ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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$K_{4}$ is bull-free and has a nontrivial module

Example

The complete graph $K_{4}$ is bull-free, and every two-vertex subset of $V (K_{4})$ is a nontrivial module.

Facts & Assumptions

Given: The complete graph $K_{4}$ .

[F1]

A bull-free graph has no induced bull (A bull-free graph).

[F2]

A module is a set to which every outside vertex is complete or anticomplete (Modules of a graph, and the trivial modules).

[F3]

In a complete graph every distinct vertex pair is adjacent (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

Verification

technique · direct

1.1

Every induced subgraph of $K_{4}$ is complete, while the bull has nonadjacent vertex pairs. So $K_{4}$ has no induced bull and is bull-free by [F1].

F1F3

2.1

Let $M = {v_{1}, v_{2}} \subseteq V (K_{4})$ . Every vertex outside $M$ is adjacent to both $v_{1}$ and $v_{2}$ by [F3], so it is complete to $M$ . Thus [F2] makes $M$ a module. Since $∣ M ∣ = 2$ and $M \neq V (K_{4})$ , it is nontrivial.

F2F3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

Maria Chudnovsky and Shmuel Safra, The Erdős-Hajnal conjecture for bull-free graphs, Section 1 (standard reference, not scraped)

K4K4​ is bull-free and has a nontrivial module