LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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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.

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A vertex set is a module of $G$ exactly when it is a module of $\overline{G}$

Statement

For every finite simple graph $G$ and every $M \subseteq V (G)$ , the set $M$ is a module of $G$ if and only if it is a module of $\overline{G}$ .

Facts & Assumptions

Given: A finite simple graph $G$ and a set $M \subseteq V (G)$ .

[F1]

$M$ is a module of $G$ when the pair $({v}, M)$ is pure for every $v \in V (G) ∖ M$ (Modules of a graph, and the trivial modules).

[F2]

The complement of $G = (V, E)$ is $\overline{G} = (V, [V]^{2} ∖ E)$ , and $\overline{\overline{G}} = G$ (Graph isomorphisms, automorphisms and graph complements).

[L1]

For disjoint vertex sets $A, B$ in a graph, complementation swaps complete pairs with anticomplete pairs and preserves pure pairs and mixed pairs (Purity is symmetric; complementation swaps complete and anticomplete pairs and preserves mixed pairs).

[F3]

A disjoint pair is pure when it is complete or anticomplete (Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs).

Proof

technique · direct

1.1

For $v \in V (G) ∖ M$ the sets ${v}$ and $M$ are disjoint, so the pair $({v}, M)$ is pure in $G$ exactly when it is pure in $\overline{G}$ .

L1F3

1.2

The graphs $G$ and $\overline{G}$ have the same vertex set, so a vertex lies outside $M$ in one exactly when it lies outside $M$ in the other.

2.1

If $M$ is a module of $G$ , then $({v}, M)$ is pure in $G$ for every vertex $v$ outside $M$ , hence pure in $\overline{G}$ for every such vertex, so $M$ is a module of $\overline{G}$ .

step 1.1step 1.2F1

3.1

Applying step 2.1 to the graph $\overline{G}$ and using $\overline{\overline{G}} = G$ gives the converse implication, so $M$ is a module of $G$ exactly when it is a module of $\overline{G}$ .

step 2.1F2∎

Depends on

Used by

Dependency tree · two levels

10 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

M. Habib and C. Paul, A Survey on Algorithmic Aspects of Modular Decomposition, sec. 2.5 (standard reference, not scraped)

A vertex set is a module of GG exactly when it is a module of G‾G