Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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Maximal proper modules need not be disjoint when the graph or its complement is disconnected

Statement refuted

In every graph, the maximal proper modules are pairwise disjoint.

Facts & Assumptions

Given: The edgeless graph E3 on vertices a,b,c.

[L1]

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

[L3]

A set is a module of a graph exactly when it is a module of the complement (A vertex set is a module of G exactly when it is a module of G).

Counterexample

technique · constructive
1.1

In the edgeless graph E3, every subset is a module, because every outside vertex is anticomplete to it.

L1givenconstruct
2.1

The sets {a,b} and {a,c} are proper modules, and each is maximal among proper modules because the only larger module containing it is the whole vertex set.

step 1.1
3.1

These two maximal proper modules meet in a, so they are not pairwise disjoint.

step 2.1
3.2

By [L3], the same two sets are also overlapping maximal proper modules in the complement graph K3, which is connected while its complement is disconnected.

step 2.1L3
4.1

Therefore the conclusion of [L2] genuinely needs the connected-and-anticonnected hypotheses.

step 3.1step 3.2L2discharge-construct

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 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