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CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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For every k, the class forbidding Pk and Pk has the Erdős–Hajnal property

Statement

For every integer k2, every finite graph with no induced Pk and no induced Pk has a clique or stable set of size at least a positive power of its order. Equivalently, the class forbidding Pk and Pk has the Erdős–Hajnal property.

Facts & Assumptions

Given: An integer k2.

[L1]

For this k, the class forbidding Pk and Pk has the strong Erdős–Hajnal property (For every k, the class forbidding Pk and Pk has the strong Erdős–Hajnal property).

[L2]

Every hereditary class with the strong Erdős–Hajnal property has the Erdős–Hajnal property (The strong Erdős–Hajnal property implies the Erdős–Hajnal property).

Proof

technique · direct
1.1

The previous theorem gives the strong Erdős–Hajnal property for the hereditary class of graphs forbidding Pk and Pk.

L1
2.1

Applying [L2] to that class yields the Erdős–Hajnal property.

step 1.1L2

Depends on

Used by

Dependency tree · two levels

18 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