Alphabeta Math
TheoremStatement: 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 mixing lemma

Statement

For any subsets S,T of a finite d-regular adjacency-slot graph on n1 vertices, let e(S,T)=uS,vTAuv count ordered slots. Then e(S,T)dSTnαdS(1S/n)T(1T/n). Overlap and loop slots are allowed.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

For the regular multigraph and spectral conventions in the stated convention, put α=M1. For n2 order the eigenvalues 1=μ1μ2μn, counting multiplicity, and put γ=1μ2. Thus α=maxj2μj, which also controls negative eigenvalues. Write cut(S)=uS,vSAuv and VS={vS:Auv>0 for some uS}. Normalized edge expansion and external vertex expansion are h=min0<Sn/2cut(S)dS,hV=min0<Sn/2VSS. For n=1, put α=0 and leave μ2,γ,h,hV undefined; cut-expansion assertions are vacuous. A bounded-degree family is an expander family when its normalized edge expansion has a positive uniform lower bound for n2. Polynomial-time constructibility means a uniform algorithm outputs the adjacency list in time polynomial in n; neighbor computation in time polynomial in logn is a stronger requirement. (Spectral edge and vertex expansion).

[F2]

For vectors u,v in a real or complex inner product space, u,vuv. Equality holds if and only if u and v are linearly dependent, including the case in which either vector is zero. (Cauchy–Schwarz: u,vuv, with equality exactly for linearly dependent vectors).

Proof

1.1

Set s=S/n, t=T/n, f=1Ss1 and g=1Tt1. Both are mean zero and have normalized squared norms s(1s) and t(1t). Since M preserves constants and their orthogonal complement, e(S,T)/(nd)st=f,Mg. This counts loops and overlap exactly as specified.

F1
2.1

Cauchy–Schwarz and the defining operator bound give f,Mgαs(1s)t(1t). Multiply by nd. Empty or full sets give zero centered vectors and equality; at n=1 all sets are of that form. No division by a set size or by α is made.

F2step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources