ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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.

A large epsilon-restricted induced subgraph gives a polynomial clique or stable set

Example

Let $ϵ : = \frac{1}{4}$ and let $S$ be an $ϵ$ -restricted induced subgraph on $16$ vertices. Take an ambient graph order $∣ G ∣ : = 2^{84}$ , so that $∣ G ∣^{1 / 42} = 4$ .

Facts & Assumptions

Given: The data in the Example.

[L1]

An $ϵ$ -restricted set is $ϵ$ -sparse in one of $G$ and $\overline{G}$ , and complementation swaps cliques with stable sets ( $c$ -sparse, $c$ -dense and $c$ -restricted vertex sets, A set is $c$ -sparse in $G$ exactly when it is $c$ -dense in $\overline{G}$ , so $c$ -restrictedness is complement-invariant, Complementation swaps cliques with stable sets, so $ω (\overline{G}) = α (G)$ ).

[L2]

For a nonempty $ϵ$ -sparse graph $H$ ,

χ (H) \leq ϵ ∣ H ∣ + 1, ∣ H ∣ \leq χ (H) α (H)

(The greedy colouring bound $χ (G) \leq Δ (G) + 1$ for every nonnull finite graph, The bounds $ω (G) \leq χ (G)$ and $∣ V (G) ∣ \leq χ (G) α (G)$ ).

Verification

technique · direct

1.1

By [L1], after taking complements if necessary we may assume that $S$ is $ϵ$ -sparse.

2.1

Applying [L2] to the $16$ -vertex graph $G [S]$ gives $α (G [S]) \geq \frac{16}{ϵ \cdot 16 + 1} = \frac{16}{5} > 3.$ Since $α (G [S])$ is an integer, $α (G [S]) \geq 4$ .

step 1.1L2algebra

3.1

Because $∣ G ∣ = 2^{84}$ , one has $∣ G ∣^{1 / 42} = 2^{2} = 4$ . Therefore the stable set from step 2.1 already has the same size as the final polynomial bound used in the A-page reduction. If the sparse side had arisen in the complement instead, [L1] would turn the same calculation into a clique of size $4$ in the original graph.

step 2.1L1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 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

Shenwei Huang, Yiao Ju, and Yidong Zhou, Erdős-Hajnal beyond the five-vertex path, Lemma 3.5 (standard reference, not scraped)