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.

In a regular pair, fewer than $ϵ ∣ X ∣$ vertices have too small a degree into a large subset, and fewer than $ϵ ∣ X ∣$ have too large a degree

Statement

Suppose $(X, Y)$ is an $ϵ$ -regular pair of density $d$ , and $Y^{'} \subseteq Y$ satisfies $∣ Y^{'} ∣ \geq ϵ ∣ Y ∣$ . Then fewer than $ϵ ∣ X ∣$ vertices $x \in X$ have $∣ N (x) \cap Y^{'} ∣ < (d - ϵ) ∣ Y^{'} ∣,$ and fewer than $ϵ ∣ X ∣$ vertices have $∣ N (x) \cap Y^{'} ∣ > (d + ϵ) ∣ Y^{'} ∣ .$

Facts & Assumptions

Given: An $ϵ$ -regular pair $(X, Y)$ of density $d$ and a set $Y^{'} \subseteq Y$ with $∣ Y^{'} ∣ \geq ϵ ∣ Y ∣$ .

[L1]

Every $A \subseteq X$ , $B \subseteq Y$ with $∣ A ∣ \geq ϵ ∣ X ∣$ and $∣ B ∣ \geq ϵ ∣ Y ∣$ satisfies $∣ d (A, B) - d (X, Y) ∣ \leq ϵ$ ( $ϵ$ -regular pairs and self-regular vertex sets).

Proof

technique · contradiction

1.1assume-contragivenchoose

Let $A$ be the set of all $x \in X$ with $∣ N (x) \cap Y^{'} ∣ < (d - ϵ) ∣ Y^{'} ∣$ and let $B$ be the set of all $x \in X$ with $∣ N (x) \cap Y^{'} ∣ > (d + ϵ) ∣ Y^{'} ∣$ . Suppose, for contradiction, that $∣ A ∣ \geq ϵ ∣ X ∣$ .

2.1step 1.1L1algebra

Summing degrees over $A$ gives $e (A, Y^{'}) < (d - ϵ) ∣ A ∣ ∣ Y^{'} ∣$ , hence $d (A, Y^{'}) < d - ϵ$ . Since $∣ A ∣ \geq ϵ ∣ X ∣$ and $∣ Y^{'} ∣ \geq ϵ ∣ Y ∣$ , this contradicts [L1]. Therefore $∣ A ∣ < ϵ ∣ X ∣$ .

3.1step 2.1L1algebra

Assume likewise that $∣ B ∣ \geq ϵ ∣ X ∣$ . Summing degrees over $B$ gives $d (B, Y^{'}) > d + ϵ$ , and the same two size conditions again contradict [L1]. Therefore $∣ B ∣ < ϵ ∣ X ∣$ .

4.1step 2.1step 3.1discharge-contradiction∎

Both exceptional sets therefore have size strictly below $ϵ ∣ X ∣$ , which is the Statement.

Depends on

$\epsilon$-regular pairs and self-regular vertex sets

Used by

Dependency tree · two levels

2 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

Y. Zhao, Graph Theory and Additive Combinatorics, Lemma 2.2.3 (standard reference, not scraped)

In a regular pair, fewer than ϵ∣X∣ϵ∣X∣ vertices have too small a degree into a large subset, and fewer than ϵ∣X∣ϵ∣X∣ have too large a degree