LemmaStatement: Literature-sourcedProof: 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.

The intersection of two modules is a module

Statement

If $M$ and $N$ are modules of a finite simple graph $G$ , then $M \cap N$ is a module of $G$ . No hypothesis relating $M$ and $N$ is needed, and the case $M \cap N = \emptyset$ is included.

Facts & Assumptions

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

[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

1.1

If $v$ is complete to a set $S$ then $v$ is complete to every subset of $S$ , and if $v$ is anticomplete to $S$ then $v$ is anticomplete to every subset of $S$ ; so purity of $({v}, S)$ passes to every subset of $S$ .

1.2

First case: $v \notin M$ . Then $({v}, M)$ is pure because $M$ is a module.

assume-case outMF1

1.3

Second case: $v \in M$ . Since $v \notin M \cap N$ , this forces $v \notin N$ , and then $({v}, N)$ is pure because $N$ is a module.

assume-case inMF1given

2.1

In the first case step 1.1 applied to $S = M \supseteq M \cap N$ makes $({v}, M \cap N)$ pure, and in the second case step 1.1 applied to $S = N \supseteq M \cap N$ does the same. The two cases exhaust the possibilities for $v$ .

step 1.1step 1.2step 1.3cases-exhaustive

3.1

Every vertex outside $M \cap N$ therefore has $({v}, M \cap N)$ pure, which is the module condition of [F1], so $M \cap N$ 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. 2.1 (standard reference, not scraped)