ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13

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.

The five-cycle is $K_{2, 2}$ -avoiding and shows the KST problem is not just a complete-bipartite construction

Example

The cycle $C_{5}$ contains no ordinary $K_{2, 2}$ , although $C_{5}$ itself is not bipartite.

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

For $n \geq 3$ , $C_{n}$ has the consecutive edges ${i, i + 1}$ and the closing edge ${n - 1, 0}$ (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

[F2]

The complete bipartite graph $K_{A, B}$ has exactly all edges joining a vertex of $A$ to a vertex of $B$ (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

Verification

technique · inspect common neighbours

1.1

Label the cycle vertices modulo $5$ . Two adjacent vertices have no common neighbour, and two nonadjacent vertices have exactly one common neighbour. Thus no pair has the two common neighbours required to form a $K_{2, 2}$ .

givenF1F2

2.1

Therefore $C_{5}$ is $K_{2, 2}$ -free. Its odd cycle is also a reminder that the host graphs in the ordinary KST problem need not themselves be bipartite.

step 1.1given∎

Depends on

Empty and complete graphs, complete bipartite graphs, and the convention that $P_n$ and $C_n$ have $n$ vertices

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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.

The five-cycle is K2,2K2,2​-avoiding and shows the KST problem is not just a complete-bipartite construction