Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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.

An irregularity witness raises the pair energy by more than ϵ4∣X∣∣Y∣/n2

Statement

Let G have order n>0, let X,Y⊆V(G) be nonempty, and suppose that A⊆X, B⊆Y witness that (X,Y) is not ϵ-regular. If A={A,X∖A} and B={B,Y∖B} after empty cells are omitted, then q(A,B)−∣X∣∣Y∣n2d(X,Y)2>ϵ4∣X∣∣Y∣n2.

Facts & Assumptions

Given: An irregular pair and witness sets as in the Statement.

[L1]

Such witnesses satisfy ∣A∣≥ϵ∣X∣, ∣B∣≥ϵ∣Y∣, and ∣d(A,B)−d(X,Y)∣>ϵ (ϵ-regular pairs and self-regular vertex sets).

[L2]

Pair energy is the product-size-weighted mean square of the densities of the refined subpairs (The mean-square density, or energy, of a vertex partition).

Proof

technique · direct
1.1givenL2algebra

Choose (x,y) uniformly from X×Y, and let Z be the density between the cells of A and B containing x and y. Double-counting gives EZ=d(X,Y), and [L2] identifies ∣X∣∣Y∣E(Z2)/n2 with q(A,B).

2.1step 1.1algebra

Therefore the energy gain in the Statement is (∣X∣∣Y∣/n2)Var⁡(Z)=(∣X∣∣Y∣/n2)E((Z−d(X,Y))2).

2.2L1step 1.1

On the event (x,y)∈A×B, which has probability ∣A∣∣B∣/(∣X∣∣Y∣)≥ϵ2 by [L1], the random variable equals d(A,B) and differs from its mean by more than ϵ.

3.1step 2.1step 2.2algebra∎

Restricting the nonnegative expectation in step 2.1 to this event gives a strict lower bound ϵ2ϵ2∣X∣∣Y∣/n2, which is the asserted boost.

Depends on

Used by

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.

Sources