CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A difference of two nested modules that is not a module

Statement refuted

If $N \subseteq M$ are modules of a graph, then $M ∖ N$ is a module.

Facts & Assumptions

Given: The graph on vertices $a, b, c, d$ with exactly the two edges $c a$ and $c b$ , together with $N = {a, b}$ and $M = {a, b, c, d}$ .

[L1]

A set is a module when every outside vertex is complete or anticomplete to it (Modules of a graph, and the trivial modules).

[L2]

The difference lemma requires overlapping modules, not merely nested ones (If two modules overlap, then each difference and their symmetric difference are modules).

Counterexample

technique · constructive

1.1

The set $N = {a, b}$ is a module: the only outside vertices are $c$ and $d$ , and $c$ is complete to $N$ while $d$ is anticomplete to $N$ .

L1givenconstruct

1.2

The set $M$ is the whole vertex set, so it is a module vacuously.

L1given

2.1

The difference $M ∖ N = {c, d}$ is not a module, because the vertex $a$ is adjacent to $c$ and not to $d$ .

step 1.1given

3.1

Here $N \subseteq M$ , so the two modules do not overlap. Thus the overlap hypothesis in [L2] cannot be weakened to inclusion.

step 1.2step 2.1L2discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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.