Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

Cycles with disjoint supports commute

Statement

Cycles with disjoint supports commute. More precisely, if X is finite and α and β are cycles in Sym⁡(X) and supp⁡(α)∩supp⁡(β)=∅, then αβ=βα.

Facts & Assumptions

Given: A finite set X and two cycles α,β∈Sym⁡(X) with disjoint supports.

[L1]

A cycle fixes every point outside its support, and two cycles are disjoint exactly when their supports are disjoint (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

Proof

technique · direct
1.1

Every x∈X lies in supp⁡(α), in supp⁡(β), or in neither support, and the first two alternatives cannot both hold.

givenL1
2.1

If x∈supp⁡(α), then β fixes both x and α(x), so αβ(x)=βα(x)=α(x); the symmetric argument applies on supp⁡(β), while outside both supports both cycles fix x. Thus the two composites agree at every point.

step 1.1L1∎

Depends on

Used by

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources