ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-01

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.

A graph is $P_{3}$ -free if and only if every connected component is complete

Statement

A finite graph is $P_{3}$ -free if and only if each of its connected components is complete. Such graphs are called cluster graphs.

Facts & Assumptions

Given: A finite graph $G$ .

[F1]

$P_{3}$ -free means having no induced copy of the three-vertex path ( $H$ -free and $F$ -free graphs under the induced-subgraph convention, Induced embeddings and induced copies of a graph).

[L1]

Connected components partition the vertices and contain every path between their vertices (The connected components of a graph partition its vertex set and are its maximal connected subgraphs).

[F2]

$P_{3}$ has two consecutive edges and no edge between its endpoints (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

Verification

technique · direct

1.1

If every component is complete, three vertices in one component induce all three edges, while vertices meeting more than one component induce edges only within their component parts and therefore at most one edge. Neither pattern is $P_{3}$ , so $G$ is $P_{3}$ -free.

L1F1F2

1.2

Conversely, suppose a component contains nonadjacent vertices $x, y$ . Choose a shortest $x$ - $y$ path $x = v_{0}, v_{1}, \dots, v_{k} = y$ , where $k \geq 2$ .

chooseL1

2.1

Minimality of the path gives $v_{0} v_{2} \notin E (G)$ , so $G [{v_{0}, v_{1}, v_{2}}] ≅ P_{3}$ .

step 1.2F2

3.1

Therefore a $P_{3}$ -free graph cannot have a noncomplete component. Combining both directions proves the equivalence.

step 1.1step 2.1F1∎

Remarks


\node[vertex] (d) at (4.1,.65) {}; \node[vertex] (e) at (5.55,.65) {}; \draw[edge] (d)--(e); \node[caption] at (4.825,-.65) {complete component $K_2$};

\node[caption] at (2.85,-1.25) {$K_3\mathbin{\dot\cup}K_2$ is a cluster graph and is $P_3$-free}; \end{tikzpicture} ```

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.

Sources

ISGCI, Cluster graphs (standard reference, not scraped)

A graph is P3P3​-free if and only if every connected component is complete