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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A module of $G [M]$ is a module of $G$ whenever $M$ is a module of $G$

Statement

Let $M$ be a module of a finite simple graph $G$ and let $X \subseteq M$ be a module of the induced subgraph $G [M]$ . Then $X$ is a module of $G$ .

Facts & Assumptions

Given: A module $M$ of a finite simple graph $G$ , a module $X$ of $G [M]$ , and a vertex $v \in V (G) ∖ X$ .

[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]

$G [M] = (M, E (G) \cap [M]^{2})$ , so two vertices of $M$ are adjacent in $G [M]$ exactly when they are adjacent in $G$ (Subgraphs, induced subgraphs and spanning subgraphs).

[F3]

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

1.1

First case: $v \in M ∖ X$ . Since $X$ is a module of $G [M]$ and $v$ is a vertex of $G [M]$ outside $X$ , the pair $({v}, X)$ is pure in $G [M]$ ; as $v$ and the vertices of $X$ all lie in $M$ , the same adjacencies hold in $G$ , so $({v}, X)$ is pure in $G$ .

assume-case inMF1F2F3

1.2

Second case: $v \notin M$ . Then $({v}, M)$ is pure in $G$ because $M$ is a module of $G$ , and $X \subseteq M$ , so $({v}, X)$ is pure in $G$ .

assume-case outMF1F3

2.1

A vertex $v \notin X$ lies in $M$ or outside $M$ , so the two cases are exhaustive and $({v}, X)$ is pure in $G$ for every $v \in V (G) ∖ X$ .

step 1.1step 1.2givencases-exhaustive

3.1

That is the module condition of [F1] for $X$ in $G$ , so $X$ is a module of $G$ .

step 2.1F1∎

Depends on

Used by

Nothing in the library uses this result yet.

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. 2.3 (standard reference, not scraped)

A module of G[M]G[M] is a module of GG whenever MM is a module of GG