LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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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.

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Deleting the low-degree vertices of a $γ$ -self-regular set of density $d$ leaves more than $(1 - γ)$ of it, and that remainder is $((1 - d + 2 γ) / (1 - γ))$ -dense

Statement

Let $0 < γ < 1$ , let $W \subseteq V (G)$ be nonempty, and suppose $(W, W)$ is a $γ$ -regular pair of density $d$ . Then there is a subset $W^{'} \subseteq W$ with $∣ W^{'} ∣ > (1 - γ) ∣ W ∣$ such that $W^{'}$ is $((1 - d + 2 γ) / (1 - γ))$ -dense.

Facts & Assumptions

Given: A finite simple graph $G$ , a real $0 < γ < 1$ , a nonempty set $W \subseteq V (G)$ , and a density $d = d_{G} (W, W)$ such that $(W, W)$ is $γ$ -regular.

[L1]

In a $γ$ -regular pair $(X, Y)$ , all but fewer than $γ ∣ X ∣$ vertices of $X$ have at least $(d (X, Y) - γ) ∣ Y^{'} ∣$ neighbours in any subset $Y^{'} \subseteq Y$ of size at least $γ ∣ Y ∣$ (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, $ϵ$ -regular pairs and self-regular vertex sets).

[L2]

A set is $c$ -dense exactly when every vertex has at most $c ∣ X ∣$ non-neighbours inside it other than itself ( $c$ -sparse, $c$ -dense and $c$ -restricted vertex sets).

Proof

technique · direct

1.1L1

Apply [L1] to the pair $(W, W)$ with $Y^{'} = W$ . Fewer than $γ ∣ W ∣$ vertices of $W$ then have fewer than $(d - γ) ∣ W ∣$ neighbours in $W$ .

2.1step 1.1choose

Let $W^{'}$ be the remaining vertices. Then $∣ W^{'} ∣ > (1 - γ) ∣ W ∣$ .

3.1step 2.1algebra

For $x \in W^{'}$ , at most $γ ∣ W ∣$ of its neighbours lie outside $W^{'}$ , so $x$ has at least $(d - 2 γ) ∣ W ∣$ neighbours in $W^{'}$ .

4.1step 3.1L2algebra∎

Hence $x$ has at most $∣ W^{'} ∣ - 1 - (d - 2 γ) ∣ W ∣ < ((1 - d + 2 γ) / (1 - γ)) ∣ W^{'} ∣$ non-neighbours inside $W^{'}$ . By [L2], the set $W^{'}$ is $((1 - d + 2 γ) / (1 - γ))$ -dense.

Depends on

Used by

Nikiforov: for every H and every ε∈(0,1/2) there is δ>0 such that every graph G with ind_H(G)<(δ|V(G)|)^|V(H)| has an ε-restricted vertex set of size at least δ|V(G)| Theorem

Dependency tree · two levels

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Sources

Y. Zhao, Graph Theory and Additive Combinatorics, Lemma 2.2.3 and Remark 2.3.2 (standard reference, not scraped)

Deleting the low-degree vertices of a γγ-self-regular set of density dd leaves more than (1−γ)(1−γ) of it, and that remainder is ((1−d+2γ)/(1−γ))((1−d+2γ)/(1−γ))-dense