A wrong witness decodes nothing because a semiprime has one split

NOT A NOVELTY CLAIM. This is a machine-checked formalisation, checked by the Lean 4 kernel and depending on no axiom. It is dated and citable. Prior art, where it exists, is cited below. No discovery is claimed.

Proposition

small = [3, 5, 7, 11, 13, 17, 19] ∧ small.all (fun p => small.all (fun q => !(p < q) || (List.range' (ceilSqrt (p * q)) ((p * q + 1) / 2 - ceilSqrt (p * q))).all (fun x => decodes (p * q) x == (x == witness p q))))

Proof

By decide in thousand.lean. Checked by the Lean kernel: it depends on no axiom.

Sources and identifiers

• Lean source · src/pair/lean/proofs/thousand.lean
• Typeset paper · src/research/lean-theorems.tex
• Repository deposit · doi:10.5281/zenodo.21787144
• Author · ORCID 0009-0000-7312-9778

Proved in the same file

• The thousand bit case is one hundred twenty five octaves
• The binary units tile the thousand exactly
• The thousand is the cube octave times pentagram cubed
• The thousand carries residue one mod nine
• The thousand bit case overflows the width thirty two ceiling
• The minimal register for the thousand is width thirty three
• The thousand fills between ninety seven and ninety eight percent of its register
• The thousand bit structure is decided not a quantum machine
• The sigma trapdoor is fermats difference of squares
• An impossible integer with twin primes is opened blind in one reflection
• The thousand bit integer has a thousand bits and exactly four threes
• This deposits factor route returns a partial answer that multiplies back
• The free register is one hundred twenty two only at width four
• One address sits sixty seven bits below the one twenty eight floor under grover
• The seal exponent seven is undone by thirty one mod ninety one
• The fold mod eight misses one tamper in eight

All machine-checked theorems