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.

Two disjoint nonempty modules form a complete or an anticomplete pair

Statement

Let $M$ and $N$ be disjoint nonempty modules (Modules of a graph, and the trivial modules) of a finite simple graph $G$ . Then the pair $(M, N)$ is complete or anticomplete, and it is not both.

Facts & Assumptions

Given: Disjoint nonempty modules $M, N$ of a finite simple graph $G$ .

[L1]

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).

[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 · direct

1.1

Fix $x_{0} \in M$ and $y_{0} \in N$ , which exist because both sets are nonempty.

givenchoose

1.2

Let $x \in M$ and $y \in N$ . Since $y \notin M$ and $x, x_{0} \in M$ , applying [L1] to the module $M$ gives that $x y \in E (G)$ if and only if $x_{0} y \in E (G)$ .

L1given

1.3

Since $x_{0} \notin N$ and $y, y_{0} \in N$ , applying [L1] to the module $N$ gives that $x_{0} y \in E (G)$ if and only if $x_{0} y_{0} \in E (G)$ .

L1given

2.1

Combining steps 1.2 and 1.3, every $x \in M$ and $y \in N$ satisfy: $x y \in E (G)$ if and only if $x_{0} y_{0} \in E (G)$ .

step 1.1step 1.2step 1.3

3.1

If $x_{0} y_{0} \in E (G)$ then step 2.1 makes every cross pair an edge, so $(M, N)$ is complete; otherwise step 2.1 makes no cross pair an edge, so $(M, N)$ is anticomplete. The two cannot both hold, since the single pair $x_{0} y_{0}$ would then be both an edge and not an edge.

step 2.1F2∎

Depends on

Used by

Dependency tree · two levels

8 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)