TheoremStatement: 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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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A graph is recovered from any modular partition by the induced subgraphs on the parts together with the quotient graph

Statement

Let $P$ be a modular partition of a finite simple graph $G$ , and let $x, y$ be distinct vertices of $G$ , lying in the parts $M, N \in P$ respectively. Then

if $M \neq N$ : ${x, y} \in E (G)$ if and only if ${M, N} \in E (G / P)$ ;
if $M = N$ : ${x, y} \in E (G)$ if and only if ${x, y} \in E (G [M])$ .

Consequently $E (G)$ is determined by $G / P$ together with the induced subgraphs $G [M]$ for $M \in P$ . In particular, if $P = {M_{1}, M_{2}}$ has exactly the two parts $M_{1}, M_{2}$ and $t \in M_{2}$ , then $G = G [M_{1} \cup {t}] [t \to G [M_{2}]]$ .

Facts & Assumptions

Given: A modular partition $P$ of a finite simple graph $G$ , and distinct vertices $x \in M$ and $y \in N$ with $M, N \in P$ .

[F1]

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).

[F2]

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).

[F3]

$G [W] = (W, E (G) \cap [W]^{2})$ (Subgraphs, induced subgraphs and spanning subgraphs).

[F4]

The vertex set of $G_{1} [a \to G_{2}]$ is $(V (G_{1}) ∖ {a}) \cup V (G_{2})$ , a disjoint union; two vertices of $V (G_{1}) ∖ {a}$ are adjacent there exactly when they are adjacent in $G_{1}$ , two vertices of $V (G_{2})$ exactly when they are adjacent in $G_{2}$ , and $u \in V (G_{1}) ∖ {a}$ is adjacent to $x \in V (G_{2})$ exactly when $u$ is adjacent to $a$ in $G_{1}$ (Substituting one graph for a vertex of another).

[L2]

For a module $M$ of $G$ : for all $x, y \in M$ and all $v \in V (G) ∖ M$ , $v x \in E (G)$ if and only if $v y \in E (G)$ (Three equivalent descriptions of a module: purity of every outside vertex, equality of outside neighbourhoods, and indistinguishability of the members).

[F5]

$M$ 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).

Proof

technique · cases

1.1

First case: $M \neq N$ . The parts are disjoint nonempty modules, so $(M, N)$ is complete or anticomplete and not both; if it is complete then ${x, y} \in E (G)$ , and if it is anticomplete then ${x, y} \notin E (G)$ . Since ${M, N} \in E (G / P)$ says exactly that $(M, N)$ is complete, the two conditions agree.

assume-case acrossF1L1F2

1.2

Second case: $M = N$ . Then $x, y$ are distinct vertices of $M$ , and the edges of $G [M]$ are the edges of $G$ with both ends in $M$ , so ${x, y} \in E (G)$ if and only if ${x, y} \in E (G [M])$ .

assume-case insideF3

1.3

Suppose now that $P = {M_{1}, M_{2}}$ with $M_{1} \neq M_{2}$ , fix $t \in M_{2}$ and put $G_{1} = G [M_{1} \cup {t}]$ and $G_{2} = G [M_{2}]$ . Then $V (G_{1}) ∖ {t} = M_{1}$ is disjoint from $V (G_{2}) = M_{2}$ , and $M_{2} \neq \emptyset$ , so $G_{1} [t \to G_{2}]$ is a substitution with vertex set $M_{1} \cup M_{2} = V (G)$ .

F1F3F4choose

2.1

Every pair of distinct vertices of $G$ falls into exactly one of the two cases, since each vertex lies in exactly one part, so the cases are exhaustive and steps 1.1 and 1.2 determine $E (G)$ from $G / P$ and the graphs $G [M]$ .

step 1.1step 1.2F1cases-exhaustive

2.2

In the two-part situation of step 1.3, take distinct $p, q \in V (G)$ . If $p, q \in M_{1}$ , then ${p, q}$ is an edge of $G_{1} [t \to G_{2}]$ exactly when it is an edge of $G_{1}$ , hence exactly when it is an edge of $G$ ; if $p, q \in M_{2}$ the same holds through $G_{2}$ ; and if $p \in M_{1}$ and $q \in M_{2}$ , then ${p, q}$ is an edge of $G_{1} [t \to G_{2}]$ exactly when ${p, t} \in E (G_{1})$ , that is exactly when ${p, t} \in E (G)$ , which by [L2] applied to the module $M_{2}$ with $t, q \in M_{2}$ and $p \notin M_{2}$ holds exactly when ${p, q} \in E (G)$ .

step 1.3F3F4L2F5

3.1

So in the two-part situation the graphs $G$ and $G [M_{1} \cup {t}] [t \to G [M_{2}]]$ have the same vertex set and the same edges, and are therefore equal; with step 2.1 this proves every clause of the Statement.

step 2.1step 2.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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)
T. Harju, Lecture Notes on Combinatorial Structures in Graph Theory, sec. 3 (standard reference, not scraped)