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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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Quantitative density theorem for ell divisive graphs

Statement

Let H be a nonempty finite graph that is -divisive for a subreciprocal function . There is CH,>0 such that, for 0<ϵ<1/2 and δ=2CH,log2(1/ϵ)2/log2(ϵ), every nonempty finite graph G with indH(G)(δG)H has a nonempty SV(G) of size at least δG with e(G[S])ϵ(S2) or e(G[S])ϵ(S2).

Facts & Assumptions

Given: A nonempty -divisive pattern H, a subreciprocal , and a nonempty host satisfying the copy bound for the fraction specified at each stage below.

[F1]

For the stated subreciprocal function and witnesses, the parameter construction has p>1, 0<η<1, δ<xdηt, and t the least natural with ptϵ2. (Admissible parameters for the density recursion).

[F2]

With the parameter setup of the amplification lemma, a fixed nonempty G satisfying indH(G)(δG)H and δG>1 has βs1(u,v)ηmin{βs(pu,v),βs(u,pv)} for 1st and u,vϵ. (Ell divisibility amplifies through a blockade).

[F3]

From Qid finite density recursion profile: If a1 or b1, every full F qualifies and βs(a,b)=1. At s=0, the only qualifying set is V(G), so β0(a,b)=ρG(a,b).

Proof

1.1

Choose divisibility witnesses c,d, put z=(c)1/2, b=2log2(1z)>2, and C0=20bd. First take 0<ϵ<c, and define δ0 by the displayed formula with C0. If δ0G1, any singleton S has the required size and zero edges in both graphs. Otherwise the parameter construction [F1] supplies x,p,η,t and δ0<xdηt<ηt; all hypotheses of [F2] hold with δ=δ0.

F1given
2.1

For each integer 0rt, the recurrence implies β0(ϵ,ϵ)ηrmin0irβr(piϵ,priϵ). At r=0 this is equality. To pass from r<t to r+1, apply [F2] with s=r+1 to every pair (piϵ,priϵ); both coordinates are at least ϵ because p>1. The two children have exponent pairs (i+1,ri) and (i,r+1i), whose union over i is exactly all pairs summing to r+1. Taking their finite minimum proves the induction step.

F2step 1.1algebra
3.1

At r=t, the product of the two arguments in every terminal pair is ptϵ21 by the least-natural property in [F1]. At least one argument is therefore at least 1. Each terminal profile value equals 1 by [F3]. Hence ρG(ϵ,ϵ)=β0(ϵ,ϵ)ηt>δ0. The attained maximum defining ρG supplies a nonempty set of at least δ0G vertices with one of the required edge bounds. Together with the singleton case, this proves the theorem on (0,c) with constant C0.

F1F3step 2.1
4.1

For the full interval set a=log2(1/c)>1 and CH,=a2C0. Given 0<ϵ<1/2, put ϵ=ϵa<c and let δ be the small-interval fraction at ϵ with constant C0. Since ϵϵ and is nonincreasing, log2(ϵ)log2(ϵ)>0. Thus log2(1/δ)=C0a2log2(1/ϵ)2/log2(ϵ)log2(1/δ), so δδ.

step 1.1step 3.1algebra
5.1

The original hypothesis implies indH(G)(δG)H. Apply the small-interval result to ϵ: its set has size at least δGδG, and its edge bound with ϵ implies that with ϵ. This establishes the claimed constant on the entire open interval.

step 3.1step 4.1algebra

Source notes

Proof/convention locator: Bucic, Nguyen, Scott and Seymour, Induced subgraph density I, 5.2 complete proof.

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