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

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

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 $((d + γ) / (1 - γ))$ -sparse.

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]

If $(X, Y)$ is a $γ$ -regular pair of density $d$ and a fixed set $Y^{'} \subseteq Y$ has $∣ Y^{'} ∣ \geq γ ∣ Y ∣$ , then fewer than $γ ∣ X ∣$ vertices of $X$ have more than $(d + γ) ∣ Y^{'} ∣$ neighbours in $Y^{'}$ (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 $c$ -sparse exactly when every vertex of the induced graph on it has degree at most $c$ times its size (A set is $c$ -sparse exactly when the maximum degree of the graph it induces is at most $c$ times its size, $c$ -sparse, $c$ -dense and $c$ -restricted vertex sets).

Proof

technique · direct

1.1L1

Apply [L1] to the pair $(W, W)$ with $Y^{'} = W$ . Since $∣ W ∣ \geq γ ∣ W ∣$ , fewer than $γ ∣ W ∣$ vertices of $W$ have more than $(d + γ) ∣ W ∣$ neighbours in $W$ .

2.1step 1.1choose

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

3.1step 2.1L2algebra∎

For every $x \in W^{'}$ , one has $\deg_{G [W^{'}]} (x) \leq \deg_{G [W]} (x) \leq (d + γ) ∣ W ∣ < ((d + γ) / (1 - γ)) ∣ W^{'} ∣$ . Therefore [L2] makes $W^{'}$ $((d + γ) / (1 - γ))$ -sparse.

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

11 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 and Remark 2.3.2 (standard reference, not scraped)

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