LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The star-expansion of $K_{3}$ contains the hatted five-cycle

Statement

The star-expansion of $K_{3}$ contains an induced hatted five-cycle.

Facts & Assumptions

Given: The star-expansion of $K_{3}$ with triangle vertices $b_{1}, b_{2}, b_{3}$ , teeth $a_{1}, a_{2}, a_{3}$ , and root $v$ .

[L1]

In the star-expansion, the only new edges are $v a_{i}$ and $a_{i} b_{i}$ (The star-expansion of a graph).

Proof

technique · direct finite check

1.1

The five vertices $v, a_{1}, b_{1}, b_{2}, a_{2}$ induce the cycle $v - a_{1} - b_{1} - b_{2} - a_{2} - v$ . Indeed, [L1] supplies the four new edges and the edge $b_{1} b_{2}$ comes from the triangle. No other edge among these five vertices is present.

2.1

The remaining triangle vertex $b_{3}$ is adjacent to $b_{1}$ and $b_{2}$ but to neither $v, a_{1}, a_{2}$ in the chosen five-vertex set. Thus adjoining $b_{3}$ to the cycle from step 1.1 yields a hat vertex adjacent to two adjacent cycle vertices.

step 1.1L1

3.1

Therefore the star-expansion of $K_{3}$ contains an induced hatted five-cycle.

step 2.1∎

Depends on

Used by

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

Maria Chudnovsky, Alex Scott, Paul Seymour, and Sophie Spirkl, Erdős-Hajnal for graphs with no 5-hole, proof of Theorem 8.1 (standard reference, not scraped)

The star-expansion of K3K3​ contains the hatted five-cycle