Alphabeta Math
CorollaryStatement: AI-generatedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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 c-sparse set X satisfies α(G[X])≥∣X∣/(c∣X∣+1), and a c-dense set satisfies ω(G[X])≥∣X∣/(c∣X∣+1)

Statement

Let G be a finite simple graph, let c≥0, and let X⊆V(G) be nonempty.

  1. If X is c-sparse, then α(G[X])≥∣X∣/(c∣X∣+1).
  2. If X is c-dense, then ω(G[X])≥∣X∣/(c∣X∣+1).

Facts & Assumptions

Given: A finite simple graph G, a real c≥0, and a nonempty set X⊆V(G).

[L4]

Proof

technique · direct
1.1L1L2

In the sparse case, [L1] gives Δ(G[X])≤c∣X∣, so [L2] gives χ(G[X])≤c∣X∣+1.

1.2L4

The two published definitions of α and ω agree by [L4], so the complement statement can be read with the same symbols.

2.1step 1.1L3algebra

Applying [L3] to G[X] yields ∣X∣≤χ(G[X])α(G[X]), hence α(G[X])≥∣X∣/(c∣X∣+1).

3.1step 2.1step 1.2L5∎

If X is c-dense, then [L5] makes X c-sparse in G‾, so step 2.1 applied there gives a stable set of size at least ∣X∣/(c∣X∣+1). Reading that set back in G via [L5] gives a clique of the same size.

Depends on

Used by

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Dependency tree · two levels

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