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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Every $H_{5}$ -overlap class is connected

Statement

Every $H_{5}$ -overlap class induces a connected graph.

Facts & Assumptions

Given: An $H_{5}$ -overlap class $C$ in one comb block.

[F1]

Two vertices in $C$ are joined by a finite chain of induced $H_{5}$ copies with successive copies sharing a vertex (The $H_{5}$ -overlap-chain relation in one comb block).

[F2]

The graph $H_{5}$ is connected (The graphs $H_{0}, H_{1}, \dots, H_{5}$ ).

Proof

technique · direct

1.1

Let $r, s \in C$ . By [F1], choose an overlap chain from a copy containing $r$ to a copy containing $s$ . Within each copy, [F2] gives paths from its entering vertex to its shared vertex and then to its exiting vertex.

F1F2choose

2.1

Concatenating these paths at the shared vertices is a walk in $G [C]$ from $r$ to $s$ , and deleting repetitions gives a path. Thus every two vertices of $C$ are connected.

step 1.1F2algebra

3.1

Hence $G [C]$ is connected.

step 2.1∎

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

8 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 Lemma 6.4 (standard reference, not scraped)

Every H5H5​-overlap class is connected