ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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.

Boundary expansion in the free group

Example

In the free group $F (a, b)$ with symmetric generating set $S = {a, a^{- 1}, b, b^{- 1}}$ , the balls $B_{n}$ satisfy

$\frac{∣ S B_{n} ∖ B_{n} ∣}{∣ B_{n} ∣} ⟶ 2.$

In particular their one-sided generator boundary proportion stays uniformly positive.

Facts & Assumptions

Given: The free group $F (a, b)$ with the symmetric generating set $S = {a, a^{- 1}, b, b^{- 1}}$ .

[L1]

The rank-two free group is nonamenable (The free group of rank two is nonamenable).

Verification

technique · direct

1.1

For every $n \geq 0$ , one has $∣ B_{n} ∣ = 1 + 4 \sum_{k = 1}^{n} 3^{k - 1} = 2 \cdot 3^{n} - 1$ . Every element of $S B_{n} ∖ B_{n}$ has reduced-word length exactly $n + 1$ , and every reduced word of length $n + 1$ is obtained by taking its length- $n$ prefix in $B_{n}$ and multiplying by its final letter in $S$ . Hence $S B_{n} = B_{n + 1}$ and $∣ S B_{n} ∖ B_{n} ∣ = ∣ B_{n + 1} ∣ - ∣ B_{n} ∣ = 4 \cdot 3^{n} .$ Therefore $\frac{∣ S B_{n} ∖ B_{n} ∣}{∣ B_{n} ∣} = \frac{4 \cdot 3^{n}}{2 \cdot 3^{n} - 1} ⟶ 2.$

givenalgebra

2.1

In particular the one-sided generator boundary ratio of the balls never approaches $0$ . This explicit boundary expansion is the geometric obstruction behind the nonamenability recorded in [L1].

L1step 1.1∎

Depends on

The free group of rank two is nonamenable

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.