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

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For a modular partition, a set of parts is a module of the quotient exactly when the union of those parts is a module of the graph

Statement

Let $P$ be a modular partition of a finite simple graph $G$ , let $X \subseteq P$ , and let $U = ⋃_{M \in X} M$ . Then $X$ is a module of $G / P$ if and only if $U$ is a module of $G$ .

Facts & Assumptions

Given: A modular partition $P$ of a finite simple graph $G$ , a subset $X \subseteq P$ , and the union $U = ⋃_{M \in X} M$ .

[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]

A modular partition of $G$ is a set of nonempty, pairwise disjoint modules of $G$ whose union is $V (G)$ ; the quotient $G / P$ has vertex set $P$ , with distinct parts $M, N$ adjacent exactly when $(M, N)$ is a complete pair in $G$ (Modular partitions and the quotient graph they define).

[L1]

Two disjoint nonempty modules of $G$ form a complete or an anticomplete pair, and not both (Two disjoint nonempty modules form a complete or an anticomplete pair).

[F3]

A disjoint pair is complete when every cross pair is an edge, anticomplete when no cross pair is an edge, and pure when it is complete or anticomplete (Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs).

Proof

technique · direct

1.1

Since the parts are nonempty, pairwise disjoint and cover $V (G)$ , a vertex $v$ lies outside $U$ exactly when the unique part $N$ containing it lies outside $X$ ; and $N$ is then disjoint from every $M \in X$ .

1.2

For $N \in P ∖ X$ , $M \in X$ and $v \in N$ , the pair $(N, M)$ is complete or anticomplete by [L1]; it is complete exactly when $v$ is adjacent to every vertex of $M$ , and anticomplete exactly when $v$ is adjacent to no vertex of $M$ , because $M$ and $N$ are nonempty and the alternative is excluded.

L1F3F2

2.1

For the forward direction, assume $X$ is a module of $G / P$ and let $v \in V (G) ∖ U$ , lying in the part $N \notin X$ of step 1.1. Then $N$ is adjacent in $G / P$ to every member of $X$ or to none. In the first case every $(N, M)$ with $M \in X$ is complete, so by step 1.2 the vertex $v$ is adjacent to every vertex of $U$ ; in the second case every such $(N, M)$ is anticomplete, so $v$ is adjacent to no vertex of $U$ .

step 1.1step 1.2F1F2

2.2

For the converse direction, assume $U$ is a module of $G$ and let $N \in P ∖ X$ , which is a vertex of $G / P$ outside $X$ . Choose $v \in N$ ; then $v \notin U$ by step 1.1, so $v$ is adjacent to every vertex of $U$ or to no vertex of $U$ . In the first case step 1.2 makes every pair $(N, M)$ with $M \in X$ complete, so $N$ is adjacent in $G / P$ to every member of $X$ ; in the second case every such pair is anticomplete, so $N$ is adjacent to none of them.

step 1.1step 1.2F1F2choose

3.1

Step 2.1 makes $({v}, U)$ pure for every $v \in V (G) ∖ U$ , so $U$ is a module of $G$ , and step 2.2 makes $({N}, X)$ pure in $G / P$ for every part $N$ outside $X$ , so $X$ is a module of $G / P$ ; together these are the two directions of the equivalence.

step 2.1step 2.2F1F3∎

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. 2.3 (standard reference, not scraped)