CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-31

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 comb can have an edge between two blocks

Statement refuted

Every comb has pairwise anticomplete blocks.

Facts & Assumptions

Given: The five-vertex graph from A cross-edge in a rooted stable-tooth comb creates an induced five-cycle, with teeth $a_{1}, a_{2}$ and singleton blocks ${b_{1}}, {b_{2}}$ .

[L1]

The definition of a comb only requires each tooth to be adjacent to its own block and anticomplete to the other blocks; it does not impose any condition on edges between different blocks (Combs in a graph).

Counterexample

technique · direct

1.1

In the given graph, $a_{1}$ is adjacent to $b_{1}$ and not to $b_{2}$ , while $a_{2}$ is adjacent to $b_{2}$ and not to $b_{1}$ . The teeth are distinct and the singleton blocks are disjoint. Therefore [L1] shows that $((a_{1}, {b_{1}}), (a_{2}, {b_{2}}))$ is a comb.

L1given

2.1

The two blocks are not anticomplete, because the graph was built with the edge $b_{1} b_{2}$ . So this comb refutes the statement that every comb has pairwise anticomplete blocks.

step 1.1given∎

Depends on

Used by

FALSE: every comb has pairwise anticomplete blocks False statement

Dependency tree · two levels

8 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

Maria Chudnovsky, Alex Scott, Paul Seymour, and Sophie Spirkl, Erdős-Hajnal for graphs with no 5-hole, proof of Theorem 4.4 (standard reference, not scraped)