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PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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A graph is bull-free if and only if its complement is bull-free

Statement

A finite simple graph G is bull-free if and only if its complement G is bull-free.

Facts & Assumptions

Given: A finite simple graph G.

[F1]

The bull has vertices {x1,x2,x3,y,z} and edges x1x2, x2x3, x1x3, x1y, and x2z (The bull graph).

[F2]

In the complement graph, two distinct vertices are adjacent exactly when they are nonadjacent in the original graph (Graph isomorphisms, automorphisms and graph complements).

[F3]

A graph is bull-free exactly when it has no induced bull (A bull-free graph).

Proof

technique · direct
1.1

By [F1] and [F2], the complement of the bull is again a bull: the bijection x1y, x2z, x3x3, yx2, zx1 sends nonedges of the bull to edges of the bull.

F1F2algebra
2.1

If G contains an induced bull on a vertex set S, then G[S] is the complement of that bull, hence another bull by step 1.1. The same argument with G and G interchanged proves the converse implication.

step 1.1F2
3.1

Therefore G has an induced bull exactly when G does, so [F3] gives the equivalence of bull-freeness.

step 2.1F3

Depends on

Used by

Dependency tree · two levels

6 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