Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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.

Loglog quantitative induced density bound

Statement

For every nonempty finite graph H there is CH>0 such that, for 0<x<1/2 and δ=2−CH(log⁡2(1/x))2/log⁡2log⁡2(1/x), every nonempty finite graph G with ind⁡H(G)≤(δ∣G∣)∣H∣ has a nonempty S⊆V(G) of size at least δ∣G∣ with e(G[S])≤x(∣S∣2) or e(G‾[S])≤x(∣S∣2). Every H-free host qualifies. The strict few-copy version covers the null pattern vacuously.

Facts & Assumptions

Given: Nonempty H,G, 0<x<1/2, and the copy hypothesis with the displayed fraction after CH is chosen.

[F1]

From Qid logarithmic and constant divisibility: Every nonempty finite graph H is ℓ-divisive for each of ℓ(x)=log⁡2(1/x) and ℓ(x)=2. Both functions are subreciprocal on (0,1/2).

[F2]

For nonempty ℓ-divisive H and subreciprocal ℓ, some CH,ℓ>0 gives the fraction δ=2−CH,ℓlog⁡2(1/ϵ)2/log⁡2ℓ(ϵ) on 0<ϵ<1/2; a nonempty host with at most (δ∣G∣)∣H∣ embeddings has the asserted nonempty sparse-or-dense set of size at least δ∣G∣. (Quantitative density theorem for ell divisive graphs).

Proof

1.1F1F2

Take ℓ(x)=log⁡2(1/x). The hypotheses on the function and on the nonempty H required by [F2] hold by [F1]. Since 0<x<1/2, we have ℓ(x)>1 and hence log⁡2ℓ(x)>0. Substitution into [F2] gives exactly the displayed fraction and the required set, with CH=CH,ℓ.

2.1step 1.1algebra∎

An H-free host has ind⁡H(G)=0≤(δ∣G∣)∣H∣. For null H, there is exactly one induced embedding, the empty function; (δ∣G∣)0=1, so the strict inequality is impossible. This gives the stated boundary clauses without applying the formula at x=1/2.

Source notes

Proof/convention locator: Bucic, Nguyen, Scott and Seymour, Induced subgraph density I, 1.8; 5.1 and 5.2.

Depends on

Used by

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Sources