TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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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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Induced counting lemma: regular edge and nonedge pairs force many induced copies

Statement

Let $H$ be a graph on labelled vertices $[h]$ and let $0 < η < 1 / 2$ . There are $γ = γ (H, η) > 0$ , $c = c (H, η) > 0$ , and $N = N (H, η)$ such that the following holds. Let $W_{1}, \dots, W_{h}$ be vertex sets of size at least $N$ , with repetitions allowed. Assume every $(W_{i}, W_{j})$ , including those with $W_{i} = W_{j}$ , is $γ$ -regular. If $d (W_{i}, W_{j}) \geq η for i j \in E (H), d (W_{i}, W_{j}) \leq 1 - η for i j \notin E (H),$ then there are at least $c \prod_{i} ∣ W_{i} ∣$ injective maps $ϕ$ with $ϕ (i) \in W_{i}$ that induce $H$ .

Facts & Assumptions

Given: $H, η$ and regular host sets as in the Statement.

[L1]

Regular dense pairs support a greedy count of many part-respecting injective edge-preserving maps of any fixed graph (Counting lemma: regular dense pairs contain many part-respecting copies of every fixed graph).

[L2]

For disjoint sets, complementation preserves the regularity parameter and replaces density $d$ by $1 - d$ (Complementation sends a disjoint $ϵ$ -regular pair of density $d$ to one of density $1 - d$ ).

[L3]

An induced embedding is injective and preserves both adjacency and nonadjacency; its labelled count is ${ind}_{H} (G)$ (Induced embeddings and induced copies of a graph, The induced-embedding count ${ind}_{H} (G)$ ).

Proof

technique · direct

1.1givenL3

Colour each unordered pair $i j$ of pattern vertices by whether it is an edge or a nonedge of $H$ . On an edge pair retain adjacency in the host; on a nonedge pair regard nonadjacency as the required relation.

2.1step 1.1L2algebra

If $W_{i}$ and $W_{j}$ are disjoint, [L2] turns the latter relation into a $γ$ -regular pair of complementary density at least $η$ . If the host sets coincide or overlap, the same conclusion for candidate degrees follows directly from the defining regularity inequalities: replace each density $d (A, B)$ by the proportion of distinct ordered pairs in $A \times B$ that are nonedges. For candidate sets $A \subseteq W_{i}$ and $B \subseteq W_{j}$ the coinciding pairs number $∣ A \cap B ∣$ , so this replacement changes the proportion by at most $∣ A \cap B ∣ / (∣ A ∣ ∣ B ∣) \leq 1 / \max (∣ A ∣, ∣ B ∣)$ — a bound in the current candidate sizes, not in $\min_{i} ∣ W_{i} ∣$ , since candidate sets shrink as the greedy argument proceeds.

3.1step 2.1L1choose

Choose $γ$ small enough for the greedy argument underlying [L1] with density threshold $η / 2$ and all $(\binom{h}{2})$ coloured constraints, and let $ρ = ρ (H, η) > 0$ be the fraction of its host set that every candidate set provably retains throughout that argument, so that every candidate set met in step 2.1 has size at least $ρ N$ . Choose $N$ large enough that $1 / (ρ N) < η / 4$ ; then the diagonal error of step 2.1 stays below $η / 4$ at every stage.

4.1step 3.1L1induction

Run that greedy proof with the required relation for each pair. At each stage the typical-degree exclusions occupy only a controlled fraction of the current candidate set, so at least $c_{0} \prod_{i} ∣ W_{i} ∣$ relation-preserving maps remain for a constant $c_{0} = c_{0} (H, η) > 0$ .

5.1step 4.1algebrachoose

At most $(\binom{h}{2}) N^{- 1} \prod_{i} ∣ W_{i} ∣$ of these maps have a collision. Enlarge $N$ so this is below $c_{0} \prod_{i} ∣ W_{i} ∣ / 2$ and put $c = c_{0} / 2$ .

6.1step 1.1step 5.1L3∎

Every surviving map is injective and realizes adjacency exactly on the edges of $H$ , so it is an induced embedding by [L3]. This proves the claimed bound.

Depends on

Used by

Dependency tree · two levels

14 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, Theorem 2.6.2 with Remark 2.6.3(b) (standard reference, not scraped)
D. Conlon and J. Fox, Graph removal lemmas, sec. 2.3 (standard reference, not scraped)