ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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.

A concrete relator set with its pieces and a direct C prime(1/6) check

Example

Let

R = {a b c d e f g, a h i j k l m n} .

Then the only nontrivial pieces in the symmetrised set are the one-letter words $a$ and $a^{- 1}$ , so this relator set satisfies $C^{'} (1 / 6)$ .

Facts & Assumptions

Given: The two relators $r_{1} = a b c d e f g$ and $r_{2} = a h i j k l m n$ .

[L1]

Pieces are common initial segments of distinct symmetrised occurrences (A piece is a common initial segment occurring in two distinct places of a symmetrised relator set).

[L2]

$C^{'} (1 / 6)$ requires every piece to have length less than one sixth of the relator containing it (The small-cancellation conditions C(lambda) and C prime(lambda)).

Verification

technique · direct

1.1

The words $r_{1}$ and $r_{2}$ start with the same letter $a$ and then immediately diverge, so $a$ is a piece by [L1]. Their inverse words have cyclic conjugates starting with $a^{- 1}$ and then immediately diverging, so $a^{- 1}$ is also a piece. Every other letter occurs in only one cyclic position among the two relators and their inverses, so there is no other nontrivial piece and no piece of length greater than $1$ .

L1given

2.1

The relevant relator lengths are $∣ r_{1} ∣ = 7$ and $∣ r_{2} ∣ = 8$ , while both pieces have length $1$ . Since $1 < 7 / 6$ and $1 < 8 / 6$ , [L2] shows that the $C^{'} (1 / 6)$ inequalities hold for every symmetrised occurrence.

L2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

GAP SmallCancellation manual, Chapter 1: Small Cancellation Theory — the classical conditions (standard reference, not scraped)
Jay Williams, Universal Countable Borel Quasi-Orders (standard reference, not scraped)
Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, Section 3.5 (standard reference, not scraped)
Clara Löh, Geometric Group Theory: An Introduction, Section 7.4.1 (standard reference, not scraped)