ExampleConstruction: Literature-sourcedVerification: 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.

Every vertex set is a module of a complete graph and of an edgeless graph

Example

If $G$ is a complete graph or an edgeless graph, then every vertex subset of $G$ is a module (Modules of a graph, and the trivial modules).

Facts & Assumptions

Given: An integer $n \geq 1$ , the complete graph $K_{n}$ , the edgeless graph $\overline{K_{n}}$ , and a subset $M$ of their common vertex set.

[L1]

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

[L2]

The graphs $K_{n}$ and $\overline{K_{n}}$ are, respectively, the complete and edgeless graphs on $n$ vertices (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

[L3]

A graph is prime when its only modules are the trivial ones (Prime graphs: those whose only modules are the trivial ones).

Verification

technique · direct

1.1

In $K_{n}$ , every vertex outside $M$ is adjacent to every member of $M$ , so [L1] makes $M$ a module.

L1L2

1.2

In $\overline{K_{n}}$ , every vertex outside $M$ is adjacent to no member of $M$ , so [L1] again makes $M$ a module.

L1L2

2.1

If $n \geq 3$ , then $K_{n}$ and $\overline{K_{n}}$ have nontrivial modules by steps 1.1 and 1.2, so [L3] shows that neither is prime. For $n = 1$ and $n = 2$ there is no subset of size between $2$ and $n - 1$ , so every module is trivial and both graphs are prime.

step 1.1step 1.2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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)