LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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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.
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The union of two modules with a common vertex is a module

Statement

Let $M$ and $N$ be modules of a finite simple graph $G$ with $M \cap N \neq \emptyset$ . Then $M \cup N$ is a module of $G$ .

Facts & Assumptions

Given: Modules $M, N$ of a finite simple graph $G$ with $M \cap N \neq \emptyset$ , and a vertex $v \in V (G) ∖ (M \cup N)$ .

[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 pair $(A, B)$ of disjoint sets is complete when every $a \in A$ is adjacent to every $b \in B$ , anticomplete when no $a \in A$ is adjacent to any $b \in B$ , and pure when it is complete or anticomplete (Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs).

Proof

technique · cases

1.1givenF1choose

Fix $u \in M \cap N$ . Since $v \notin M \cup N$ , the vertex $v$ lies outside $M$ and outside $N$ , so both $({v}, M)$ and $({v}, N)$ are pure.

1.2assume-case adjF2

First case: $u v \in E (G)$ . Then $({v}, M)$ is not anticomplete, since $u \in M$ , so it is complete; and likewise $({v}, N)$ is complete.

1.3assume-case nonadjF2

Second case: $u v \notin E (G)$ . Then $({v}, M)$ is not complete, since $u \in M$ , so it is anticomplete; and likewise $({v}, N)$ is anticomplete.

2.1step 1.1step 1.2step 1.3F2cases-exhaustive

In the first case $v$ is adjacent to every vertex of $M$ and to every vertex of $N$ , hence to every vertex of $M \cup N$ ; in the second case $v$ is adjacent to no vertex of $M$ and to no vertex of $N$ , hence to no vertex of $M \cup N$ . The two cases exhaust the possibilities.

3.1step 2.1F1∎

So $({v}, M \cup N)$ is pure for every vertex $v$ outside $M \cup N$ , which is the module condition of [F1].

Depends on

Used by

Dependency tree · two levels

7 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. 2.1 (standard reference, not scraped)