ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Two different transposition factorisations of the same permutation have the same parity

Example

In $S_3$ , the three-cycle $(0\,1\,2)$ has the two factorizations

$(0\,1\,2)=(0\,2)(0\,1)=(0\,2)(0\,1)(1\,2)(1\,2).$

Their lengths are $2$ and $4$ , so both are even, as the parity theorem predicts.

Facts & Assumptions

Given: Products in $S_3$ act from right to left.

[L1]

Every transposition factorisation of a permutation has parity equal to its inversion sign (Every transposition factorisation of $\sigma$ has parity $(-1)^{\operatorname{inv}(\sigma)}$ ).

[L2]

Inversion sign is $(-1)$ raised to the number of decreasing pairs in one-line notation (Inversions, inversion number, the sign $\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}$ , and even and odd permutations).

Verification

technique · direct

1.1

The product $(0\,2)(0\,1)$ sends $0\mapsto1\mapsto2\mapsto0$ , hence equals $(0\,1\,2)$ , and the final pair $(1\,2)(1\,2)$ is the identity, so the four-factor product is the same permutation.

givenL1

2.1

The one-line form of $(0\,1\,2)$ is $[1,2,0]$ , with two inversions and sign $+1$ by [L2]. The factor counts $2$ and $4$ are both even and therefore both have parity $+1$ , in agreement with [L1].

step 1.1L1L2∎

Depends on

Used by

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