LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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 quotient-level mixed-block witness descends to two mixed member blocks

Statement

Let $L$ be a blockade in a graph $G$ , and suppose that every block of $L$ is connected or every block is anticonnected. Let $D_{1}, D_{2}$ be distinct mixed blocks of the quotient blockade $L / M$ . Assume there are vertices $x, y, u \notin D_{1} \cup D_{2}$ such that:

$x$ and $y$ are nonadjacent and both are complete to $D_{1} \cup D_{2}$ ;
$u \in N (x) ∖ N (y)$ , with $u$ complete to $D_{1}$ and anticomplete to $D_{2}$ ; and
no vertex of $D_{1}$ is mixed on $D_{2}$ .

Then there are mixed original blocks $A_{1}, A_{2}$ of $L$ , both contained in $D_{1}$ , and vertices $x^{'}, y^{'}, u^{'} \notin A_{1} \cup A_{2}$ such that:

$x^{'}$ and $y^{'}$ are nonadjacent and both are complete to $A_{1} \cup A_{2}$ ; and
$u^{'} \in N (x^{'}) ∖ N (y^{'})$ , with $u^{'}$ complete to $A_{1}$ and anticomplete to $A_{2}$ .

Facts & Assumptions

Given: The hypotheses of the Statement.

[L1]

Distinct original blocks lying in different quotient blocks are pure to each other (Blocks from distinct mixed-block classes are pure to each other).

[L2]

If a vertex outside a quotient block is mixed on that quotient block but pure to each original block inside it, then two mixed original member blocks witness opposite adjacency to that vertex (A vertex mixed on a quotient block but pure on each member block yields two mixed member blocks with opposite adjacency).

Proof

technique · direct

1.1

Since $D_{1}$ and $D_{2}$ are mixed as quotient blocks, there is a vertex in one of them that is mixed on the other. Hypothesis 3 excludes the possibility that a vertex of $D_{1}$ is mixed on $D_{2}$ , so choose a vertex $b_{2} \in D_{2}$ that is mixed on $D_{1}$ .

givenchoose

2.1

The quotient blocks $D_{1}$ and $D_{2}$ are distinct. Therefore [L1] implies that every original block of $L$ contained in $D_{2}$ is pure to every original block of $L$ contained in $D_{1}$ . In particular, if $B$ is the original block of $L$ containing $b_{2}$ , then $b_{2}$ is pure to every original block contained in $D_{1}$ .

step 1.1L1

3.1

Now $b_{2}$ is outside $D_{1}$ , is mixed on $D_{1}$ by step 1.1, and is pure to every original block inside $D_{1}$ by step 2.1. Applying [L2], choose mixed original blocks $A_{1}, A_{2} \subseteq D_{1}$ such that $b_{2}$ is complete to $A_{1}$ and anticomplete to $A_{2}$ .

step 2.1L2choose

4.1

Set $x^{'} : = y$ , $y^{'} : = u$ , and $u^{'} : = b_{2}$ . Because $y$ is complete to $D_{1} \cup D_{2}$ , it is complete to $A_{1} \cup A_{2}$ and adjacent to $b_{2}$ . Because $u$ is complete to $D_{1}$ and anticomplete to $D_{2}$ , it is complete to $A_{1} \cup A_{2}$ and nonadjacent to $b_{2}$ . Hypothesis 2 gives $u \in N (x) ∖ N (y)$ , so $y$ and $u$ are nonadjacent. Therefore $u^{'} = b_{2} \in N (x^{'}) ∖ N (y^{'})$ , while $u^{'}$ is complete to $A_{1}$ and anticomplete to $A_{2}$ by step 3.1. This is exactly the required witness.

step 3.1givenalgebra∎

Depends on

Used by

The quotient-witness reduction in a three-block configuration Example

Dependency tree · two levels

10 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, Lemma 6.2 (standard reference, not scraped)