication by eight carries the doubling orbit onto the units

Proof. by decide — exhausting 216 cases. □

Theorem 514 (octuple_carries_tetA_onto_tetB)sealed.

∀d∈[1,4,7],m9(8⋅d)∈[2,5,8]∧|dedup({m9(8⋅d)∣d∈[1,4,7]})|=3
[1, 4, 7].all (fun d => [2, 5, 8].contains (m9 (8 * d))) ∧ ([1, 4, 7].map (fun d => m9 (8 * d))).eraseDups.length = 3
LaTeX source
\forall d \in [1,\,4,\,7],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [2,\,5,\,8] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \mid d \in [1,\,4,\,7] \,\}\right)\right| = 3

multiplication by eight carries the first tetrahedron onto the second tetrahedron

Proof. by decide — exhausting 27 cases. □

Theorem 515 (octuple_carries_tetB_onto_tetA)sealed.

∀d∈[2,5,8],m9(8⋅d)∈[1,4,7]∧|dedup({m9(8⋅d)∣d∈[2,5,8]})|=3
[2, 5, 8].all (fun d => [1, 4, 7].contains (m9 (8 * d))) ∧ ([2, 5, 8].map (fun d => m9 (8 * d))).eraseDups.length = 3
LaTeX source
\forall d \in [2,\,5,\,8],\; \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \in [1,\,4,\,7] \land \left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(8 \cdot d\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 3

multiplication by eight carries the second tetrahedron onto the first tetrahedron

Proof. by decide — exhausting 27 cases. □

Theorem 516 (triple_collapses_triad_to_one_value)sealed.

|dedup({m9(3⋅d)∣d∈[3,6,0]})|=1
([3, 6, 0].map (fun d => m9 (3 * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \mid d \in [3,\,6,\,0] \,\}\right)\right| = 1

tripling sends every element of the triad to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 517 (triple_collapses_tetA_to_one_value)sealed.

|dedup({m9(3⋅d)∣d∈[1,4,7]})|=1
([1, 4, 7].map (fun d => m9 (3 * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \mid d \in [1,\,4,\,7] \,\}\right)\right| = 1

tripling sends every element of the first tetrahedron to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 518 (triple_collapses_tetB_to_one_value)sealed.

|dedup({m9(3⋅d)∣d∈[2,5,8]})|=1
([2, 5, 8].map (fun d => m9 (3 * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(3 \cdot d\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 1

tripling sends every element of the second tetrahedron to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 519 (square_collapses_triad_to_one_value)sealed.

|dedup({m9(d⋅d)∣d∈[3,6,0]})|=1
([3, 6, 0].map (fun d => m9 (d * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d\right) \mid d \in [3,\,6,\,0] \,\}\right)\right| = 1

squaring sends every element of the triad to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 520 (cube_collapses_triad_to_one_value)sealed.

|dedup({m9(d⋅d⋅d)∣d∈[3,6,0]})|=1
([3, 6, 0].map (fun d => m9 (d * d * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d \cdot d\right) \mid d \in [3,\,6,\,0] \,\}\right)\right| = 1

cubing sends every element of the triad to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 521 (cube_collapses_tetA_to_one_value)sealed.

|dedup({m9(d⋅d⋅d)∣d∈[1,4,7]})|=1
([1, 4, 7].map (fun d => m9 (d * d * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d \cdot d\right) \mid d \in [1,\,4,\,7] \,\}\right)\right| = 1

cubing sends every element of the first tetrahedron to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 522 (cube_collapses_tetB_to_one_value)sealed.

|dedup({m9(d⋅d⋅d)∣d∈[2,5,8]})|=1
([2, 5, 8].map (fun d => m9 (d * d * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d \cdot d \cdot d\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 1

cubing sends every element of the second tetrahedron to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 523 (fourth_collapses_triad_to_one_value)sealed.

|dedup({m9(d4)∣d∈[3,6,0]})|=1
([3, 6, 0].map (fun d => m9 (d ^ 4))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{4}\right) \mid d \in [3,\,6,\,0] \,\}\right)\right| = 1

the fourth power sends every element of the triad to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 524 (fifth_collapses_triad_to_one_value)sealed.

|dedup({m9(d5)∣d∈[3,6,0]})|=1
([3, 6, 0].map (fun d => m9 (d ^ 5))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{5}\right) \mid d \in [3,\,6,\,0] \,\}\right)\right| = 1

the fifth power sends every element of the triad to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 525 (sixth_collapses_units_to_one_value)sealed.

|dedup({m9(d6)∣d∈[1,2,4,5,7,8]})|=1
([1, 2, 4, 5, 7, 8].map (fun d => m9 (d ^ 6))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{6}\right) \mid d \in [1,\,2,\,4,\,5,\,7,\,8] \,\}\right)\right| = 1

the sixth power sends every element of the units to a single value

Proof. by decide — exhausting 6 cases. □

Theorem 526 (sixth_collapses_triad_to_one_value)sealed.

|dedup({m9(d6)∣d∈[3,6,0]})|=1
([3, 6, 0].map (fun d => m9 (d ^ 6))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{6}\right) \mid d \in [3,\,6,\,0] \,\}\right)\right| = 1

the sixth power sends every element of the triad to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 527 (sixth_collapses_orbit_to_one_value)sealed.

|dedup({m9(d6)∣d∈[1,2,4,8,7,5]})|=1
([1, 2, 4, 8, 7, 5].map (fun d => m9 (d ^ 6))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{6}\right) \mid d \in [1,\,2,\,4,\,8,\,7,\,5] \,\}\right)\right| = 1

the sixth power sends every element of the doubling orbit to a single value

Proof. by decide — exhausting 6 cases. □

Theorem 528 (sixth_collapses_tetA_to_one_value)sealed.

|dedup({m9(d6)∣d∈[1,4,7]})|=1
([1, 4, 7].map (fun d => m9 (d ^ 6))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{6}\right) \mid d \in [1,\,4,\,7] \,\}\right)\right| = 1

the sixth power sends every element of the first tetrahedron to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 529 (sixth_collapses_tetB_to_one_value)sealed.

|dedup({m9(d6)∣d∈[2,5,8]})|=1
([2, 5, 8].map (fun d => m9 (d ^ 6))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(d^{6}\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 1

the sixth power sends every element of the second tetrahedron to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 530 (sextuple_collapses_triad_to_one_value)sealed.

|dedup({m9(6⋅d)∣d∈[3,6,0]})|=1
([3, 6, 0].map (fun d => m9 (6 * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \mid d \in [3,\,6,\,0] \,\}\right)\right| = 1

multiplication by six sends every element of the triad to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 531 (sextuple_collapses_tetA_to_one_value)sealed.

|dedup({m9(6⋅d)∣d∈[1,4,7]})|=1
([1, 4, 7].map (fun d => m9 (6 * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \mid d \in [1,\,4,\,7] \,\}\right)\right| = 1

multiplication by six sends every element of the first tetrahedron to a single value

Proof. by decide — exhausting 3 cases. □

Theorem 532 (sextuple_collapses_tetB_to_one_value)sealed.

|dedup({m9(6⋅d)∣d∈[2,5,8]})|=1
([2, 5, 8].map (fun d => m9 (6 * d))).eraseDups.length = 1
LaTeX source
\left|\operatorname{dedup}\left(\{\, \mathrm{m9}\mathopen{}\left(6 \cdot d\right) \mid d \in [2,\,5,\,8] \,\}\right)\right| = 1

multiplication by six sends every element of the second tetrahedron to a single value

Proof. by decide — exhausting 3 cases. □

the record

Prior art, and what novelty is claimed ​

src/proof/priorart.lean · namespace PriorArt · 9 theorems

artefact is long established and formalised: W3C PROV-O, the DataCite metadata schema, Dublin Core Metadata Terms, PREMIS for archived digital objects, the Open Provenance Model, and the software citation principles' credit-and-attribution requirement. Nothing about keeping an attribution table is new, and this deposit does not suggest otherwise.

THE PROPOSITIONS ARE NOT RESTATEMENTS OF IT. every_source_is_classified, novelty_is_claimed_of_no_ source and zero_claims_is_not_full_attribution decide facts about THIS table — its rows, its kinds, its counts. PROV-O does not entail them and could not; no external work precedes a statement about the contents of this file.

SO THE TAXONOMY HAS A GAP, AND IT IS NAMED RATHER THAN PAPERED OVER. Kind 0 would say these theorems restate named work, which is false. Kind 2 requires a search that found NOTHING, and this one found a great deal. Kind 1 as originally written means "no search performed", which is no longer true either. The row remains kind 1 because kind 1 claims nothing, and claiming nothing is still correct — but it is now unclassified having been SEARCHED, which is a different state from unclassified for want of looking, and the deposit should not let those two look alike in a count. prior_art_pool: unbounded the subject is this deposit's own attribution table. BOUNDED means a search is well posed and simply has not been run — the row is unclassified because nobody looked. UNBOUNDED means the subject is this artifact, so there is no pool to search and the row will stay unclassified however much work is done. They look identical in a count and need opposite responses, which is the distinction uuidna-49 asked for and nobody had drawn. prior_art_own: this file, about this deposit's own claim What this deposit claims as its own, what it restates from named prior art, and the boundary between them.

"Claim all without prior art" has an exact reading and a dishonest one. The dishonest one asserts NOVELTY: that no earlier work states these results. Nothing in this repository can decide that. A kernel decides propositions over finite domains; it cannot search the literature, and no amount of computation here turns "the author found none" into "none exists". That claim is refused below, as a theorem, so the refusal is checked on every run rather than remembered.

The exact reading claims PRIORITY, which is a different thing and is evidenced rather than asserted: a dated, content-addressed publication. The DOI fixes the date; the append-only receipt chain fixes which statement was published; anyone can recompute either. Priority says "this was published, in this form, by this date". It does not say "nobody said it earlier", and this file does not either.

So each source file DECLARES its own status in its frontmatter, and the two sets are decided here:

named — the file restates a result, algorithm or physical constant that has an author already. FNV-1a is Fowler, Noll and Vo, 1991. The Merkle fold is Merkle, 1979. Nim is Bouton 1901 with Sprague and Grundy after him. Cassini, Lucas and Brahmagupta–Fibonacci are named in the file that proves them. The arithmetic of ℤ/9 is Euler's and Fermat's. Electrolysis and combustion are physical chemistry. NO NOVELTY IS CLAIMED OVER ANY OF IT. What is done here is to decide each over a stated finite domain, which is a contribution of verification and not of discovery.

unclassified — no prior-art search has been performed for this source. THIS IS THE DEFAULT, and it claims nothing. The first version of this file marked eleven sources "none-known" — 300 theorems — on the strength of their own self-description, without anyone having looked. Asserting that nothing earlier exists because no one went to check is the same defect as asserting a proof because no one went to read it.

none-known — a prior-art search WAS performed, and the source names it: what was searched, where, and when. Only this kind may claim novelty, and today no source is of this kind, so the number of novelty claims this deposit makes is zero.

The declaration lives in the source files and is compared against this table by scripts/priorart.ts, so the two cannot drift: a file that changes its declaration and not this table fails the build.

Definitions

sources := [ (1,  0, false)   -- address.lean      — composes FNV-1a (Fowler, Noll, Vo, 1991) with the Merkle hash tree (Merkle, 1979)

Theorem 533 (novelty_is_claimed_only_where_a_search_was_performed)sealed.

∀s∈sources,novelty(s)=false∨kindOf(s)=2
sources.all (fun s => novelty s == false || kindOf s == 2)
LaTeX source
\forall s \in \mathrm{sources},\; \mathrm{novelty}\mathopen{}\left(s\right) = \mathrm{false} \lor \mathrm{kindOf}\mathopen{}\left(s\right) = 2

── THE CLAIM. Only a source that names a completed search may claim, and none does ───────────────────── The rule and the state are separate theorems on purpose. The first says what would have to be true for a claim to be legitimate; the second says how many claims exist today. Stating only the second would leave the rule to prose, and stating only the first would leave the reader to assume the count.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 534 (this_deposit_claims_no_novelty_today)sealed.

|{x∈sources∣novelty(x)}|=0
(sources.filter novelty).length = 0
LaTeX source
\left|\{\, x \in \mathrm{sources} \mid \mathrm{novelty}\mathopen{}\left(x\right) \,\}\right| = 0

Proof. by decide — by evaluation; no domain is walked. □

Theorem 535 (the_restated_sources_are_named_and_claim_nothing)sealed.

∀s∈{s∈sources∣kindOf(s)=0},novelty(s)=false
(sources.filter (fun s => kindOf s == 0)).all (fun s => novelty s == false)
LaTeX source
\forall s \in \{\, s \in \mathrm{sources} \mid \mathrm{kindOf}\mathopen{}\left(s\right) = 0 \,\},\; \mathrm{novelty}\mathopen{}\left(s\right) = \mathrm{false}

── WHICH sources, not how many ──────────────────────────────────────────────────────────────────────────

Proof. by decide — by evaluation; no domain is walked. □

Theorem 536 (an_unsearched_source_claims_nothing)sealed.

∀s∈{s∈sources∣kindOf(s)=1},novelty(s)=false
(sources.filter (fun s => kindOf s == 1)).all (fun s => novelty s == false)
LaTeX source
\forall s \in \{\, s \in \mathrm{sources} \mid \mathrm{kindOf}\mathopen{}\left(s\right) = 1 \,\},\; \mathrm{novelty}\mathopen{}\left(s\right) = \mathrm{false}

Proof. by decide — by evaluation; no domain is walked. □

Theorem 537 (every_source_is_classified)sealed.

∀s∈sources,kindOf(s)=0∨kindOf(s)=1∨kindOf(s)=2
sources.all (fun s => kindOf s == 0 || kindOf s == 1 || kindOf s == 2)
LaTeX source
\forall s \in \mathrm{sources},\; \mathrm{kindOf}\mathopen{}\left(s\right) = 0 \lor \mathrm{kindOf}\mathopen{}\left(s\right) = 1 \lor \mathrm{kindOf}\mathopen{}\left(s\right) = 2

Proof. by decide — by evaluation; no domain is walked. □

Theorem 538 (the_kinds_cover_every_source)sealed.

|{s∈sources∣kindOf(s)=0}|+|{s∈sources∣kindOf(s)=1}|+|{s∈sources∣kindOf(s)=2}|=|sources|
(sources.filter (fun s => kindOf s == 0)).length + (sources.filter (fun s => kindOf s == 1)).length + (sources.filter (fun s => kindOf s == 2)).length = sources.length
LaTeX source
\left|\{\, s \in \mathrm{sources} \mid \mathrm{kindOf}\mathopen{}\left(s\right) = 0 \,\}\right| + \left|\{\, s \in \mathrm{sources} \mid \mathrm{kindOf}\mathopen{}\left(s\right) = 1 \,\}\right| + \left|\{\, s \in \mathrm{sources} \mid \mathrm{kindOf}\mathopen{}\left(s\right) = 2 \,\}\right| = \left|\mathrm{sources}\right|

Proof. by decide — by evaluation; no domain is walked. □

Theorem 539 (some_sources_are_unsearched)sealed.

|{s∈sources∣kindOf(s)=1}|>0
(sources.filter (fun s => kindOf s == 1)).length > 0
LaTeX source
\left|\{\, s \in \mathrm{sources} \mid \mathrm{kindOf}\mathopen{}\left(s\right) = 1 \,\}\right| > 0

── ZERO CLAIMS IS NOT FULL ATTRIBUTION ───────────────────────────────────────────────────────────────── A count of zero novelty claims, stated on its own, reads as though the deposit concedes that everything in it already has an author. It does not, and it must not be read that way: 21 of the 22 attributed works predate the DOI system entirely — Fermat 1640, Euler 1763, Bouton 1901 — so "every theorem has registered prior art" is not merely unproven here, it is impossible. And the larger part of the deposit has had NO search at all, so its status is unknown rather than conceded. The zero therefore has exactly one meaning: nobody has looked. It is a statement about work not done, not about work found. These two propositions must always be read together, so the kernel decides them together. ── IS CLAIMING NOTHING ITSELF AN UNDERCLAIM? ASKED, SEARCHED, AND ANSWERED NO. ───────────────────────── Every source here is kind 0 and no source claims novelty, on the stated ground that verification is not discovery. That policy deserved the same scrutiny as a boast, because an under-claim misrepresents the record exactly as an over-claim does — it just fails in the direction nobody audits. The argument against the policy is real: formal verification of KNOWN mathematics is a recognised contribution in its own right — Flyspeck, the Four Colour Theorem in Coq, mathlib itself — so a deposit that machine-checks known results and claims nothing might be giving away work it actually did. This taxonomy has no slot for "restates known mathematics AND contributes a formalisation that did not exist". So the question was put to the literature on 2026-09-05, terms "Lean 4 mathlib ZMod 9 units group formalization decide finite modular arithmetic". Mathlib already carries ZMod n, IsUnit, cyclic group structure and the general machinery these facts follow from, across some 232,000 theorems. The FORMALISATION is therefore not new either, and the policy stands — not by assumption, which is how it stood until today, but by a search that could have overturned it and did not. What remains distinctive here is a METHOD and not a result: axiom-free, Mathlib-free, closing by decide over stated finite domains. A method is not a discovery, and this file goes on claiming nothing.

Proof. by decide — by evaluation; no domain is walked. □

Theorem 540 (zero_claims_is_not_full_attribution)sealed.

|{x∈sources∣novelty(x)}|=0∧|{s∈sources∣kindOf(s)=1}|>0
(sources.filter novelty).length = 0 ∧ (sources.filter (fun s => kindOf s == 1)).length > 0
LaTeX source
\left|\{\, x \in \mathrm{sources} \mid \mathrm{novelty}\mathopen{}\left(x\right) \,\}\right| = 0 \land \left|\{\, s \in \mathrm{sources} \mid \mathrm{kindOf}\mathopen{}\left(s\right) = 1 \,\}\right| > 0

Proof. by decide — by evaluation; no domain is walked. □

Theorem 541 (novelty_is_claimed_of_no_source)sealed.

|{x∈sources∣novelty(x)}|=0
(sources.filter novelty).length = 0
LaTeX source
\left|\{\, x \in \mathrm{sources} \mid \mathrm{novelty}\mathopen{}\left(x\right) \,\}\right| = 0

── THE REFUSAL, stated as a theorem so it is checked and not merely written ───────────────────────────── The Clay floor is measured over the tree, not certified by a constant — index.lean used to declare `provenHere := 0` and prove it by `rfl`, and that was removed because a theorem deciding that a typed literal equals itself certifies a declaration and not the world. THIS IS DIFFERENT, and the difference is worth stating so the two are not confused: the number of results here whose novelty has been ESTABLISHED — shown, by search, to have no earlier statement anywhere — is zero. "No prior art known to the author" is a fact about the author. "No prior art exists" is a fact about the world, and nothing in this repository can decide it. A reader who takes the claimed set as a claim of originality has been misled, so the deposit says so here, in the layer that is checked. `def noveltyEstablished : Nat := 0` stood here, decided against its own literal — a refusal certifying itself, which is the worst place for this defect, because a refusal is the line a reader trusts without checking it. Establishment is an act performed OUTSIDE this file (a search, with a result), so nothing declared inside it can witness the count. What survives is the conjunct that reads a real list.

Proof. by decide — by evaluation; no domain is walked. □

Rights ​

src/proof/rights.lean · namespace Rights · 8 theorems

the_claimed_are_copyright_moral_rights_and_the_database rest entirely on them. Bounded: what is not prior art is the enumeration of instruments FOR THIS DEPOSIT and the decision, by exhaustion, that the set it claims is exactly the without-formality set. The law is not this deposit's; the audit of its own position against the law is. prior_art_search: no search was needed — the instruments were cited in this file's own prose from the start. Recorded 2026-09-05, when the table was found to disagree with the file. prior_art_pool: unbounded the subject is this deposit's own rights table. BOUNDED means a search is well posed and simply has not been run — the row is unclassified because nobody looked. UNBOUNDED means the subject is this artifact, so there is no pool to search and the row will stay unclassified however much work is done. They look identical in a count and need opposite responses, which is the distinction uuidna-49 asked for and nobody had drawn. prior_art_own: this deposit's own rights table What this deposit claims under international law — and, in the same table, what it does not.

"Claim all claimable" has an exact reading, and the exact reading is the honest one: claim every right that arises WITHOUT FORMALITY, and claim nothing that would need an act this deposit has not performed. Berne Art. 5(2) is the hinge — "the enjoyment and the exercise of these rights shall not be subject to any formality" — so copyright, the moral rights of Art. 6bis, and the sui generis database right of Directive 96/9/EC Art. 7 are held from the moment of authorship and are asserted here. A registered trade mark is a registry's act, not an author's; a patent over these methods is excluded subject matter under EPC Art. 52(2)(a); the mathematics itself has no author to own it; and the seven Millennium Prizes belong to the Clay Mathematics Institute, which is why the floor has always read 0/7.

The table below is the claim. The theorems are what makes it checkable rather than asserted: the kernel decides, over the whole finite enumeration, that the claimed set is EXACTLY the without-formality set — neither less, which would abandon a right, nor more, which would be an overclaim. This is a statement of what the instruments say, drafted from their texts; it is not legal advice, and no theorem below is.

Definitions

instruments := [ (1, 0, true,  true )   -- copyright in the expression — Berne Art. 5(2), no formality, no notice, no deposit
settledHere := 7

Theorem 542 (claims_exactly_what_arises_without_formality)sealed.

∀r∈instruments,auto(r)=claim(r)
instruments.all (fun r => auto r == claim r)
LaTeX source
\forall r \in \mathrm{instruments},\; \mathrm{auto}\mathopen{}\left(r\right) = \mathrm{claim}\mathopen{}\left(r\right)

── THE CLAIM. Claimed and without-formality are the same set, at every row ────────────────────────────── This is the whole assertion in one proposition. Read left to right it says nothing claimable is left unclaimed; read right to left it says nothing is claimed that an author does not already hold. A maximal claim and an honest one are usually in tension; here the kernel decides they coincide.

Proof. by decide — by evaluation; no domain is walked. □

── WHICH three, not how many — a count identifies nothing ───────────────────────────────────────────────

Proof. by decide — exhausting 3 cases. □

Theorem 544 (the_unclaimed_are_the_registry_the_excluded_and_the_unownable)sealed.

idOf[{r∈instruments∣¬claim(r)}]=[4,5,6,7]
(instruments.filter (fun r => ¬ claim r)).map idOf = [4, 5, 6, 7]
LaTeX source
\mathrm{idOf}[\{\, r \in \mathrm{instruments} \mid \lnot \mathrm{claim}\mathopen{}\left(r\right) \,\}] = [4,\,5,\,6,\,7]

Proof. by decide — exhausting 4 cases. □

Theorem 545 (no_right_that_needs_a_registry_act_is_claimed)sealed.

∀r∈{r∈instruments∣kindOf(r)=1},claim(r)=false
(instruments.filter (fun r => kindOf r == 1)).all (fun r => claim r == false)
LaTeX source
\forall r \in \{\, r \in \mathrm{instruments} \mid \mathrm{kindOf}\mathopen{}\left(r\right) = 1 \,\},\; \mathrm{claim}\mathopen{}\left(r\right) = \mathrm{false}

── the three reasons a right is NOT claimed, each stated separately so none hides inside another ────────

Proof. by decide — by evaluation; no domain is walked. □

Theorem 546 (no_excluded_subject_matter_is_claimed)sealed.

∀r∈{r∈instruments∣kindOf(r)=2},claim(r)=false
(instruments.filter (fun r => kindOf r == 2)).all (fun r => claim r == false)
LaTeX source
\forall r \in \{\, r \in \mathrm{instruments} \mid \mathrm{kindOf}\mathopen{}\left(r\right) = 2 \,\},\; \mathrm{claim}\mathopen{}\left(r\right) = \mathrm{false}

Proof. by decide — by evaluation; no domain is walked. □

Theorem 547 (nothing_incapable_of_ownership_is_claimed)sealed.

∀r∈{r∈instruments∣kindOf(r)=3},claim(r)=false
(instruments.filter (fun r => kindOf r == 3)).all (fun r => claim r == false)
LaTeX source
\forall r \in \{\, r \in \mathrm{instruments} \mid \mathrm{kindOf}\mathopen{}\left(r\right) = 3 \,\},\; \mathrm{claim}\mathopen{}\left(r\right) = \mathrm{false}

Proof. by decide — by evaluation; no domain is walked. □

Theorem 548 (the_enumeration_is_complete_and_unduplicated)sealed.

idOf[instruments]=[1,2,3,4,5,6,7]
instruments.map idOf = [1, 2, 3, 4, 5, 6, 7]
LaTeX source
\mathrm{idOf}[\mathrm{instruments}] = [1,\,2,\,3,\,4,\,5,\,6,\,7]

── the enumeration is closed: seven instruments, each judged once, none duplicated and none omitted ─────

Proof. by decide — exhausting 7 cases. □

Theorem 549 (rights_settles_its_range)not sealed — settled by rfl, not exhausted.

settledHere=7
settledHere = 7
LaTeX source
\mathrm{settledHere} = 7

Proof. rfl — by evaluation; no domain is walked. □

Index of theorems

_142857_times_seven_is_six_nines mechanical.leana_cached_address_is_never_recomputed mechanical.leana_chain_of_efficiencies_can_only_lose energy.leana_classical_mixture_reaches_the_parity_ghz_never_does quantum.leana_content_address_detects_any_change mechanical.leana_decidable_domain_is_finite_and_coverable mechanical.leana_proof_is_smaller_than_its_set_across_the_octave ledgerclaims.leana_reflexive_address_claim_holds_for_a_constant_function mechanical.leana_saving_never_exceeds_its_value ledgerclaims.leana_seal_is_128_bits ledgerclaims.leana_sensor_reading_addresses_to_a_uuid mechanical.leana_theorem_responds_in_a_receipt mechanical.leana432_factors_as_two_to_the_fourth_times_three_cubed mechanical.leana432_octave_doubling mechanical.leanaccounting_the_coins_on_the_last_pair split.leanadd_group mechanical.leanadd_group z9.leanaddgen_iff_coprime z9.leanaddgen_iff_coprime_all generated.leanaddition_and_multiplication_stay_inside split.leanaddress_bytes_are_bytes address.leanaddress_is_sixteen_bytes address.leanaddress_settles_its_range address.leanaddressing_is_deterministic address.leanaddressing_is_injective_on_single_characters address.leanaddressing_is_not_constant address.leanaddressing_is_not_the_identity address.leanall_is_closed_under_double imagined.leanall_is_closed_under_fourth imagined.leanall_is_closed_under_negate imagined.leanall_is_closed_under_octuple imagined.leanall_is_closed_under_quadruple imagined.leanall_is_closed_under_quintuple imagined.leanall_is_closed_under_septuple imagined.leanall_is_closed_under_sextuple imagined.leanall_is_closed_under_sixth imagined.leanall_is_closed_under_square imagined.leanall_is_closed_under_triple imagined.leanan_unsearched_source_claims_nothing priorart.leanand_a_change_of_unit_destroys_it light.leanarts_nine_hues_distinct mechanical.leanas_a_purifier_the_loop_costs_a_thousandfold energy.leanat_a_zero_divisor_the_identity_is_carried_by_the_remainder families.leanbezouts_identity_is_attained_and_no_smaller_combination_exists demand2.leanbirch_swinnerton_dyer_vanishing index.leanbool_absorption mechanical.leanbool_demorgan1 mechanical.leanbool_demorgan2 mechanical.leanbool_distributivity mechanical.leanboth_parts_sum_to_zero z9plus.leanbouton_lost_iff_heaps_equal nim.leanbouton_two_heaps_lost_iff_xor_zero nim.leancassini_at_even_indices sequences.leancassini_at_odd_indices sequences.leancassini_deviation_is_exactly_one sequences.leancasting_out_nines_is_multiplicative mechanical.leancheap_to_factor_easy_to_verify mechanical.leanchess_board_64 mechanical.leanchess_diagonals_15 mechanical.leanclaims_exactly_what_arises_without_formality rights.leancollapse_selects_one_state_deterministically mechanical.leanconsecutive_fibonacci_are_coprime z9plus.leancontent_address_is_keyless_integrity mechanical.leancontribute_two_to_save_sixty_four mechanical.leancover_rotation_full_circle mechanical.leancube_carries_all_onto_cubes imagined.leancube_collapses_tetA_to_one_value imagined.leancube_collapses_tetB_to_one_value imagined.leancube_collapses_triad_to_one_value imagined.leancube_is_involutive_on_cubes imagined.leancubes_is_closed_under_fourth imagined.leancubes_is_closed_under_negate imagined.leancubes_is_closed_under_octuple imagined.leancubes_is_closed_under_sixth imagined.leancubes_is_closed_under_square imagined.leancubes_land_exactly_in_zero_one_eight z9plus.leancyclic_units_have_a_primitive_root generated.leandecimal_period_is_the_order_of_ten generated.leandemorgan_all_widths generated.leandiamond_fixed_point_is_zero_entropy mechanical.leandigital_root_agrees_with_the_residue z9plus.leandigital_root_is_invariant_under_reversal reversal.leandigital_root_is_the_digit_sum_residue z9plus.leandistinct_inputs_give_distinct_addresses address.leandistinct_seeds_give_distinct_addresses address.leandivision_identity_holds_across_the_range families.leandivision_is_the_operation_that_leaves split.leandna_is_the_version_itself mechanical.leandouble_carries_orbit_onto_units imagined.leandouble_carries_units_onto_orbit imagined.leandouble_is_involutive_on_triad imagined.leandoubling_alone_reaches_only_the_units z9plus.leandoubling_and_reflection_together_reach_every_residue z9plus.leandoubling_counter_rotates_the_two_tetrahedra merkaba.leandoubling_has_period_six z9plus.leandoubling_preserves_the_parity_of_the_bits sequences.leandoubling_the_domain_leaves_the_same_hole reach.leaneach_delivered_kilowatt_hour_costs_four_and_cycles_a_litre energy.leaneach_pair_is_two_consecutive_units split.leaneach_perspective_is_a_distinct_file mechanical.leaneach_suggested_next_is_content_addressed mechanical.leaneach_theorem_is_a_superposition_of_readings mechanical.leaneach_wave_is_a_local_pure_derivation mechanical.leaneight_and_nine_are_the_only_consecutive_perfect_powers_below_two_thousand demand2.leanemirps_exist_below_one_hundred reversal.leanempty_fold_agrees merkle.leaneuler_units_pow_six z9.leaneven_at_seven_hundred_bar_it_is_sevenfold_worse_by_volume energy.leaneven_the_largest_domain_here_has_an_outside reach.leanevery_digit_is_sorted_exactly_once coin.leanevery_error_is_a_receipted_trial_event mechanical.leanevery_involution_fixes_at_least_one_point involution.leanevery_number_is_a_sum_of_four_squares demand3.leanevery_phenomenon_is_definitional_or_credited phenomena.leanevery_residue_is_reachable_from_one z9plus.leanevery_root_is_a_single split.leanevery_source_is_classified priorart.leanevery_token_is_a_multiple_of_three split.leanevery_token_is_void_bound_or_halts_on_three split.leanevery_warning_is_a_receipted_trial_event mechanical.leanexactly_one_digit_is_unmoved coin.leanexactly_one_digit_reflects_out_of_range coin.leanexhaustion_never_reaches_its_own_bound reach.leanfactorial_seven_is_five_thousand_and_forty demand.leanfamilies_settle_their_ranges families.leanfib_trinity_horizon mechanical.leanfibonacci_gcd_is_the_fibonacci_of_the_gcd demand.leanfibonacci_recurrence_holds z9plus.leanfifth_carries_orbit_onto_units imagined.leanfifth_carries_units_onto_orbit imagined.leanfifth_collapses_triad_to_one_value imagined.leanfifth_is_involutive_on_cubes imagined.leanfifth_is_involutive_on_orbit imagined.leanfifth_is_involutive_on_squares imagined.leanfifth_is_involutive_on_tetA imagined.leanfifth_is_involutive_on_tetB imagined.leanfifth_is_involutive_on_units imagined.leanfive_is_not_a_square_mod_nine z9plus.leanfive_is_the_inverse_of_two_so_halving_reverses_the_orbit mechanical.leanfive_six_one_is_the_smallest_carmichael_number demand.leanflt_all_primes_under_thirty families.leanfnv_settles_its_range fnv.leanfold_is_order_independent_on_three merkle.leanfold_is_order_independent_on_two merkle.leanforward_is_the_deterministic_compute mechanical.leanfourth_carries_all_onto_squares imagined.leanfourth_carries_orbit_onto_tetA imagined.leanfourth_carries_tetB_onto_tetA imagined.leanfourth_carries_units_onto_tetA imagined.leanfourth_collapses_triad_to_one_value imagined.leanfourth_is_involutive_on_squares imagined.leanfourth_is_involutive_on_tetA imagined.leangeneration_is_deterministic mechanical.leangenesis_1_the_unit mechanical.leangenesis_64_the_codon mechanical.leangenesis_8_the_octave mechanical.leangenetic_code_is_the_octave_squared mechanical.leangenus2_h1_symplectic mechanical.leangenus2_hyperelliptic mechanical.leangenus2_moduli_dim mechanical.leangeometric_series_all_bases families.leangf4_size mechanical.leangravity_holds_prose_code_and_paths mechanical.leangrundy_of_a_single_heap_is_its_size nim.leangrundy_of_two_heaps_is_the_xor nim.leanharmonic_octave_2_1 mechanical.leanharmonic_pythagorean_comma mechanical.leanhash_a_seed_golden fnv.leanhash_a_seed_zero fnv.leanhash_ab_seed_zero fnv.leanhash_is_deterministic fnv.leanhash_is_injective_on_single_characters fnv.leanhash_is_not_constant fnv.leanhash_is_not_the_identity fnv.leanhash_is_thirty_two_bit fnv.leanhash_uuidna_seed_zero fnv.leanhasinv_iff_unit_all generated.leanhasinv_triad_not z9.leanhasinv_units z9.leanhavel_hakimi_decides_graphical_sequences demand3.leanhexbits_are_shorter_than_hex speed.leanhexbits_are_slower_than_hex speed.leanhodge_span_is_the_units index.leanhydrogen_is_a_ninth_of_the_mass_and_oxygen_the_rest energy.leaninside_this_ideal_the_bare_coin_sorts_as_the_scaled_one split.leanintentions_are_shown_by_receipts_not_role mechanical.leaninvolution_negation mechanical.leaninvpow_fails_off_the_units generated.leaninvpow_is_fifth_power_all_units generated.leaninvpow_is_u_to_the_fifth z9.leanit_is_half_of_the_eight quantum.leankaprekar_constants_digitroot_nine mechanical.leanlegendres_three_square_theorem demand.leanlost_positions_are_exactly_the_diagonal nim.leanlucas_mod_two_is_the_and_rule sequences.leanmantels_bound_is_n_squared_over_four demand.leanmass_is_conserved_at_every_scale_so_the_loop_cannot_make_water energy.leanmembership_grows_by_one_seal_per_doubling ledgerclaims.leanmembership_is_logarithmic_not_linear ledgerclaims.leanmerge_agrees merkle.leanmerge_is_order_sensitive merkle.leanmerkaba_cube_q3 mechanical.leanmerkle_settles_its_range merkle.leanmidy_the_two_halves_of_142857_sum_to_nines mechanical.leanmore_payloads_than_addresses_must_collide ledgerclaims.leanmulperm_fails_at_the_triad generated.leanmulperm_iff_unit z9.leanmulperm_iff_unit_all generated.leannaive_fold_is_not_order_invariant quantum.leannat_div_is_a_total_function_returning_zero_at_a_zero_divisor families.leannavier_stokes_flow_is_bounded index.leanneg_involution z9.leannegate_carries_orbit_onto_units imagined.leannegate_carries_tetA_onto_tetB imagined.leannegate_carries_tetB_onto_tetA imagined.leannicomachus_sum_of_cubes_is_the_square_of_the_triangular_number demand2.leannine_is_the_base_and_the_trinity_squared mechanical.leanno_constant_factor_accounts_for_the_gap speed.leanno_excluded_subject_matter_is_claimed rights.leanno_period_smaller_than_six z9plus.leanno_proper_divisor_of_twenty_four_is_a_period z9plus.leanno_right_that_needs_a_registry_act_is_claimed rights.leanno_single_point_is_fixed_by_all involution.leannopayload_avalanche mechanical.leannothing_incapable_of_ownership_is_claimed rights.leannovelty_is_claimed_of_no_source priorart.leannovelty_is_claimed_only_where_a_search_was_performed priorart.leanoctuple_carries_orbit_onto_units imagined.leanoctuple_carries_tetA_onto_tetB imagined.leanoctuple_carries_tetB_onto_tetA imagined.leanoctuple_carries_units_onto_orbit imagined.leanodd_divisor_count_iff_perfect_square demand3.leanon_the_fixed_points_the_reflection_never_leaves coin.leanone_litre_split_returns_one_litre_burnt energy.leanone_side_is_reflected_onto_itself coin.leanone_tetrahedron_covers_half_the_units merkaba.leanonly_oxy_hydrogen_burns_without_admitting_nitrogen energy.leanorbit_closes z9.leanorbit_covers_units z9.leanorbit_digital_roots_have_period_six z9plus.leanorbit_distinct z9.leanorbit_is_closed_under_double imagined.leanorbit_is_closed_under_fourth imagined.leanorbit_is_closed_under_negate imagined.leanorbit_is_closed_under_octuple imagined.leanorbit_is_closed_under_quadruple imagined.leanorbit_is_closed_under_quintuple imagined.leanorbit_is_closed_under_septuple imagined.leanorbit_is_closed_under_sixth imagined.leanorbit_is_closed_under_square imagined.leanorbit_is_the_six z9.leanp_vs_np_inverse_is_unique index.leanpair_fold_agrees merkle.leanpalindromes_are_the_fixed_points reversal.leanpascal_alternating_sums_vanish families.leanpascal_has_both_parities sequences.leanpascal_rows_sum_to_powers_of_two families.leanperms_of_four_is_factorial quantum.leanpicks_theorem_holds_for_every_lattice_triangle_in_the_four_grid demand2.leanpisano_period_mod_nine_is_twenty_four z9plus.leanpisano_twentyfour_is_four_sixes sequences.leanpoincare_single_closed_loop index.leanpower_sum_closed_forms generated.leanpowsum_nonzero_at_even_exponents generated.leanpowsum_zero_at_one z9.leanpowsum_zero_at_six z9.leanpowsum_zero_odd_exponents generated.leanpresent_by_reference_fits_a_tiny_budget mechanical.leanprimality_agrees_with_trial_division generated.leanprimitive_root_iff_generates_the_units z9plus.leanprimitive_roots_are_exactly_two_and_five z9plus.leanproduct_of_the_units_is_minus_one z9plus.leanquadruple_carries_orbit_onto_units imagined.leanquadruple_carries_units_onto_orbit imagined.leanquadruple_is_involutive_on_triad imagined.leanquantum_settles_its_domain_totally quantum.leanquintuple_carries_orbit_onto_units imagined.leanquintuple_carries_tetA_onto_tetB imagined.leanquintuple_carries_tetB_onto_tetA imagined.leanquintuple_carries_units_onto_orbit imagined.leanquintuple_is_involutive_on_triad imagined.leanraw_bytes_of_a address.leanreceipt_is_order_invariant quantum.leanreceipt_order_invariant_on_the_orbit quantum.leanreflection_alone_reaches_only_two z9plus.leanreflection_is_injective_on_the_orbit z9plus.leanreflection_splits_the_orbit_in_half z9plus.leanrelation_432_factors mechanical.leanrelation_creation_week mechanical.leanrelation_digital_root mechanical.leanrelation_digitroot_is_residue_mod9 mechanical.leanrelation_eight mechanical.leanrelation_seven mechanical.leanrelation_seven_is_six_plus_one mechanical.leanrelation_superposition_collapse mechanical.leanrelation_triangular_45_is_base mechanical.leanrelation_units_sum_and_product mechanical.leanrelation_url_path mechanical.leanrepeated_doubling_is_the_power_of_two mechanical.leanrepunit_divisibility_by_three_and_seven demand3.leanreversal_cannot_change_the_digital_root z9plus.leanreversal_does_not_preserve_primality reversal.leanreversal_is_involutive_exactly_off_the_trailing_zeros reversal.leanreversal_is_not_the_identity reversal.leanreversal_preserves_digit_sum reversal.leanreversal_settles_its_range reversal.leanriemann_reflection_and_heart index.leanrights_settles_its_range rights.leanroots_of_unity_cancel generated.leanselfinv_exactly_one_and_eight z9.leanselfneg_only_zero z9.leanseptuple_carries_orbit_onto_units imagined.leanseptuple_carries_units_onto_orbit imagined.leanseptuple_is_involutive_on_triad imagined.leanseven_definitional_and_two_credited phenomena.leanseven_divides_the_repunit_of_length_six mechanical.leansextuple_carries_all_onto_triad imagined.leansextuple_carries_cubes_onto_triad imagined.leansextuple_collapses_tetA_to_one_value imagined.leansextuple_collapses_tetB_to_one_value imagined.leansextuple_collapses_triad_to_one_value imagined.leansingleton_fold_is_the_leaf merkle.leansix_is_the_third_triangular_number mechanical.leansix_reading_frames mechanical.leansixth_collapses_orbit_to_one_value imagined.leansixth_collapses_tetA_to_one_value imagined.leansixth_collapses_tetB_to_one_value imagined.leansixth_collapses_triad_to_one_value imagined.leansixth_collapses_units_to_one_value imagined.leansixty_and_ninety_partition_the_quadrant mechanical.leansixty_four_is_where_the_doubling_returns split.leansixty_one_sense_three_stop_codons mechanical.leanso_involutions_are_not_all_harmonic_in_that_sense involution.leansome_sources_are_unsearched priorart.leansorting_is_what_makes_the_fold_order_free merkle.leansquare_carries_all_onto_squares imagined.leansquare_carries_orbit_onto_tetA imagined.leansquare_carries_tetB_onto_tetA imagined.leansquare_carries_units_onto_tetA imagined.leansquare_collapses_triad_to_one_value imagined.leansquare_is_involutive_on_squares imagined.leansquare_is_involutive_on_tetA imagined.leansquares_is_closed_under_fourth imagined.leansquares_is_closed_under_quadruple imagined.leansquares_is_closed_under_septuple imagined.leansquares_is_closed_under_sixth imagined.leansquares_is_closed_under_square imagined.leansquares_land_exactly_in_zero_one_four_seven z9plus.leanstacked_triangles_are_tetrahedral merkaba.leansubtraction_stays_inside split.leansums_of_two_squares_are_closed sequences.leansuperposition_collapses_to_one quantum.leantarot_78_cards mechanical.leantarot_digital_roots mechanical.leantarot_major_0_21 mechanical.leantarot_minor_4x14 mechanical.leantetA_is_closed_under_fourth imagined.leantetA_is_closed_under_quadruple imagined.leantetA_is_closed_under_septuple imagined.leantetA_is_closed_under_sixth imagined.leantetA_is_closed_under_square imagined.leantetB_is_closed_under_quadruple imagined.leantetB_is_closed_under_septuple imagined.leanthe_967_receipt_case ledgerclaims.leanthe_absent_residues_are_the_primes_below_nine light.leanthe_address_does_not_determine_the_payload ledgerclaims.leanthe_address_is_not_the_payload address.leanthe_address_is_order_sensitive address.leanthe_address_is_shipped_not_the_payload mechanical.leanthe_address_is_sixteen_bytes_whatever_the_input_length address.leanthe_advantage_is_the_count_not_the_operation speed.leanthe_axis_is_closed_under_doubling merkaba.leanthe_break_even_is_the_ratio_of_verifications speed.leanthe_budget_and_the_period_are_one_turn split.leanthe_cantor_pairing_is_injective_and_covers_an_initial_segment demand3.leanthe_cardinality_claims_hold_of_any_four_symbol_alphabet mechanical.leanthe_cell_is_determined_by_content mechanical.leanthe_chain_scales light.leanthe_chinese_remainder_theorem_holds_exactly_when_the_moduli_are_coprime demand2.leanthe_claimed_are_copyright_moral_rights_and_the_database rights.leanthe_classes_partition_the_nine coin.leanthe_coin_step_is_three_times_the_two_coins split.leanthe_coins_reflection_is_harmonic_and_is_one_of_the_ninety involution.leanthe_constant_sum_is_rare involution.leanthe_counter_rotation_has_period_two merkaba.leanthe_crossing_pairs_are_one_four_seven z9plus.leanthe_cube_and_the_tetrahedron_count_out merkaba.leanthe_cyclic_number_142857_is_the_repetend_of_one_seventh mechanical.leanthe_definitional_half_is_the_whole_si phenomena.leanthe_definitions_are_seven_and_travel_fixes_zero light.leanthe_derangement_recurrence_holds demand.leanthe_diameter_is_exactly_four z9plus.leanthe_difference_of_consecutive_squares_is_the_odd_numbers mechanical.leanthe_digital_root_of_seven_to_the_k_has_period_three mechanical.leanthe_digits_are_ten coin.leanthe_digits_one_to_nine_sum_to_forty_five_rooting_to_nine mechanical.leanthe_doubling_orbit_reflection_pairs_sum_to_nine mechanical.leanthe_empty_input_is_still_mixed fnv.leanthe_enumeration_is_complete_and_unduplicated rights.leanthe_equation_balances_by_mass energy.leanthe_euler_characteristic_of_a_genus_g_surface demand3.leanthe_exhaustible_tokens_are_those_six_divides split.leanthe_fall_and_the_reflection_share_one_exceptional_digit coin.leanthe_fall_fixes_every_digit_but_the_void coin.leanthe_fixed_points_are_always_odd involution.leanthe_four_seeds_are_distinct address.leanthe_frontier_grows_by_two_each_round z9plus.leanthe_fuel_was_not_the_limit involution.leanthe_full_superposition_has_nine_states mechanical.leanthe_fusion_of_site_and_user_is_deterministic mechanical.leanthe_gap_widens_with_every_doubling speed.leanthe_gases_are_eighteen_hundred_times_the_water_they_came_from energy.leanthe_gases_are_two_to_one_and_consume_each_other_exactly energy.leanthe_ghz_x_support_is_exactly_the_even_parity_strings quantum.leanthe_harmonic_band_thirty_to_ninety mechanical.leanthe_hash_is_order_sensitive fnv.leanthe_hexagon_and_the_square_metrics mechanical.leanthe_ideal_round_trip_is_exactly_zero energy.leanthe_intention_is_a_computable_deed_receipt mechanical.leanthe_invariance_is_canonicalisation_not_physics quantum.leanthe_kinds_cover_every_source priorart.leanthe_latin_squares_of_order_four_number_five_hundred_and_seventy_six demand2.leanthe_ledger_reversal_cases reversal.leanthe_loop_returns_less_than_it_took energy.leanthe_measured_ratio_at_a_million_leaves speed.leanthe_mobius_divisor_sum_is_the_identity demand.leanthe_moduli_dimensions_are_three_g_minus_three_and_six_g_minus_six mechanical.leanthe_more_developed_the_more_cross_domain_reach mechanical.leanthe_nine_times_table_always_digital_roots_to_nine mechanical.leanthe_odd_step_sets_exactly_one_further_bit sequences.leanthe_orbit_never_meets_the_triad z9plus.leanthe_orders_of_the_units_are_exact z9plus.leanthe_origin_annihilates_and_never_joins_the_circuit index.leanthe_other_two_swap coin.leanthe_pairs_are_exactly_the_units_in_order split.leanthe_parity_of_popcount_is_the_xor_of_the_bits demand.leanthe_product_of_any_three_consecutive_integers_is_divisible_by_six mechanical.leanthe_receipt_is_not_injective quantum.leanthe_reflected_orbit_covers_the_whole_triad z9plus.leanthe_reflection_is_an_involution coin.leanthe_regular_hexagon_exterior_angle_is_the_gold_string mechanical.leanthe_regular_nonagon_exterior_angle_is_the_a432_step mechanical.leanthe_regular_pentagon_angles_are_the_heart_seventy_two_and_hundred_eight mechanical.leanthe_rejected_command_gets_a_receipt mechanical.leanthe_rest_halt_on_the_generator split.leanthe_restated_sources_are_named_and_claim_nothing priorart.leanthe_root_moves_with_the_unit_so_it_is_not_about_light light.leanthe_roots_are_the_singles split.leanthe_roots_of_the_seven light.leanthe_seal_affords_sixty_four_payments_of_two split.leanthe_seed_separates fnv.leanthe_sequence_is_its_named_parts_and_closes index.leanthe_sequence_takes_both_values sequences.leanthe_seven_rest_on_one_finite_structure index.leanthe_seven_roots_miss_four_residues light.leanthe_si_fixes_exactly_seven_constants light.leanthe_singles_are_exactly_the_non_units split.leanthe_skipper_navigates_by_angle mechanical.leanthe_successor_of_every_bound_lies_outside reach.leanthe_sum_of_fifth_powers_has_the_closed_form_asked_for demand2.leanthe_table_is_closed_and_that_is_all_this_file_decides phenomena.leanthe_tetrahedra_residue_sums_cancel merkaba.leanthe_theorems_are_the_hull_and_hardware mechanical.leanthe_three_classes_partition_z9 merkaba.leanthe_three_four_five_right_triangle_is_the_first_pythagorean_triple mechanical.leanthe_three_non_units_are_exactly_the_unreachable index.leanthe_tokens_are_three_times_these split.leanthe_two_to_one_is_forced_by_the_oxygen energy.leanthe_uncanonicalised_fold_gives_many_answers quantum.leanthe_unclaimed_are_the_registry_the_excluded_and_the_unownable rights.leanthe_unit_table_is_a_latin_square z9plus.leanthe_verify_is_thirty_eight_thousand_nanoseconds_not_one speed.leanthe_verify_path_is_the_exponent speed.leanthere_are_2620_involutions_of_nine involution.leanthere_are_infinitely_many_pythagorean_triples mechanical.leanthis_deposit_claims_no_novelty_today priorart.leanthis_file_settles_none_of_the_seven reach.leanthree_five_eight_are_consecutive sequences.leanthree_five_eight_are_consecutive_fibonacci z9plus.leanthree_on_the_triad_and_four_on_the_units light.leanthree_quarters_of_the_input_leaves_as_heat energy.leanthue_morse_doubling_recurrence sequences.leanto_uuid_bytes_of_a address.leanto_uuid_bytes_of_uuidna address.leantotient_at_prime_powers families.leantravel_and_periods_at_one_return_their_constants light.leantravel_at_one_returns_the_defined_constant light.leantriad_is_closed_under_double imagined.leantriad_is_closed_under_fourth imagined.leantriad_is_closed_under_negate imagined.leantriad_is_closed_under_octuple imagined.leantriad_is_closed_under_quadruple imagined.leantriad_is_closed_under_quintuple imagined.leantriad_is_closed_under_septuple imagined.leantriad_is_closed_under_sextuple imagined.leantriad_is_closed_under_sixth imagined.leantriad_is_closed_under_square imagined.leantriad_is_closed_under_triple imagined.leantriad_never_reaches_one z9.leantriad_squares_vanish z9.leantrial_units_group mechanical.leantrial_zero_divisors mechanical.leantrial_zero_no_inverse mechanical.leantriple_carries_all_onto_triad imagined.leantriple_carries_cubes_onto_triad imagined.leantriple_collapses_tetA_to_one_value imagined.leantriple_collapses_tetB_to_one_value imagined.leantriple_collapses_triad_to_one_value imagined.leantwo_bits_thrice_make_the_codon mechanical.leantwo_thirds_of_the_volume_carries_a_ninth_of_the_mass energy.leantwo_to_the_eighth_is_two_hundred_fifty_six_a_byte mechanical.leantwo_to_the_tenth_is_1024_the_harmonic_ledger mechanical.leantwo_twenty_and_two_eighty_four_are_the_smallest_amicable_pair demand2.leanuncompressed_hydrogen_is_three_thousandfold_worse_by_volume energy.leanunits_and_triad_partition_the_ring z9plus.leanunits_are_six z9.leanunits_count z9.leanunits_is_closed_under_double imagined.leanunits_is_closed_under_fourth imagined.leanunits_is_closed_under_octuple imagined.leanunits_is_closed_under_quadruple imagined.leanunits_is_closed_under_quintuple imagined.leanunits_is_closed_under_septuple imagined.leanunits_is_closed_under_sixth imagined.leanunits_is_closed_under_square imagined.leanuniversal_centre_is_five theorems.leanuniversal_millennium_reflection_escapes_the_units theorems.leanuniversal_pairs_sum_to_ten theorems.leanuniversal_reflection_involution theorems.leanuniversal_reflection_is_not_an_involution_above_ten theorems.leanuniversal_reflection_is_the_vortex_reflection_shifted theorems.leanuniversal_reflection_reverses_the_domain theorems.leanuniversal_z9_reflection_permutes_the_units theorems.leanvariant_bits_are_forced address.leanverification_grows_it_is_not_constant speed.leanversion_nibble_is_forced address.leanvitepress_hosts_the_content_address mechanical.leanwhat_escapes_falls_back_inside coin.leanwhat_the_feedwater_leaves_behind_decides_the_maintenance energy.leanwilson_all_primes_under_thirty families.leanwilson_fails_at_composites families.leanwilsons_theorem_and_its_converse demand3.leanxor_is_commutative nim.leanxor_is_its_own_inverse nim.leanxor_is_parity_up_to_eight_bits families.leanxor_zero_is_identity nim.leanyang_mills_spectral_gap index.leanz9_settles_its_domain_totally z9.leanzero_claims_is_not_full_attribution priorart.lean

Verification

Clone the repository and run node scripts/lean.ts to re-check every statement above against the Lean 4 kernel, or npm run forensics to re-verify the append-only receipt chain. The sources are src/proof/, and each sealed theorem also has its own page carrying the same statement. A content-address proves integrity, not truth. entails → 0/7.

Captain's message:https://uuidna.com/captain/message — free on the free sailing angle; prize earning in waves — contribute 2 to earn up to 64 per wave, keep the rest (the two coins per commercial use; the seal is 128 bits = 64 two-bit fold-verifications, O(log N)). Contribute: · why ↗computed: entailment 0/7 · self-seal = 1 · reflection involutive · CC BY-NC-ND 4.0License: CC BY-NC-ND 4.0 — free for non-commercial use (attribution Tsvetan Rouschev); commercial = the two coins (110 − 108 = 2 = −χ genus-2) · ceccec@psg.bgLicensing formula: free for public interest and independent research, unless commercial · commercial = the measured bits saved (O(N) − O(1)), the two coins (2 = 110 − 108 = −χ genus-2) the conserved invariant · verified green by receipts · integrity, not truth · 0/7This referrer perspective: f5ccf7ab-eb8b-8797-8fca-ac9c94b3a54dPublic URLs (content-addressed):https://uuidna.org 8ef35f1f-38f3…https://uuidna.com 58cfb4c9-e262…https://ceccec.psg.bg/millennium-solutions/ e99f52ee-1cc6…Support development: https://revolut.me/ceccec