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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07
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The family consisting of H5 and co-E has the Erdős–Hajnal property

Statement

The finite forbidden family {H5,co-E} has the Erdős–Hajnal property.

Facts & Assumptions

Given: The graphs H0,,H5 and the graph co-E.

[F1]

The graph H0 has the Erdős–Hajnal property (The graph H0 has the Erdős-Hajnal property).

[F2]

The graph P5 has the Erdős–Hajnal property (The five-vertex path and its complement have the Erdős-Hajnal property).

[F3]

The Erdős–Hajnal property passes to a hereditary subclass (The Erdős–Hajnal property and each of its constants pass to hereditary subclasses).

[F4]

Huang--Ju--Zhou, Corollary 1.8, states the following leaf/co-leaf transfer. Let F be a finite family, let H1F have a leaf v, and let H2F have a co-leaf w. If both families obtained from F by replacing, respectively, H1 by H1{v} and H2 by H2{w} have the Erdős--Hajnal property, then F has the Erdős--Hajnal property.

Proof

technique · induction
1.1

The class of {H0,co-E}-free graphs is a hereditary subclass of the class of H0-free graphs, and the class of {Hi,P5}-free graphs is a hereditary subclass of the class of P5-free graphs. Thus [F1]--[F3] give the Erdős–Hajnal property for both families, for every i[5].

F1F2F3base
1.2

Fix i[5] and suppose that {Hi1,co-E} has the property. In F={Hi,co-E}, deleting the leaf vi of Hi gives Hi1. The vertex q is a leaf of E, so it is a co-leaf of co-E, and deleting it from co-E leaves P5. Hence the two modified families in [F4] are exactly {Hi1,co-E} and {Hi,P5}.

F4ih
2.1

Step 1.2, the induction hypothesis, and the second base family from step 1.1 let [F4] yield the property for {Hi,co-E}.

step 1.1step 1.2F4
3.1

Starting with i=1 and repeating step 2.1 through i=5 proves the property for {H5,co-E}.

step 1.1step 2.1discharge-induction

Depends on

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