Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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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.

Expander walk bad edge return

Statement

Let F be a nonempty set of nonloop ordinary edges of a reverse-paired d-regular graph, and put ε=F/E. In a stationary walk, condition on some edge being in F. For i1, the probability that the edge i positions later belongs to F is at most ε+αi1. Interpret α0=1.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

A walk that at each step chooses one of the d ports uniformly has transition matrix M and stationary uniform law u=1/n. For any initial probability vector p and integer t0, using the ordinary Euclidean norm, Mtpu2αtpu2,TV(Mtp,u)n2αt. For t=0 the factor α0 is interpreted as one. For t1, the adjacency-slot power has nontrivial norm αt. Here total variation means half the 1 distance. (Expander walk contraction).

Proof

1.1

The conditioned edge is uniform in F and its orientation is uniform. Its terminal vertex therefore has law xv=degF(v)/(2F). The next-step probability of using F from v is yv=degF(v)/d=(2F/d)xv. Also maxxvd/(2F) and xv=1, whence x22d/(2F). This conditioning is legitimate because F.

F1algebra
2.1

Between that terminal vertex and the later tested edge there are i1 transitions, so the probability is yTMi1x. Its constant part is yTu=2F/(dn)=ε. For the other part, spectral contraction and Cauchy–Schwarz give absolute value at most (2F/d)xu22αi1αi1; here y=(2F/d)x allows its constant component to be removed in that inner product. This proves the result, including adjacent edges i=1. The nonloop condition ensures the stated two-endpoint count.

F1step 1.1algebra

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Sources