LemmaStatement: Literature-sourcedProof: 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.

The $E$ -graph and the Bird graph are wonderful

Statement

The singleton families ${E}$ and ${B i r d}$ are wonderful. Equivalently, the $E$ -graph and the Bird graph are wonderful.

Facts & Assumptions

Given: The $E$ -graph, the Bird graph, and the wonderfulness criterion.

[L1]

A finite family is wonderful if it satisfies either the star-subdivision obstruction or the special-vertex obstruction from the previous criterion (Star and special-vertex obstructions force wonderfulness).

[L2]

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

[L3]

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).

[L4]

The Bird graph and co-Bird are complements of one another, and $H^{+}$ is the graph obtained by adding a new vertex adjacent exactly to the two distinguished vertices (The Bird graph and co-Bird, The graphs $H^{+}$ and $H^{-}$ for two distinguished vertices).

[A1]

Let $H$ be the graph on vertices $v_{1}, \dots, v_{6}$ with edge set

{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}} .

Its distinguished vertices are $v_{1}$ and $v_{2}$ .

Proof

technique · verify the two criterion inputs explicitly

1.1

For the $E$ -graph, take the $1$ -subdivision of $K_{1, 3}$ with center $c$ , subdivision vertices $s_{1}, s_{2}, s_{3}$ , and leaves $t_{1}, t_{2}, t_{3}$ . On the six-vertex subset ${t_{1}, s_{1}, c, s_{2}, t_{2}, s_{3}}$ , the induced edges are $t_{1} s_{1}$ , $s_{1} c$ , $c s_{2}$ , $s_{2} t_{2}$ , and $c s_{3}$ , which is exactly the edge set of the $E$ -graph from The $E$ -graph and co- $E$ . Thus $E$ is an induced subgraph of the $1$ -subdivision of $K_{1, 3}$ , so [L1] makes ${E}$ wonderful.

L1givenconstruct

1.2

In the graph $H$ from [A1], the vertices $v_{2}$ and $v_{5}$ are adjacent to each other and both have the same neighbourhood outside ${v_{2}, v_{5}}$ , namely ${v_{1}, v_{3}, v_{4}}$ . Hence ${v_{2}, v_{5}}$ is a homogeneous clique. Let $Q$ be the five-vertex graph on ${x, v_{1}, v_{3}, v_{4}, v_{6}}$ with edge set ${x v_{1}, x v_{3}, x v_{4}, v_{1} v_{3}, v_{1} v_{6}, v_{4} v_{6}}$ . Then $H$ is obtained from $Q$ by substituting $K_{2}$ for the vertex $x$ . By [L2], both $Q$ and $K_{2}$ have the Erdős-Hajnal property, so [L3] gives the Erdős-Hajnal property for $H$ .

A1L2L3

1.3

Form $H^{+}$ from [A1] by adjoining a new vertex $v$ adjacent to $v_{1}$ and $v_{2}$ , and delete $v_{3}$ . On the remaining six vertices ${v, v_{1}, v_{2}, v_{4}, v_{5}, v_{6}}$ the edge set is ${v v_{1}, v v_{2}, v_{1} v_{2}, v_{1} v_{5}, v_{1} v_{6}, v_{2} v_{4}, v_{2} v_{5}, v_{4} v_{5}, v_{4} v_{6}}$ . Under the relabelling $x_{1} = v$ , $x_{2} = v_{6}$ , $x_{3} = v_{5}$ , $y = v_{4}$ , $z = v_{2}$ , and $w = v_{1}$ , the six missing edges are exactly $x_{1} x_{2}$ , $x_{1} x_{3}$ , $x_{2} x_{3}$ , $x_{1} y$ , $x_{2} z$ , and $y w$ , which are precisely the Bird edges. Therefore $H^{+} - v_{3}$ is co-Bird, so $H^{+}$ is not co-Bird-free.

A1L4algebra

2.1

Step 1.2 shows that ${H} \cup {c o - B i r d}$ has the Erdős-Hajnal property, and step 1.3 shows that $H^{+}$ is not co-Bird-free. Therefore [L1] applies to the singleton family ${B i r d}$ and proves that Bird is wonderful. Together with step 1.1, this proves the statement.

L1step 1.1step 1.2step 1.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 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, Lemma 2.2 and Figure 6 (standard reference, not scraped)

The EE-graph and the Bird graph are wonderful