ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-09-07

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An $H_{5}$ -overlap class and its terminal quotient

Example

Fix a comb block $B_{i}$ whose induced graph consists of two labeled copies of $H_{5}$ sharing precisely their rim vertex $v_{1}$ , with no other cross edges.

Facts & Assumptions

Given: The comb block in the Example.

[F1]

Two vertices in $X_{i}$ are related when a finite vertex sequence joins them with each consecutive pair contained in one induced $H_{5}$ inside $B_{i}$ (The $H_{5}$ -overlap-chain relation in one comb block).

Proof

technique · direct

1.1

Every vertex of $B_{i}$ belongs to one of the two induced copies, so $X_{i} = B_{i}$ . For any $d, d^{'} \in B_{i}$ , the vertex sequence $d, v_{1}, d^{'}$ has each consecutive pair in one of those copies; omit repeated consecutive vertices if necessary. Thus [F1] gives $d \sim_{H_{5}} d^{'}$ , and $B_{i}$ is the unique overlap class.

2.1

The initial overlap blockade is therefore $(B_{i})$ . There is no pair of distinct blocks, so it is pure vacuously. Its mixed-block reachability relation has just the singleton class ${B_{i}}$ ; replacing that class by its union returns $(B_{i})$ . Every iterate is consequently $(B_{i})$ , already terminal at the first stage.

step 1.1

3.1

This gives the claimed overlap class and terminal quotient.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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)

An H5H5​-overlap class and its terminal quotient