LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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.

If $M$ is a module of $G$ and $W \subseteq V (G)$ , then $M \cap W$ is a module of $G [W]$

Statement

Let $M$ be a module of a finite simple graph $G$ and let $W \subseteq V (G)$ . Then $M \cap W$ is a module of the induced subgraph $G [W]$ .

Facts & Assumptions

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

[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]

$G [W] = (W, E (G) \cap [W]^{2})$ , so two vertices of $W$ are adjacent in $G [W]$ 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 · direct

1.1F1given

Let $v \in W ∖ (M \cap W)$ . Since $v \in W$ , this gives $v \notin M$ , so $({v}, M)$ is pure in $G$ .

2.1F2

The vertex set of $G [W]$ is $W$ , so the vertices outside $M \cap W$ in $G [W]$ are exactly the vertices $v$ of step 1.1.

2.2step 1.1F3

If $({v}, M)$ is complete in $G$ then $v$ is adjacent in $G$ to every vertex of $M \cap W \subseteq M$ , and if it is anticomplete then $v$ is adjacent in $G$ to no vertex of $M \cap W$ .

3.1step 2.2F2F3

Both $v$ and the vertices of $M \cap W$ lie in $W$ , so those adjacencies are the same in $G [W]$ as in $G$ ; hence $({v}, M \cap W)$ is pure in $G [W]$ .

4.1step 2.1step 3.1F1∎

Every vertex of $G [W]$ outside $M \cap W$ therefore satisfies the module condition of [F1] in $G [W]$ , so $M \cap W$ is a module of $G [W]$ .

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)

If MM is a module of GG and W⊆V(G)W⊆V(G), then M∩WM∩W is a module of G[W]G[W]