Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06
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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.

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A wide integral geometric layer forces the complete-or-anticomplete property-(*) blockade

Statement

Assume the structural comb-partition hypothesis and let c(0,1] be a common Erdős–Hajnal constant for F1-free and F2-free graphs. In one decreasing structural partition, let Cr be an integral geometric layer with r<q. If every block of Cr has size at least w/5r/2, then G has a complete or anticomplete (k,w/k10/c)-blockade for some kcr/4.

Facts & Assumptions

Given: r<q, a wide layer Cr, and a common constant c(0,1].

[F1]

The first mr structural blocks form an induced subgraph of the F2-free pattern graph (The structural comb-partition hypothesis).

[F3]

An Erdős–Hajnal constant c supplies a pattern clique or stable set of size at least mrc in a nonempty F2-free graph (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).

[F4]

A clique or stable set in a pure-blockade pattern lifts to a complete or anticomplete blockade with the same selected width (Homogeneous sets in pure-blockade patterns lift to complete or anticomplete blockades).

Proof

technique · direct
1.1

The induced pattern on the first mr blocks is F2-free: an induced forbidden copy there would also be one in the full pattern. By [F1] and [F3], it has a clique or stable set S of cardinality kmrc.

F1F3
2.1

From [F2] and step 1.1, k(r/4)c=cr/4 by [F5].

F2F5step 1.1
2.2

The blocks indexed by S lie among the first mr blocks and therefore in layers through Cr; decreasing block sizes and the width assumption on Cr give them size at least w/5r/2. By [F4] they form a complete or anticomplete blockade of length k and at least that width.

F4step 1.1
3.1

Step 2.1 and [F5] give k10/c5r/2, so w/5r/2w/k10/c. Together with step 2.2 this proves the claim.

step 2.1step 2.2F5algebra

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Sources