LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Purity on every induced $H_{5}$ propagates along an $H_{5}$ -overlap class

Statement

Let $C$ be an $H_{5}$ -overlap class and let $u \notin C$ . If $u$ is pure to every induced $H_{5}$ contained in $C$ , then $u$ is pure to $C$ .

Facts & Assumptions

Given: $C, u$ as in the Statement.

[F1]

A chain of induced $H_{5}$ copies links the copies meeting any two vertices of $C$ (The $H_{5}$ -overlap-chain relation in one comb block).

[F2]

A vertex pure to a nonempty set is either complete or anticomplete to it (Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs).

Proof

technique · direct

1.1

Two consecutive copies in an overlap chain share a vertex. By [F2], $u$ cannot be complete to one and anticomplete to the other, since their shared vertex would then be both adjacent and nonadjacent to $u$ .

F1F2

2.1

Thus the complete/anticomplete label is constant along every overlap chain. By [F1], every vertex of $C$ lies in a copy reached from any fixed copy, so all vertices of $C$ receive one label.

step 1.1F1

3.1

Hence $u$ is complete or anticomplete to $C$ , as claimed.

step 2.1F2∎

Depends on

Used by

In a special-vertex comb of a co-E-free graph, vertices in other comb blocks remain pure to every H₅-overlap quotient block Lemma

Dependency tree · two levels

6 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

Huang, Ju, and Zhou, Erdős-Hajnal beyond the five-vertex path, proof of Claim 6.4.3 (standard reference, not scraped)

Purity on every induced H5H5​ propagates along an H5H5​-overlap class