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✓ 11 results · all verified · 7 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Property (*) and Comb Outcomes

1 · Prerequisites

2 · Summary

This page packages the second reduction stage in Huang-Ju-Zhou. It starts from the comb trigger recorded as property (*), combines it with the earlier sparse comb and leaf-reduction lemmas, and then runs the same iterative-sparsification pattern used on the generalized-niceness page.

The later items separate the two load-bearing transfer claims from the final theorem. First the page reaches a constant-scale four-outcome theorem, then Rödl initialization removes that scale assumption, and finally the local pure-or-sparse blockade hypothesis is fed into the published blockade-to-nice theorem to recover generalized niceness.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-05Open item page →

Property (*) for a finite graph family

Definition

Let F be a finite family of finite graphs. We say that F has property (∗) if there exist constants c1,c2,c3>0 such that the following holds for every F‾-free graph G, where F‾:={H‾:H∈F} is the family of graph complements (Graph isomorphisms, automorphisms and graph complements) (H-free and F-free graphs under the induced-subgraph convention).

Suppose there is an (ℓ,w)-comb ((ai,Bi):i∈[ℓ]) in G (Combs in a graph) with ℓ,w≥4, and suppose there is a vertex v∈V(G)∖({ai:i∈[ℓ]}∪⋃i=1ℓBi) such that v is complete to ⋃iBi and anticomplete to {ai:i∈[ℓ]}. Then at least one of the following holds:

  1. G has a clique or stable set of size at least wc1 (Cliques, stable sets, the clique number ω(G) and stability number α(G));
  2. G has a complete or anticomplete (k,w/kc2)-blockade for some real k≥ℓc3, where the real length threshold k means that the blockade's integral length is at least ⌈k⌉ (Blockades, their length, their width, and their support, Complete, anticomplete, pure, weakly sparse, and x-sparse blockades);
  3. G has a pure (ℓ,w/ℓ2)-blockade.

This condition records exactly the three ways the special-vertex comb trigger can terminate the second sparsification round.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-05Open item page →

Property (*) and leaf reducibility yield five comb outcomes in a restricted graph

Statement

Suppose that F has property (∗) and that F is leaf-reducible. Then there exist constants c1,c2,c3>0 and c4,c5≥4 such that for every 0<x≤y≤2−4c5 and every y3-restricted F‾-free graph G, at least one of the following holds:

  1. there are disjoint sets X,Y⊆V(G) with ∣X∣≥yc4∣G∣,∣Y∣≥(1−c5y)∣G∣, and Y is x-sparse or complete to X;
  2. G has a 2y4-restricted induced subgraph with at least yc4∣G∣ vertices;
  3. G has a clique or stable set of size at least (x9∣G∣)c1;
  4. G has a complete or anticomplete (k,∣G∣/kc2+6/c3)-blockade for some real k≥y−c3;
  5. G has a pure (ℓ,∣G∣/ℓ8)-blockade for some real ℓ∈[y−1,x−2].

Facts & Assumptions

Given: A finite family F with property (∗) and leaf-reducible, parameters 0<x≤y≤2−4c5, and a y3-restricted F‾-free graph G.

[L1]

Because F has property (∗), there exist constants c1,c2,c3>0 such that every special-vertex (ℓ,w)-comb with ℓ,w≥4 in an F‾-free graph yields either a clique or stable set of size wc1, or a complete or anticomplete (k,w/kc2)-blockade with k≥ℓc3, or a pure (ℓ,w/ℓ2)-blockade (Property (*) for a finite graph family).

[L2]

Since F is leaf-reducible, there exist constants d>0 and h≥1 such that every y3-sparse F-free graph has either a large anticomplete pair or a y12-restricted induced subgraph of size at least (y3)4d+1∣G∣=y12d+3∣G∣ (Leaf-reducible families yield a large anticomplete pair or a deeper restricted induced subgraph).

[L3]

If G is y3-sparse and ∣G∣≥y−4, then either there are disjoint sets X,Y⊆V(G) with ∣X∣≥y4∣G∣, ∣Y∣≥(1−4y)∣G∣, and Y x-sparse to X, or G is 2y4-sparse, or G contains a special-vertex comb with parameters ℓ∈[y−1,x−2] and width w=y4∣G∣/ℓ2 (A sparse graph either sparsifies further or yields a comb or a large sparse pair).

Proof

technique · treat the dense side by applying the leaf-reducible lemma to $\overline G$, and otherwise apply the sparse comb lemma to $G$ and feed the comb branch into property $(*)$
1.1L1L2choose

Let c1,c2,c3 be the constants from [L1]. Let d>0 and h≥1 be the constants from [L2], and set c4:=max⁡{12d+3,4},c5:=max⁡{h,4}.

1.2

[assume-case dense-side] Suppose first that G‾ is y3-sparse. Because G is F‾-free, the complement graph G‾ is F-free. Applying [L2] to G‾ with the parameter y3 and b=4, we obtain either:

  1. disjoint sets X,Y⊆V(G) with ∣X∣≥y12d+3∣G∣, ∣Y∣≥(1−hy)∣G∣, and Y complete to X in G; or
  2. a y12-restricted induced subgraph of G‾ with at least y12d+3∣G∣ vertices.

In the first branch, c4≥12d+3 and c5≥h, so outcome 1 holds. In the second branch, [L4] transfers restrictedness back to G, and because y12≤2y4 for 0<y<1, outcome 2 holds. [step 1.1, L2, L4, given, algebra]

2.1step 1.1step 1.2given

[assume-case sparse-side] We may therefore assume that G itself is y3-sparse. If ∣G∣≤x−9, then (x9∣G∣)c1≤1, so any vertex of G already gives outcome 3. Hence we may further assume that ∣G∣≥x−9≥y−4.

3.1step 2.1L3cases

Apply [L3] to the sparse graph G. If its first branch holds, then outcome 1 holds immediately. If its second branch holds, then outcome 2 holds immediately. So only the comb branch remains.

4.1step 1.1step 2.1step 3.1algebra

In that comb branch, [L3] gives an integer ℓ0∈[y−1,⌈x−2⌉], a width w:=y4∣G∣/ℓ02, an (ℓ0,w)-comb ((ai,Bi):i∈[ℓ0]), and a vertex v complete to ⋃iBi and anticomplete to the teeth. Since x≤y≤2−4c5 and c5≥4, one has x≤2−16 and hence ℓ0≥y−1≥4. Also ⌈x−2⌉≤x−2+1≤2x−2, so w=y4∣G∣ℓ02≥y4∣G∣(2x−2)2≥x8∣G∣4≥x9∣G∣, because y≥x and x≤14. Using ∣G∣≥x−9 from step 2.1 and x≤2−16 again, this also gives w≥x−1/4≥4.

5.1step 1.1step 4.1L1algebra

Apply [L1] to this special-vertex comb. If it yields a clique or stable set of size at least wc1, then step 4.1 gives wc1≥(x9∣G∣)c1, so outcome 3 holds.

5.2step 4.1L1algebra

If [L1] yields a complete or anticomplete (k,w/kc2)-blockade with k≥ℓ0c3, then wkc2≥∣G∣ℓ06kc2≥∣G∣kc2+6/c3, and also k≥ℓ0c3≥y−c3. So outcome 4 holds.

5.3step 4.1L1algebra

If [L1] yields a pure (ℓ0,w/ℓ02)-blockade, set ℓ:=ℓ0/y. Because ℓ0≥y−1, one has ℓ≥y−1 and also ℓ0≥ℓ, so the same blockade has length at least ℓ. Since ℓ0≤2x−2 and y≥x, ℓ2=ℓ0/y≤2x−2/x=2x−3≤x−4, because x≤12, and therefore ℓ≤x−2. Finally, wℓ02=y4∣G∣ℓ04=∣G∣ℓ8. Hence outcome 5 holds.

6.1step 1.2step 2.1step 3.1step 5.1step 5.2step 5.3cases-exhaustive∎

Steps 1.2, 2.1, 3.1, 5.1, 5.2, and 5.3 exhaust all cases, so one of the five stated outcomes always holds.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Property (*) and leaf reducibility yield a long x-sparse or complete blockade, or a better outcome

Statement

Suppose that F has property (∗) and that F is leaf-reducible. Then there exist constants c1,c2,c3>0 and c4,c5≥4 such that, with c:=2−4c5, for every 0<x≤y≤c and every cy3-restricted F‾-free graph G, at least one of the following holds:

  1. G has an x-sparse or complete blockade of length at least y−1 and width at least yc4+2∣G∣;
  2. G has a 2y4-restricted induced subgraph with at least yc4+2∣G∣ vertices;
  3. G has a clique or stable set of size at least (x10∣G∣)c1;
  4. G has a complete or anticomplete (k,∣G∣/kc2+7/c3)-blockade for some real k≥y−c3;
  5. G has a pure (ℓ,∣G∣/ℓ9)-blockade for some real ℓ∈[y−1,x−2].

Facts & Assumptions

Given: A finite family F with property (∗) and leaf-reducible, parameters 0<x≤y≤c, and a cy3-restricted F‾-free graph G.

[L1]

The five-outcome lemma provides constants c1,c2,c3>0 and c4,c5≥4 for y3-restricted graphs (Property (*) and leaf reducibility yield five comb outcomes in a restricted graph).

[L2]

If every induced subgraph F of G with ∣F∣≥c∣G∣ has disjoint sets X,Y with ∣X∣≥yc4∣F∣, ∣Y∣≥(1−c5y)∣F∣, and Y x-sparse or complete to X, then G has an x-sparse or complete blockade of length at least y−1 and width at least yc4+2∣G∣ (Large sparse-pair hypotheses yield an x-sparse or complete blockade).

[L3]

If a graph is cy3-restricted and F is an induced subgraph with ∣F∣≥c∣G∣, then F is y3-restricted (c-sparse, c-dense and c-restricted vertex sets).

Proof

technique · either every large induced subgraph satisfies the large pair hypothesis of [L2], or choose a counterexample $F$ and apply the five-outcome lemma inside it
1.1L1choose

Let c1,c2,c3>0 and c4,c5≥4 be the constants from [L1], and put c:=2−4c5.

2.1step 1.1givenalgebracases

If ∣G∣≤y−(c4+2), then yc4+2∣G∣≤1. Any single vertex spans an induced subgraph that is 0-restricted, hence 2y4-restricted, so outcome 2 holds. Therefore we may assume ∣G∣≥y−(c4+2).

2.2step 1.1L2

[assume-case universal-pair] Suppose that every induced subgraph F of G with ∣F∣≥c∣G∣ has disjoint sets X,Y with ∣X∣≥yc4∣F∣, ∣Y∣≥(1−c5y)∣F∣, and Y x-sparse or complete to X. Then [L2] gives outcome 1.

2.3step 1.1L1L3

[assume-case obstruction] Assume instead that there is an induced subgraph F of G with ∣F∣≥c∣G∣ for which no such pair X,Y exists. By [L3], the graph F is y3-restricted, so [L1] applies to F. Because the first outcome of [L1] fails for this specific F, one of the remaining four outcomes of [L1] holds inside F.

3.1step 2.3L1algebra

If [L1] gives a 2y4-restricted induced subgraph of F with at least yc4∣F∣ vertices, then that subgraph has at least yc4c∣G∣=yc4+1∣G∣≥yc4+2∣G∣ vertices because y≤c<1. Hence outcome 2 holds in G.

3.2step 2.3L1algebra

If [L1] gives a clique or stable set of size at least (x9∣F∣)c1, then (x9∣F∣)c1≥(x9⋅c∣G∣)c1≥(x9⋅x∣G∣)c1=(x10∣G∣)c1, because x≤y≤c. So outcome 3 holds.

3.3step 2.3L1algebra

If [L1] gives a complete or anticomplete (k,∣F∣/kc2+6/c3)-blockade with k≥y−c3, then ∣F∣kc2+6/c3≥c∣G∣kc2+6/c3≥y∣G∣kc2+6/c3≥∣G∣kc2+7/c3, because k≥y−c3 implies k1/c3≥y−1. Hence outcome 4 holds.

3.4step 2.3L1algebra

If [L1] gives a pure (ℓ,∣F∣/ℓ8)-blockade with ℓ∈[y−1,x−2], then ∣F∣ℓ8≥c∣G∣ℓ8≥y∣G∣ℓ8≥∣G∣ℓ9, again because ℓ≥y−1. Thus outcome 5 holds.

4.1step 2.2step 2.3step 3.1step 3.2step 3.3step 3.4cases-exhaustive∎

The exhaustive alternatives 2.1 and 2.2, together with steps 3.1-3.4, show that one of the five stated outcomes always holds.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Under failure of the global outcomes, a large y^(10/3)-restricted induced subgraph forces a y^(11/3)-restricted induced subgraph

Statement

Let F have property (∗) and be leaf-reducible, and let c1,c2,c3>0, c4,c5≥4, and c:=2−4c5 be the constants from Property (*) and leaf reducibility yield a long x-sparse or complete blockade, or a better outcome. Fix x∈(0,c10], and let G be a c10-restricted F‾-free graph for which none of the following holds:

  1. G has an x-restricted induced subgraph with at least x22c4∣G∣ vertices;
  2. G has a clique or stable set of size at least (x30c4∣G∣)c1;
  3. G has a complete or anticomplete (k,∣G∣/kc2+27c4/c3)-blockade for some real k≥2;
  4. G has a pure or x-sparse (ℓ,∣G∣/ℓ29c4)-blockade for some real ℓ∈[c−1,x−2].

Then for every y∈[x,c3] and every y10/3-restricted induced subgraph F of G with ∣F∣≥y10(c4+2)∣G∣, there exists a y11/3-restricted induced subgraph of F with at least yc4+2∣F∣ vertices.

Facts & Assumptions

Given: The data and failure hypotheses in the Statement, together with a parameter y∈[x,c3] and an induced subgraph F of G that is y10/3-restricted and satisfies ∣F∣≥y10(c4+2)∣G∣.

[L1]

The previous lemma says that every cy3-restricted F‾-free graph satisfies one of five outcomes: a long x-sparse or complete blockade, a 2y4-restricted induced subgraph, a clique or stable set, a complete or anticomplete blockade, or a pure blockade (Property (*) and leaf reducibility yield a long x-sparse or complete blockade, or a better outcome).

[L2]

If a set is η-restricted, then it is also η′-restricted for every η′≥η (c-sparse, c-dense and c-restricted vertex sets).

Proof

technique · apply the previous lemma to $F$ with parameter $y$ and show that every outcome except the deeper restricted-set outcome contradicts one of the assumed global failures
1.1givenL2algebra

Because y≤c3, one has y10/3=y3y1/3≤cy3. Thus [L2] upgrades the hypothesis that F is y10/3-restricted to the statement that F is cy3-restricted.

2.1step 1.1L1

Apply [L1] to the graph F with the original parameter x and the current parameter y. One of the five outcomes of [L1] holds for F.

3.1step 2.1algebra

Suppose the first outcome of [L1] holds for F: there is an x-sparse or complete blockade in F of length at least y−1 and width at least yc4+2∣F∣. Since y−1∈[c−1,x−1]⊆[c−1,x−2] and yc4+2∣F∣≥y11(c4+2)∣G∣≥y29c4∣G∣, this produces the forbidden global outcome 4.

3.2step 2.1algebra

Suppose the second outcome of [L1] holds for F. Because 2y4≤y11/3 for 0<y<1, the resulting induced subgraph is already the desired y11/3-restricted induced subgraph of size at least yc4+2∣F∣.

3.3step 2.1algebra

Suppose the third outcome of [L1] holds for F. Then (x10∣F∣)c1≥(x10⋅y10(c4+2)∣G∣)c1≥(x10c4+30∣G∣)c1≥(x30c4∣G∣)c1, because y≥x and c4≥4. This contradicts the failure of global outcome 2.

3.4step 2.1algebra

Suppose the fourth outcome of [L1] holds for F: there is a complete or anticomplete (k,∣F∣/kc2+7/c3)-blockade with k≥y−c3. If k≥2, then ∣F∣kc2+7/c3≥y10(c4+2)∣G∣kc2+7/c3≥∣G∣kc2+7/c3+10(c4+2)/c3≥∣G∣kc2+27c4/c3, so global outcome 3 holds, a contradiction. If instead 1<k<2, then the blockade has at least two blocks, and ∣F∣kc2+7/c3≥∣F∣2c2+7/c3≥∣G∣2c2+7/c3+10(c4+2)/c3≥∣G∣2c2+27c4/c3, because ∣F∣≥y10(c4+2)∣G∣, the inequality y−c3≤k<2 implies y10(c4+2)≥2−10(c4+2)/c3, and 7+10(c4+2)≤27c4 for c4≥4. Thus G has a complete or anticomplete (2,∣G∣/2c2+27c4/c3)-blockade, again contradicting the failure of global outcome 3.

3.5step 2.1algebra

Suppose the fifth outcome of [L1] holds for F: there is a pure (ℓ,∣F∣/ℓ9)-blockade with ℓ∈[y−1,x−2]. Then ∣F∣ℓ9≥y10(c4+2)∣G∣ℓ9≥∣G∣ℓ29c4, because ℓ≥y−1. This again gives the forbidden global outcome 4.

4.1step 3.1step 3.2step 3.3step 3.4step 3.5∎

The first, third, fourth, and fifth cases are impossible under the standing global failure hypotheses. Therefore the second case, recorded in step 3.2, must hold, which is exactly the desired conclusion.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-05Open item page →

Constant-scale restricted property (*) yields a restricted subgraph, a polynomial clique or stable set, or two blockade alternatives

Statement

Suppose that F has property (∗) and is leaf-reducible. Then there exist constants c1,c2,c3>0, c4,c5≥4, and c:=2−4c5 such that for every x∈(0,c10] and every c10-restricted F‾-free graph G, at least one of the following holds:

  1. G has an x-restricted induced subgraph with at least x22c4∣G∣ vertices;
  2. G has a clique or stable set of size at least (x30c4∣G∣)c1;
  3. G has a complete or anticomplete (k,∣G∣/kc2+27c4/c3)-blockade for some real k≥2;
  4. G has a pure or x-sparse (ℓ,∣G∣/ℓ29c4)-blockade for some real ℓ∈[c−1,x−2].

Facts & Assumptions

Given: A finite family F with property (∗) and leaf-reducible, an x∈(0,c10], and a c10-restricted F‾-free graph G.

[L1]

The previous claim says that, under the failure of outcomes 2-4, every y10/3-restricted induced subgraph of sufficiently large relative size has a deeper y11/3-restricted induced subgraph (Under failure of the global outcomes, a large y^(10/3)-restricted induced subgraph forces a y^(11/3)-restricted induced subgraph).

[L2]

If a graph has a c10-restricted induced subgraph of size at least (c10)3(c4+2)∣G∣ and every λ-restricted induced subgraph of size at least λ3(c4+2)∣G∣ contains a λ11/10-restricted induced subgraph of size at least λ3(c4+2)/10 times as many vertices, then the graph has an x10/3-restricted induced subgraph with at least x11(c4+2)∣G∣ vertices (Iterated restricted sparsification reaches the target scale).

[L3]

If a set is x10/3-restricted, then it is x-restricted (c-sparse, c-dense and c-restricted vertex sets).

Proof

technique · if outcomes 2-4 fail, verify the hypotheses of the iterative restricted-sparsification lemma with $b_1=\tfrac{11}{10}$, $b_2=3(c_4+2)$, and $b_3=\tfrac{3(c_4+2)}{10}$
1.1L1choose

Let c1,c2,c3>0, c4,c5≥4, and c:=2−4c5 be the constants from Under failure of the global outcomes, a large y^(10/3)-restricted induced subgraph forces a y^(11/3)-restricted induced subgraph, and set b1:=11/10,b2:=3(c4+2),b3:=3(c4+2)/10.

1.2givenassume-contra

Suppose outcomes 2, 3, and 4 all fail. We will show that outcome 1 then holds.

2.1step 1.1L2givenalgebra

Hypothesis 1 of [L2] is immediate: the graph G itself is c10-restricted and has size ∣G∣≥(c10)b2∣G∣ because b2>0 and c10<1.

2.2step 1.1step 1.2L1L2algebra

Let λ∈[x10/3,c10] and let F be a λ-restricted induced subgraph of G with ∣F∣≥λb2∣G∣. Write λ=y10/3, so y=λ3/10∈[x,c3]. Then ∣F∣≥y10(c4+2)∣G∣. Since outcomes 2-4 fail globally, [L1] applied with this y gives a y11/3=λ11/10-restricted induced subgraph of F with at least yc4+2∣F∣=λb3∣F∣ vertices.

2.3step 1.1L2algebra

The exponent condition for [L2] holds because b1b2=1110⋅3(c4+2)=33(c4+2)10≥30(c4+2)10=b2+b3.

3.1step 2.1step 2.2step 2.3L2L3algebra

Therefore [L2] yields an x10/3-restricted induced subgraph S⊆G with at least (x10/3)b1b2∣G∣=x11(c4+2)∣G∣ vertices. Since c4≥4, the exponent satisfies 11(c4+2)≤22c4, so ∣S∣≥x22c4∣G∣. By [L3], the subgraph S is x-restricted. Hence outcome 1 holds.

4.1step 1.2step 3.1discharge-contradiction∎

Outcome 1 follows whenever outcomes 2-4 fail. Hence at least one of the four stated outcomes always holds.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Rödl initialization removes the constant-scale restriction in the property (*) four-outcome theorem

Statement

Suppose that F has property (∗) and is leaf-reducible. Then there exist constants c1>0, c4≥4, and d≥58c4 such that for every x∈(0,2−d) and every F‾-free graph G with ∣G∣≥x−d, at least one of the following holds:

  1. G has an x-restricted induced subgraph with at least x23c4∣G∣ vertices;
  2. G has a pure or x-sparse (k,∣G∣/kd)-blockade for some integer k∈[2,x−1];
  3. G has a clique or stable set of size at least (x31c4∣G∣)c1;
  4. G has a complete or anticomplete (k,∣G∣/kd)-blockade for some real k≥x−1.

Facts & Assumptions

Given: A finite family F with property (∗) and leaf-reducible, a parameter x∈(0,2−d), and an F‾-free graph G with ∣G∣≥x−d.

[L1]

The constant-scale four-outcome theorem gives constants c1,c2,c3>0, c4,c5≥4, and c:=2−4c5 such that every c10-restricted F‾-free graph satisfies one of the four outcomes on the current page (Constant-scale restricted property (*) yields a restricted subgraph, a polynomial clique or stable set, or two blockade alternatives).

[L2]

For ξ:=c10, every F‾-free graph has a ξ-restricted induced subgraph of size at least δ∣G∣ for some δ>0 (Rödl: for every H and every ϵ∈(0,12) there is δ>0 such that every nonempty H-free graph has an ϵ-restricted vertex set of size at least δ∣V(G)∣).

Proof

technique · apply Rödl at the fixed scale $\xi=c^{10}$, then transfer each outcome of the constant-scale theorem back to $G$ by choosing $d$ large enough
1.1L1L2choose

Let c1,c2,c3>0, c4,c5≥4, and c:=2−4c5 be the constants from [L1], and set ξ:=c10. Let δ>0 be the constant from [L2] for the family F‾ and the parameter ξ.

2.1step 1.1choose

Choose d so large that d≥58c4,2−d<c10,δ≥2116c4−d,δ≥2c2+27c4/c3−d.

3.1step 2.1L1L2

By [L2], the graph G has a ξ-restricted induced subgraph F⊆G with ∣F∣≥δ∣G∣. Since x<2−d<ξ, the parameter x lies in the range allowed by [L1], so [L1] applies to F.

4.1step 2.1step 3.1L1algebra

If [L1] gives an x-restricted induced subgraph of F with at least x22c4∣F∣ vertices, then x22c4∣F∣≥x22c4δ∣G∣≥x22c4258c4−d∣G∣≥x23c4∣G∣, because x<2−d implies xc4≤258c4−d. Thus outcome 1 holds.

4.2step 2.1step 3.1L1algebra

If [L1] gives a clique or stable set of size at least (x30c4∣F∣)c1, then the same estimate yields (x30c4∣F∣)c1≥(x31c4∣G∣)c1, so outcome 3 holds.

4.3step 2.1step 3.1L1algebrachoosecases

If [L1] gives a complete or anticomplete (k,∣F∣/kc2+27c4/c3)-blockade with k≥2, let j:=⌈k⌉. Its actual length is integral and at least k, hence at least j, while ∣F∣kc2+27c4/c3≥δ∣G∣kc2+27c4/c3≥∣G∣kd≥∣G∣jd. Thus the same blocks form a complete or anticomplete (j,∣G∣/jd)-blockade. If j≥x−1 this is outcome 4; if j<x−1, then the integer j∈[2,x−1] gives outcome 2.

4.4step 2.1step 3.1L1algebrachoose

If [L1] gives a pure or x-sparse (ℓ,∣F∣/ℓ29c4)-blockade with ℓ∈[c−1,x−2], set k:=⌊ℓ⌋. Then k is an integer in [2,x−1], because ℓ≥c−1≥24 and ℓ≤x−1. The blockade has length at least ℓ≥k, and ℓ<(k+1)2≤4k2, so ∣F∣ℓ29c4≥∣F∣(4k2)29c4=∣F∣258c4k58c4≥δ∣G∣258c4k58c4≥∣G∣kd, because k≥2 and step 2.1 gives δ≥2116c4−d. Hence outcome 2 holds.

5.1step 3.1step 4.1step 4.2step 4.3step 4.4cases-exhaustive∎

The four branches 4.1-4.4 exhaust the conclusion of [L1], so one of the stated outcomes always holds for G.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Large induced subgraphs in the property (*) four-outcome theorem contain a pure or x-sparse polynomial blockade

Statement

Let F have property (∗) and be leaf-reducible, and let c1′>0, c4′≥4, and d≥58c4′ be the constants from Rödl initialization removes the constant-scale restriction in the property (*) four-outcome theorem. Fix ϵ∈(0,12), put x:=ϵ5d, and let G be an F‾-free graph with ∣G∣≥ϵ−10d2 such that

  1. G has no clique or stable set of size at least (ϵ156c4′d∣G∣)c1′;
  2. G has no complete or anticomplete (k,∣G∣/k2d)-blockade with k≥ϵ−5d;
  3. G has no ϵ5d-restricted induced subgraph with at least ϵ116c4′d∣G∣ vertices.

Then every induced subgraph F of G with ∣F∣≥ϵd∣G∣ has a pure or x-sparse (k,∣F∣/kd)-blockade for some integer k∈[2,x−1].

Facts & Assumptions

Given: The data and hypotheses in the Statement, together with an induced subgraph F of G satisfying ∣F∣≥ϵd∣G∣.

[L1]

The previous lemma says that every F‾-free graph of size at least x−d satisfies one of four outcomes: an x-restricted induced subgraph of size at least x23c4′ times the ambient order, a pure or x-sparse (k,∣F∣/kd)-blockade for some integer k∈[2,x−1], a clique or stable set of size at least (x31c4′∣F∣)c1′, or a complete or anticomplete polynomial blockade (Rödl initialization removes the constant-scale restriction in the property (*) four-outcome theorem).

[L2]

If ∣G∣≥ϵ−10d2 and ∣F∣≥ϵd∣G∣, then ∣F∣≥ϵ−5d2=x−d because d≥1.

[L3]

If k≥x−1=ϵ−5d, then ϵd=x1/5≥k−1/5.

Proof

technique · apply the Rödl-initialized four-outcome theorem to $F$ and use the three standing failure hypotheses to rule out every branch except the pure-or-sparse blockade branch
1.1givenL1algebra

The size hypothesis on G and the bound ∣F∣≥ϵd∣G∣ imply ∣F∣≥ϵdϵ−10d2=ϵ−10d2+d≥ϵ−5d2=x−d, because d≥1. Thus [L1] applies to F.

2.1step 1.1L1

Apply [L1] to the induced subgraph F. One of its four outcomes holds.

3.1step 2.1algebra

If [L1] yields an x-restricted induced subgraph S of F with at least x23c4′∣F∣ vertices, then ∣S∣≥x23c4′ϵd∣G∣=ϵ115c4′d+d∣G∣≥ϵ116c4′d∣G∣, because c4′≥1. This contradicts standing hypothesis 3.

3.2step 2.1algebra

If [L1] yields a clique or stable set of size at least (x31c4′∣F∣)c1′, then x31c4′∣F∣≥x31c4′ϵd∣G∣=ϵ155c4′d+d∣G∣≥ϵ156c4′d∣G∣, so standing hypothesis 1 is contradicted.

3.3step 2.1L3algebra

If [L1] yields a complete or anticomplete (k,∣F∣/kd)-blockade with k≥x−1, then [L3] gives ϵd≥k−1/5, and therefore ∣F∣kd≥ϵd∣G∣kd≥∣G∣k2d. Since k≥x−1=ϵ−5d, this contradicts standing hypothesis 2.

4.1step 3.1step 3.2step 3.3∎

The first three branches are impossible, so the remaining branch of [L1] must hold: F has a pure or x-sparse (k,∣F∣/kd)-blockade for some integer k∈[2,x−1]. This is exactly the desired conclusion.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-05Open item page →

Property (*) and leaf reducibility imply generalized niceness

Statement

Let F be a finite family of graphs. If F has property (∗) and F is leaf-reducible, then F is generalized nice.

Facts & Assumptions

Given: A finite family F with property (∗) and leaf-reducible.

[L1]

There exist constants c1′>0, c4′≥4, and d≥58c4′ such that, for every x∈(0,2−d), every F‾-free graph of size at least x−d satisfies the four-outcome theorem with parameter x (Rödl initialization removes the constant-scale restriction in the property (*) four-outcome theorem).

[L2]

Under the failure of the clique/stable-set, complete-or-anticomplete blockade, and restricted-set outcomes, every induced subgraph of size at least ϵd∣G∣ has a pure or x-sparse (k,∣F∣/kd)-blockade for some integer k∈[2,x−1] when x=ϵ5d, provided ∣G∣≥ϵ−10d2 (Large induced subgraphs in the property (*) four-outcome theorem contain a pure or x-sparse polynomial blockade).

[L3]

If every induced subgraph F of G with ∣F∣≥ϵd∣G∣ has a pure or x-sparse (k,∣F∣/kd)-blockade for some k∈[2,x−1], where x=ϵ5d and ∣G∣≥ϵ−10d2, then G has an (ϵ−1,ϵ10d2∣G∣)-blockade whose distinct block pairs are pairwise complete or weakly ϵd-sparse (Local pure or x-sparse blockades yield a nice blockade).

[L4]

The definition of generalized niceness is the four-outcome schema in Generalized nice finite graph families.

Proof

technique · choose the source exponents, assume the last three generalized-nice outcomes fail, and then force the first outcome by the local pure-or-sparse blockade theorem
1.1L1choosealgebra

Let c1′>0, c4′≥4, and d≥58c4′ be the constants from [L1]. Set c1:=10d2,c2:=d,c3:=156c4′d,c4:=c1′,c5:=2d,c6:=5d,c7:=5d,c8:=116c4′d. Then c1≥3, c2≥8, c6≥1, and c7≥4.

2.1step 1.1L4assume-contra

Let G be an F‾-free graph and let ϵ∈(0,12). If outcome 2, 3, or 4 of [L4] already holds for these constants, there is nothing left to prove. So assume for contradiction that all three fail, and write x:=ϵ5d.

3.1step 1.1step 2.1L4algebracases

If ∣G∣<ϵ−1, then ϵc3∣G∣≤ϵc3−1≤1, because c3>1. Any vertex of G therefore gives a clique or stable set of size at least (ϵc3∣G∣)c4, so outcome 2 of [L4] holds. Hence we may assume that ∣G∣≥ϵ−1.

4.1step 1.1step 3.1L4choosealgebracases

If ∣G∣≤ϵ−10d2, then choose ⌈ϵ−1⌉ distinct vertices of G and make them singleton blocks. Step 3.1 makes this possible, and each singleton has size 1≥ϵ10d2∣G∣=ϵc1∣G∣. Every pair of singleton blocks is either complete or anticomplete, hence either complete or weakly ϵd-sparse. Thus outcome 1 of [L4] holds. Therefore we may assume that ∣G∣≥ϵ−10d2.

5.1step 2.1step 4.1L2

Under steps 2.1 and 4.1, [L2] applies to every induced subgraph F of G with ∣F∣≥ϵd∣G∣.

6.1step 2.1step 5.1L2algebra

If an induced subgraph F of G with ∣F∣≥ϵd∣G∣ contained a clique or stable set of size at least (x31c4′∣F∣)c1′, then x31c4′∣F∣≥x31c4′ϵd∣G∣=ϵ155c4′d+d∣G∣≥ϵ156c4′d∣G∣=ϵc3∣G∣, so outcome 2 would hold, contrary to step 2.1. Likewise, if such an F contained a complete or anticomplete (k,∣F∣/kd)-blockade with k≥x−1, then ∣F∣kd≥ϵd∣G∣kd≥∣G∣k2d, so outcome 3 would hold, again contrary to step 2.1. Therefore [L2] really does give the pure-or-x-sparse blockade alternative on every such F.

7.1step 1.1step 4.1step 5.1step 6.1L3L4

By steps 4.1 and 6.1, the hypotheses of [L3] are satisfied with the parameter d and x=ϵ5d: the graph G has order at least ϵ−10d2, and every induced subgraph F with ∣F∣≥ϵd∣G∣ has a pure or x-sparse (k,∣F∣/kd)-blockade for some integer k∈[2,x−1]. Hence G has an (ϵ−1,ϵ10d2∣G∣)-blockade whose distinct block pairs are either complete or weakly ϵd-sparse. This is exactly outcome 1 of [L4], because c1=10d2 and c2=d.

8.1step 1.1step 2.1step 7.1L4discharge-contradiction∎

Outcome 1 follows whenever outcomes 2, 3, and 4 fail, and step 1.1 records the remaining lower-bound requirements on the constants. Therefore the constants from step 1.1 satisfy Definition [L4], so F is generalized nice.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-05Open item page →

A four-tooth comb with a special vertex realizes the trigger configuration for property (*)

Example

Let G have vertices

v, a1,a2,a3,a4, bi,j (i,j∈[4])

with edges exactly aibi,j and vbi,j for i,j∈[4]. Put

Bi:={bi,1,bi,2,bi,3,bi,4}

for each i.

Facts & Assumptions

Given: The graph and the blocks displayed in the Example.

[L1]

The item Combs in a graph characterizes an (ℓ,w)-comb by the blockade conditions and the adjacency pattern of the teeth.

[L2]

If a finite family F has property (∗) and G is F‾-free, then an (ℓ,w)-comb with ℓ,w≥4 together with a vertex outside the comb that is complete to the blocks and anticomplete to the teeth is the antecedent of the three-outcome implication in Property (*) for a finite graph family.

Verification

technique · direct
1.1givenL1

The four sets B1,B2,B3,B4 are pairwise disjoint and each has 4 vertices, so (B1,B2,B3,B4) is a (4,4)-blockade.

2.1step 1.1L1

Each tooth ai is adjacent to every vertex of Bi and to no vertex of Bj for j≠i, and the teeth a1,a2,a3,a4 are distinct and lie outside the blocks. Hence ((ai,Bi):i∈[4]) is a (4,4)-comb by [L1].

3.1step 2.1L2∎

The vertex v lies outside the comb, is adjacent to every vertex in ⋃i=14Bi, and is nonadjacent to every ai. Since this comb has ℓ=w=4, the pair consisting of the comb and v realizes the geometric trigger configuration occurring in [L2]. No assertion that an unspecified family has property (∗) is being made.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The third outcome of property (*) gives a pure four-blockade

Example

Assume the trigger hypothesis of property (∗) holds for a comb of length ℓ=4 and width w. If the third outcome of property (∗) occurs, the resulting blockade is pure with width w/ℓ2.

Facts & Assumptions

Given: A special-vertex comb with ℓ=4 and width w.

[L1]

The third branch in Property (*) for a finite graph family gives a pure (ℓ,w/ℓ2)-blockade.

Verification

technique · direct
1.1L1algebra

Applying [L1] with ℓ=4 yields a pure blockade whose width is wℓ2=w16.

2.1step 1.1L1∎

Since [L1] names the blockade pure rather than complete or anticomplete, this branch keeps exactly the distinction used later on the A page: every pair of blocks is pure, but no stronger global uniformity is asserted.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

A numerical square-root rescaling identity

Example

As a standalone numerical illustration, take

c4:=4,d:=232=58c4,ℓ:=256,k:=ℓ=16,

and assume ∣F∣≥∣G∣.

Facts & Assumptions

Given: The numerical choices in the Example.

[A1]

The sample values satisfy the numerical relation d=58c4. They are not asserted to be the existential constants supplied by the source lemma.

Verification

technique · direct arithmetic
1.1A1algebra

Since k=ℓ, one has ℓ29c4=(k2)29c4=k58c4=kd.

2.1step 1.1algebra

Therefore ∣F∣ℓ29c4=∣F∣kd≥∣G∣kd, using the assumption ∣F∣≥∣G∣.

3.1step 2.1∎

This standalone calculation isolates the square-root renormalization: after replacing ℓ by k=ℓ, the width bound takes exactly the target form ∣G∣/kd when the numerical relation d=58c4 holds.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The epsilon^(5d) substitution in Claim 4.5.1 and Lemma 4.5

Example

Take

ϵ:=14,c4′:=4,d:=232=58c4′,x:=ϵ5d.

Then the four exponent comparisons in the final property-(*) reduction become explicit.

Facts & Assumptions

Given: The displayed values of ϵ, d, c4′, and x.

[A1]

Since ϵ=14<12, the powers of ϵ decrease as their exponents increase.

Verification

technique · direct arithmetic
1.1A1algebra

The defining substitution gives x=ϵ5d=2−10d<2−d. Together with d=58c4′, this verifies the two parameter inequalities required before applying the Rödl-initialized theorem; its separate graph-order hypothesis must also be checked in any application.

1.2A1algebra

For the restricted-set branch, x23c4′ϵd=ϵ115c4′d+d≥ϵ116c4′d, because 115c4′+1≤116c4′ when c4′=4.

1.3A1algebra

For the clique-or-stable-set branch, x31c4′ϵd=ϵ155c4′d+d≥ϵ156c4′d, because 155c4′+1≤156c4′.

2.1step 1.1A1algebra

If k≥x−1, then ϵd=x1/5≥k−1/5, so ϵdkd≥1k2d. This is exactly the comparison used to turn the complete-or-anticomplete blockade branch into the generalized-nice blockade outcome.

3.1step 1.2step 1.3step 2.1∎

These computations are the concrete numerical version of the four exponent transfers behind the local blockade claim and the final proof of generalized niceness.

Sources