ExampleConstruction: AI-generatedVerification: AI-generatedprecheck 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 $σ$ has parity $(- 1)^{inv (σ)}$ ).

[L2]

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

Verification

technique · direct

1.1

The product $(0 2) (0 1)$ sends $0 \mapsto 1 \mapsto 2 \mapsto 0$ , 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

Nothing in the library uses this result yet.

Dependency tree · two levels

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