CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-16

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.

Three pair densities equal to $1 / 2$ need not produce a single transversal triangle

Statement

There are three pairwise disjoint vertex sets $X, Y, Z$ for which every cross-density equals $1 / 2$ but no transversal triple spans a triangle.

Facts & Assumptions

Given: Three nonempty even-sized sets, each split equally into parts labelled $0$ and $1$ .

[L1]

Cross-density is the proportion of possible cross-pairs that are edges (Edge counts and densities between nonempty vertex sets).

[L2]

The triangle counting lemma requires regularity in addition to positive pair densities (Triangle counting lemma for three pairwise regular vertex sets).

Counterexample

technique · contradiction

1.1givenchoose

Join $X$ to $Y$ and $Y$ to $Z$ exactly when the endpoint labels agree, and join $X$ to $Z$ exactly when their labels differ.

2.1step 1.1L1algebra

For each cross-pair and each vertex, exactly half the vertices on the other side are neighbours. Thus all three densities are $1 / 2$ by [L1].

2.2assume-contrastep 1.1

Suppose, for contradiction, that $(x, y, z)$ is a transversal triangle. Its $X Y$ and $Y Z$ edges force the three labels to satisfy $x = y = z$ , while its $X Z$ edge forces $x \neq z$ .

3.1step 2.2L2discharge-contradiction∎

This contradiction shows that no transversal triangle exists. Therefore density alone does not imply the conclusion of [L2]; its regularity hypothesis carries real content.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Y. Zhao, Graph Theory and Additive Combinatorics, Theorem 2.2.1 and Remark 2.2.2 (standard reference, not scraped)

Three pair densities equal to 1/21/2 need not produce a single transversal triangle