Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck 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 Bn satisfy

∣SBn∖Bn∣∣Bn∣⟶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
1.1givenalgebra

For every n≥0, one has ∣Bn∣=1+4∑k=1n3k−1=2⋅3n−1. Every element of SBn∖Bn has reduced-word length exactly n+1, and every reduced word of length n+1 is obtained by taking its length-n prefix in Bn and multiplying by its final letter in S. Hence SBn=Bn+1 and ∣SBn∖Bn∣=∣Bn+1∣−∣Bn∣=4⋅3n. Therefore ∣SBn∖Bn∣∣Bn∣=4⋅3n2⋅3n−1⟶2.

2.1L1step 1.1∎

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].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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