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

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Two disjoint modules whose union is not a module

Statement refuted

If M and N are disjoint modules of a graph, then MN is a module.

Facts & Assumptions

Given: The path P4 with vertices 0,1,2,3 in order, together with the two sets M={0} and N={2}.

[L1]

Singletons are trivial modules (Modules of a graph, and the trivial modules).

[L2]

In the path P4, the vertices are adjacent exactly along the edges 01, 12, and 23 (Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices).

Counterexample

technique · constructive
1.1

The sets M={0} and N={2} are disjoint.

givenconstruct
1.2

By [L1], both M and N are modules of P4.

L1
2.1

The vertex 3 lies outside MN, is adjacent to 2, and is not adjacent to 0 by [L2], so it splits MN. Hence MN is not a module.

step 1.1L2
3.1

Thus P4 with M and N is a counterexample to the claim.

step 1.2step 2.1discharge-construct

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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.