ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01‡ rests on unproved material (inherited)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Strong Perfect Graph Theorem, ‡ Substituting perfect graphs preserves perfection and ‡ Weak Perfect Graph Theorem. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

A six-vertex witness graph makes the Bird criterion explicit

Example

The six-vertex graph

H : E (H) = {v_{1} v_{2}, v_{1} v_{3}, v_{1} v_{5}, v_{1} v_{6}, v_{2} v_{3}, v_{2} v_{4}, v_{2} v_{5}, v_{3} v_{5}, v_{4} v_{5}, v_{4} v_{6}}

has a homogeneous clique ${v_{2}, v_{5}}$ , so $H$ has the Erdős-Hajnal property; moreover $H^{+} - v_{3}$ and $H^{-} - v_{6}$ are both co-Bird.

Facts & Assumptions

Given: The graph $H$ above, with distinguished vertices $v_{1}$ and $v_{2}$ .

[L1]

Every graph on at most five vertices has the Erdős-Hajnal property (Every graph on at most five vertices has the Erdős-Hajnal property).

[L2]

Substitution preserves the Erdős-Hajnal property (Alon–Pach–Solymosi: if $H_{1}$ and $H_{2}$ have the Erdős–Hajnal property, so does the graph obtained from $H_{1}$ by substituting $H_{2}$ for a vertex).

[L3]

The graph $H^{+}$ adds a new vertex adjacent to the two distinguished vertices, while $H^{-}$ does the same after deleting the distinguished edge if it is present; co-Bird is the complement of Bird (The graphs $H^{+}$ and $H^{-}$ for two distinguished vertices, The Bird graph and co-Bird).

Verification

technique · direct finite check

1.1

Outside the pair ${v_{2}, v_{5}}$ , both vertices are adjacent exactly to $v_{1}, v_{3}, v_{4}$ and are nonadjacent to $v_{6}$ . Hence ${v_{2}, v_{5}}$ is a homogeneous clique. Let $Q$ be the graph on ${x, v_{1}, v_{3}, v_{4}, v_{6}}$ with edges ${x v_{1}, x v_{3}, x v_{4}, v_{1} v_{3}, v_{1} v_{6}, v_{4} v_{6}}$ . Replacing $x$ by the clique ${v_{2}, v_{5}}$ recovers all edges of $H$ , so $H = Q [x \to K_{2}]$ .

givenconstruct

1.2

In $H^{-}$ , delete $v_{6}$ . The remaining six vertices ${v, v_{1}, v_{2}, v_{3}, v_{4}, v_{5}}$ have edge set ${v v_{1}, v v_{2}, v_{1} v_{3}, v_{1} v_{5}, v_{2} v_{3}, v_{2} v_{4}, v_{2} v_{5}, v_{3} v_{5}, v_{4} v_{5}}$ . Relabel them by $x_{1} = v_{4}$ , $x_{2} = v$ , $x_{3} = v_{3}$ , $y = v_{1}$ , $z = v_{5}$ , and $w = v_{2}$ . Again the only missing edges are $x_{1} x_{2}, x_{1} x_{3}, x_{2} x_{3}, x_{1} y, x_{2} z, y w$ , so $H^{-} - v_{6}$ is also co-Bird.

L3givenalgebra

2.1

The graphs $Q$ and $K_{2}$ both have at most five vertices, so [L1] gives the Erdős-Hajnal property for each. By [L2], $H$ also has the Erdős-Hajnal property.

step 1.1L1L2

3.1

In $H^{+}$ , delete $v_{3}$ . The remaining six vertices ${v, v_{1}, v_{2}, v_{4}, v_{5}, v_{6}}$ have edge set ${v v_{1}, v v_{2}, v_{1} v_{2}, v_{1} v_{5}, v_{1} v_{6}, v_{2} v_{4}, v_{2} v_{5}, v_{4} v_{5}, v_{4} v_{6}}$ . Relabel them by $x_{1} = v$ , $x_{2} = v_{6}$ , $x_{3} = v_{5}$ , $y = v_{4}$ , $z = v_{2}$ , and $w = v_{1}$ . Then the only missing edges are $x_{1} x_{2}, x_{1} x_{3}, x_{2} x_{3}, x_{1} y, x_{2} z, y w$ , which are exactly the Bird edges. Therefore $H^{+} - v_{3}$ is co-Bird.

step 2.1L3algebra

4.1

Steps 2.1-3.1 together with step 1.2 verify the finite witness data used in the Bird route: $H$ has the Erdős-Hajnal property, and both $H^{+}$ and $H^{-}$ contain induced co-Bird subgraphs.

step 2.1step 3.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

31 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

Shenwei Huang, Yiao Ju, and Yidong Zhou, Erdős-Hajnal beyond the five-vertex path, Figure 6 and Lemma 2.2 (standard reference, not scraped)