LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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 vertex mixed on an anticonnected set yields opposite adjacency on a nonedge

Statement

Let $G$ be a finite graph, let $A \subseteq V (G)$ be anticonnected, and let $v \in V (G) ∖ A$ be mixed on $A$ . Then there exist distinct vertices $b, b^{'} \in A$ such that

$b b^{'} \notin E (G), v b \in E (G), v b^{'} \notin E (G) .$

Facts & Assumptions

Given: A finite graph $G$ , an anticonnected set $A \subseteq V (G)$ , and a vertex $v \in V (G) ∖ A$ that is mixed on $A$ .

[L1]

A set is anticonnected exactly when the induced subgraph on that set is connected in the complement graph (Anticonnected graphs and anticonnected components).

[L2]

Because $v$ is mixed on $A$ , it has at least one neighbour and at least one nonneighbour in $A$ (Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs).

Proof

technique · direct

1.1L1L2givenchoose

By [L2], choose $p, q \in A$ with $v p \in E (G)$ and $v q \notin E (G)$ . Since $A$ is anticonnected, [L1] gives a path $p = x_{0}, x_{1}, \dots, x_{m} = q$ in the complement graph $\overline{G} [A]$ .

2.1step 1.1choose

Along that path, the truth value of " $v x_{i} \in E (G)$ " changes from true at $i = 0$ to false at $i = m$ . Hence there exists $k < m$ such that $v x_{k} \in E (G)$ and $v x_{k + 1} \notin E (G)$ .

3.1step 2.1L1algebra∎

Because $x_{k} x_{k + 1}$ is an edge of $\overline{G} [A]$ , it is a nonedge of $G$ . Therefore $b : = x_{k}$ and $b^{'} : = x_{k + 1}$ satisfy $b b^{'} \notin E (G)$ , $v b \in E (G)$ , and $v b^{'} \notin E (G)$ , which is the conclusion.

Depends on

Used by

Dependency tree · two levels

7 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

Shenwei Huang, Yiao Ju, and Yidong Zhou, Erdős-Hajnal beyond the five-vertex path, proof of Lemma 2.1 (standard reference, not scraped)