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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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Every graph on at most three vertices has the Erdős–Hajnal property

Statement

Every finite graph H with V(H)3 has the Erdős–Hajnal property.

Facts & Assumptions

Given: A finite graph H with at most three vertices.

[L2]

For every t1, the class of Kt-free graphs has the Erdős–Hajnal property (For every t1, the class of Kt-free graphs has the Erdős–Hajnal property).

[L3]

Every P3-free graph G satisfies hom(G)V(G), so P3 has the property (Every P3-free graph G satisfies hom(G)V(G)).

[L4]
[L5]

The graphs Kt and P3 have the standard edge sets, and P0 is the null graph (Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices); complementation replaces the edge set by all missing pairs (Graph isomorphisms, automorphisms and graph complements).

Proof

technique · cases
1.1

[assume-case null] If V(H)=0, every graph contains the unique empty induced embedding of H, so the H-free class has no members and [L1] is vacuously satisfied by every positive exponent.

L1
1.2

[assume-case nonnull] Suppose 1V(H)3. Up to isomorphism and complementation, H is one of K1,K2,K3, or P3: this follows by the edge count for orders at most two, and for order three by separating the cases of zero, one, two, or three edges.

L5algebra
2.1

Each complete case has the property by [L2], the path case has it by [L3], and every complementary case has it by [L4].

step 1.2L2L3L4
3.1

The cases are exhaustive, so every graph on at most three vertices has the Erdős–Hajnal property.

step 1.1step 2.1cases-exhaustive

Depends on

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Sources