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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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Constant-scale restricted generalized niceness yields an x-scale restricted subgraph, a polynomial clique or stable set, or a blockade

Statement

Let F be a generalized nice, leaf-reducible, wonderful finite family. Then there exist constants c∈(0,12), a1≥1, and a2>0 such that for every x∈(0,c2) and every c2-restricted F‾-free graph G, at least one of the following holds:

  1. G has an x-restricted induced subgraph with at least xa1∣G∣ vertices;
  2. G has a clique or stable set of size at least (xa1∣G∣)a2;
  3. G has a complete or anticomplete (k,∣G∣/ka1)-blockade for some integer k∈[2,x−1].

Facts & Assumptions

Given: A generalized nice, leaf-reducible, wonderful finite family F, a parameter x∈(0,c2), and a c2-restricted F‾-free graph G.

[L1]

The three-outcome theorem provides constants c,c1,c2>0 and c3≥c4≥4 (cy-restricted generalized niceness yields three outcomes).

[L2]

Under the failure of the global clique/stable-set and blockade outcomes, every sufficiently large cy-restricted induced subgraph contains a smaller scale restricted induced subgraph (A large cy-restricted subgraph in the three-outcome theorem forces a smaller-scale restricted subgraph).

[L3]

The iterative restricted-sparsification lemma turns a constant-scale restricted starting point plus the smaller-scale hypothesis into an x-restricted induced subgraph (Iterated restricted sparsification reaches the target scale).

[L4]

A c2-restricted graph is, in particular, a valid starting point for the iterative lemma with starting constant c2 (c-sparse, c-dense and c-restricted vertex sets).

Proof

technique · if the clique/stable-set and blockade outcomes fail, use the helper claim to verify the second hypothesis of the iterative lemma
1.1L1choose

Let c,c1,c2,c3,c4 be the constants from [L1], and set a1:=c1+3c3, a2:=c2, b1:=c4/2, b2:=4c3/c4, and b3:=c3.

2.1step 1.1givenL3L4algebra

Hypothesis 1 of [L3] is automatic with starting constant c2: the graph G itself is c2-restricted by assumption, so it has a c2-restricted induced subgraph of size ∣G∣=(c2)0∣G∣, and in particular of size at least (c2)b2∣G∣ because b2>0 and c2<1.

2.2step 1.1assume-contra

Suppose outcomes 2 and 3 fail for the given graph G. We will show that outcome 1 must then hold.

3.1step 1.1step 2.2L2L3

Apply [L2] with the constants from step 1.1. It shows that for every y with cy∈[x,c2] and every cy-restricted induced subgraph F of G with ∣F∣≥(cy)4c3/c4∣G∣, there is a (cy)c4/2-restricted induced subgraph of F with at least (cy)c3∣F∣ vertices. Writing λ:=cy, this is exactly hypothesis 2 of [L3] for every λ∈[x,c2], with the starting constant c2 and the choices b1=c4/2, b2=4c3/c4, and b3=c3 from step 1.1.

4.1step 1.1step 3.1algebraL3

The inequality required by [L3] holds for these choices, because b1b2=(c4/2)⋅(4c3/c4)=2c3≥c3+4c3/c4=b3+b2, using c4≥4.

5.1step 1.1step 2.1step 3.1step 4.1L3algebra

Therefore [L3] applies and yields an x-restricted induced subgraph of G with at least xb1b2∣G∣=x2c3∣G∣ vertices. Since a1=c1+3c3≥2c3 and x∈(0,1), we have x2c3≥xa1, so outcome 1 holds.

6.1step 2.2step 5.1discharge-contradiction∎

Outcome 1 follows whenever outcomes 2 and 3 fail. Hence at least one of the three stated outcomes holds for every admissible G.

Depends on

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Sources