LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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.

Sc toolkit periodic relator overlap is a piece

Statement

Let $r = x^{m} a$ be a literal cyclically reduced relator word, with $x \neq \emptyset$ , $m \geq 2$ , and $a$ possibly empty. If $x a \neq a x$ literally, then $x^{m - 1}$ is a piece. If $x a = a x$ and $a$ is nonempty, $x$ and $a$ are positive powers of a common nonempty word, and so is $r$ . If $a$ is empty, $r$ is already a power of $x$ .

Facts & Assumptions

Given: The literal factorization $r = x^{m} a$ in a symmetrised relator set, with $x$ nonempty and $m \geq 2$ .

[F1]

Distinct symmetrised relators with a nonempty common prefix define a piece (Sc toolkit symmetrised relators and pieces).

[F2]

Nonempty literally commuting words are positive powers of a common word (Sc toolkit commuting positive words have a common root).

Proof

1.1

Moving the first copy of $x$ to the end gives another symmetrised word $s = x^{m - 1} a x$ . The equality $r = s$ holds exactly when, after cancelling the common prefix $x^{m - 1}$ , $x a = a x$ . Thus when $x a \neq a x$ , the distinct words $r, s$ have the nonempty common prefix $x^{m - 1}$ , a piece by [F1].

givenF1

2.1

When $x a = a x$ and $a \neq \emptyset$ , [F2] gives $x = c^{b}$ , $a = c^{d}$ for positive integers $b, d$ . Substitution gives $r = c^{m b + d}$ . If $a$ is empty then $r = x^{m}$ directly. This exhausts the commuting and noncommuting possibilities without mistaking equal rotations for distinct relators.

F2step 1.1algebra∎

Depends on

Used by

Periodic words: a relator root or Dehn-reduced powers Lemma

Dependency tree · two levels

5 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

Lipschutz (1964), §6 opening remark, printed p.41 (standard reference, not scraped)