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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-27
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A P4-free graph on q vertices has a homogeneous set of size at least q

Statement

If J is a P4-free finite graph with q vertices, then

hom⁡(J)≥q.

Facts & Assumptions

Given: A P4-free graph J on q vertices.

[L1]

Every P4-free graph with more than one vertex admits a partition V(J)=A⊔B with A,B≠∅ such that (A,B) is a pure pair (Chudnovsky--Scott--Seymour--Spirkl, "Erdos-Hajnal for graphs with no 5-hole", §5 Blockades, sentence immediately preceding Theorem 5.1).

Proof

technique · direct
1.1given

We prove the stronger inequality. [given] α(J)ω(J)≥q by induction on q. The cases q=0 and q=1 are immediate.

2.1step 1.1L1

Assume q≥2. By [L1], write V(J)=A⊔B with A,B≠∅ and (A,B) pure. Put. [step 1.1, L1] a=α(J[A]),b=α(J[B]),c=ω(J[A]),d=ω(J[B]). The induction hypothesis gives ac≥∣A∣ and bd≥∣B∣.

3.1step 2.1algebra

If (A,B) is complete, then. [step 2.1, algebra] ω(J)=c+d and α(J)=max⁡{a,b}, whence α(J)ω(J)=max⁡{a,b}(c+d)≥ac+bd≥∣A∣+∣B∣=q. If (A,B) is anticomplete, then α(J)=a+b and ω(J)=max⁡{c,d}, and the same calculation gives α(J)ω(J)≥ac+bd≥q.

4.1step 1.1step 3.1algebra∎

The induction closes. Since. [step 1.1, step 3.1, algebra] max⁡{α(J),ω(J)}≥α(J)ω(J), one obtains hom⁡(J)≥q.

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