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

Expander walk restricted operator

Statement

Let S have density β=S/n in a finite regular graph and let PS project onto functions supported in S. Then PSMPSα+(1α)β. For a stationary length-t walk, t0, its confinement probability is n11S,(PSMPS)t1S0, where the inner product is unnormalized.

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

For f supported in S, let Jf be its constant projection. Finite-sum Cauchy–Schwarz gives Jf2βf2. Write f=Jf+f0 with orthogonal parts. Since M fixes the constant part and contracts the other by α, the Rayleigh form is at most Jf2+αf02[α+(1α)β]f2 and at least αf2. The supported symmetric compression has an orthonormal eigenbasis, so its absolute norm has the asserted bound; outside that space it is zero.

F1algebra
2.1

Expand the matrix product: each factor PS deletes precisely the paths with a vertex outside S, and each M factor supplies its step probability. Summing endpoints with the factor 1/n gives uniform initial sampling. For t=0 the expression is S/n; if S is empty it is zero and if S=V it is one. These statements also cover n=1.

F1step 1.1

Depends on

Used by

Dependency tree · two levels

3 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