ExampleConstruction: AI-generatedVerification: 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 vertex may be mixed on a quotient block while pure on each member block

Example

Let $L = (A_{1}, A_{2})$ be a two-block blockade in which $A_{1}$ and $A_{2}$ are mixed. Then its quotient blockade has the single block $D : = A_{1} \cup A_{2}$ . If a vertex $u \notin D$ is complete to $A_{1}$ and anticomplete to $A_{2}$ , then $u$ is mixed on $D$ although it is pure to each member block separately.

Facts & Assumptions

Given: A two-block blockade $L = (A_{1}, A_{2})$ whose blocks are mixed, and a vertex $u \notin A_{1} \cup A_{2}$ that is complete to $A_{1}$ and anticomplete to $A_{2}$ .

[L1]

Mixed blocks are joined by a one-link mixed chain, and quotient blocks are the unions of mixed-reachability classes (The mixed-block reachability relation on a blockade, The quotient blockade obtained from mixed-block reachability).

[L2]

The preceding lemma characterizes exactly this situation (A vertex mixed on a quotient block but pure on each member block yields two mixed member blocks with opposite adjacency).

Verification

technique · direct

1.1

Since $A_{1}$ and $A_{2}$ are mixed, [L1] places them in the same mixed-reachability class. They are the only blocks of $L$ , so their class has union $D = A_{1} \cup A_{2}$ , the unique quotient block.

givenL1

2.1

The vertex $u$ is adjacent to every vertex of $A_{1}$ and to no vertex of $A_{2}$ . Therefore $u$ is neither complete nor anticomplete to $D$ , so $u$ is mixed on $D$ . At the same time it is pure to $A_{1}$ and pure to $A_{2}$ individually.

step 1.1given

3.1

This is exactly the phenomenon isolated by [L2]: the mixing appears only after passing to the quotient union.

step 2.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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