Alphabeta Math
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.

From one-line notation to a disjoint-cycle decomposition, with the right-hand factor acting first

Example

In S7S_7, let

α=(012)(34),β=(03)(56),σ=αβ.\alpha=(0\,1\,2)(3\,4),\qquad \beta=(0\,3)(5\,6),\qquad \sigma=\alpha\beta.

The right-hand factor acts first. The one-line form of σ\sigma is [4,2,0,1,3,6,5][4,2,0,1,3,6,5], and its disjoint-cycle decomposition is

σ=(04312)(56).\sigma=(0\,4\,3\,1\,2)(5\,6).

Facts & Assumptions

Given: The displayed permutations in S7S_7, with composition from right to left.

[L1]

A permutation of a finite set is recovered by following each unused point until its orbit closes, producing its disjoint-cycle decomposition (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation).

Verification

technique · direct
1.1

Applying β\beta and then α\alpha gives σ(0)=4\sigma(0)=4, σ(1)=2\sigma(1)=2, σ(2)=0\sigma(2)=0, σ(3)=1\sigma(3)=1, σ(4)=3\sigma(4)=3, σ(5)=6\sigma(5)=6, and σ(6)=5\sigma(6)=5, so the one-line form is [4,2,0,1,3,6,5][4,2,0,1,3,6,5].

givenL1
2.1

Starting at 00 gives 0431200\mapsto4\mapsto3\mapsto1\mapsto2\mapsto0, and the only unused points satisfy 5655\mapsto6\mapsto5. These cycles are disjoint and reproduce every value in step 1.1, so σ=(04312)(56)\sigma=(0\,4\,3\,1\,2)(5\,6).

step 1.1L1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.