LemmaStatement: 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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In a connected graph, some vertex outside a nonempty proper module is complete to it

Statement

Let $G$ be a connected finite simple graph and let $M$ be a module of $G$ with $\emptyset \neq M \neq V (G)$ . Then some vertex $v \in V (G) ∖ M$ is complete to $M$ .

Facts & Assumptions

Given: A connected finite simple graph $G$ and a module $M$ of $G$ with $\emptyset \neq M \neq V (G)$ .

[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]

A graph is connected when its vertex set is nonempty and every two of its vertices are joined by a path (Connected graphs and connected components defined by the existence of vertex paths).

[F3]

A walk of length $ℓ$ is a vertex list $(v_{0}, \dots, v_{ℓ})$ with ${v_{i - 1}, v_{i}} \in E$ for every $1 \leq i \leq ℓ$ , and a path is a walk whose vertices are distinct (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

[F4]

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).

[L1]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

Proof

technique · direct

1.1

Choose $u \in M$ and $w \in V (G) ∖ M$ ; both choices are possible because $M \neq \emptyset$ and $M \neq V (G)$ .

givenchoose

1.2

Since $G$ is connected there is a path $(v_{0}, \dots, v_{ℓ})$ with $v_{0} = u$ and $v_{ℓ} = w$ .

F2F3given

2.1

The set ${i \leq ℓ : v_{i} \notin M}$ is a nonempty subset of $N$ , since it contains $ℓ$ , so it has a least element $k$ ; and $k \geq 1$ because $v_{0} = u \in M$ .

step 1.1step 1.2L1

3.1

By minimality $v_{k - 1} \in M$ , and ${v_{k - 1}, v_{k}}$ is an edge of $G$ because consecutive vertices of a path are adjacent.

step 2.1F3

4.1

Put $v = v_{k}$ . Then $v \notin M$ , so $({v}, M)$ is pure, and it is not anticomplete because $v$ is adjacent to $v_{k - 1} \in M$ ; hence it is complete, that is, $v$ is complete to $M$ .

step 3.1F1F4∎

Depends on

Used by

Dependency tree · two levels

20 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)