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21 results · all verified · 13 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 8 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Iterative Sparsification and the Five-Vertex Path

1 · Prerequisites

2 · Summary

This draft page follows the source split that the design called out: the first half proves that P5 is nice by iteratively building pure or sparse blockades, and the second half turns a nice blockade into either a complete or anticomplete blockade or a deeper sparse subgraph until the published restricted-set theorem can close the argument.

The final two items separate the two endpoints. First the page proves the polynomial Rödl property for P5, then it converts that stronger conclusion to the Erdős-Hajnal property using the already-published Rödl-to-EH implication.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A nice graph

Definition

Let H be a finite graph. We say that H is nice if there exists a real d1 such that for every ϵ(0,12) and every H-free graph G with

V(G)ϵ10d2,

there is a (ϵ1,ϵ10d2V(G))-blockade B=(B1,,B) in G such that for all distinct i,j[] the pair (Bi,Bj) is either complete or weakly ϵd-sparse.

The complement appears because the source proves niceness for P5 through P5-free graphs. The blockade clause is exactly the local configuration that the second half of the source proof refines into complete or anticomplete blockades.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Small anticonnected components yield a complete blockade

Statement

Let G be a graph on n vertices, and let k2 be an integer. Suppose every anticonnected component of G has size less than n/k. Then G contains a complete blockade of length at least k/2 and width at least n/(2k).

Facts & Assumptions

Given: A graph G on n vertices and an integer k2 such that every anticonnected component of G has size less than n/k.

[L1]

Proof

technique · direct packing
1.1

Partition the anticonnected components into a minimum number of unions S0,,Sr, each of size less than n/k, and order them so that S0Sr. Since their union has size n, one has r+1>k.

givenchoosealgebra
2.1

For every i1, minimality gives Si1+Sin/k; otherwise these two parts could be merged. The ordering then yields Sin/(2k). Thus S1,,Sr contain at least k nonempty blocks of width at least n/(2k), and [L1] makes every cross-pair complete. Selecting any k/2 of them proves the statement.

step 1.1L1algebra
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

A dense bipartite side has a small hitting set

Statement

Let A,B be disjoint nonempty vertex sets in a graph, and let x(0,1]. Assume every vertex of B has at least xA neighbours in A. Then there is a set SA with S1/x that meets the neighbourhood in A of at least half of the vertices of B.

Facts & Assumptions

Given: Disjoint nonempty vertex sets A,B in a graph and a real x(0,1] such that every bB has at least xA neighbours in A.

[L1]

If S=A, then every neighbourhood in A is hit; otherwise a uniform m-subset of A misses a fixed b-neighbourhood with probability at most (1x)m.

Proof

technique · probabilistic existence
1.1

If 1/xA, take S:=A and every neighbourhood in A is hit. Otherwise let m:=1/x<A and choose a subset SA uniformly among all subsets of size m. For a fixed vertex bB, the probability that SN(b)= is at most (1x)mexme1<1/2.

L1givenchoosealgebra
2.1

In the first case every vertex of B is hit. In the second case the expected number of vertices of B whose neighbourhood misses S is less than B/2, so some choice of S misses fewer than half of B. Thus in either case there is a set SA with S1/x that meets the neighbourhood of at least half of the vertices of B.

step 1.1algebra
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A sparse P5-free graph has an anticomplete two-blockade

Statement

There exists η>0 such that every η-sparse P5-free graph G with V(G)η1 contains an anticomplete blockade of length 2 and width at least ηV(G).

Facts & Assumptions

Given: An η-sparse P5-free graph G with V(G)η1.

[L1]

Lemma 4.4 of the cited source proves exactly the displayed conclusion for η=25, using repeated large-component consequences of the failure of the desired anticomplete blockade.

Proof

technique · translate the cited source lemma
1.1

The cited source lemma assumes the negation of the desired blockade, first obtains a connected component of size at least V(G)/2, and then obtains two further large connected pieces outside successive neighbourhoods.

L1given
2.1

The source shows that the absence of the blockade then forces five selected vertices to induce P5, contradicting the hypothesis. Therefore the stated anticomplete two-blockade exists.

step 1.1L1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A sparse P5-free graph has a large nearly covered sparse pair

Statement

Let x,y>0 with xy28, and let G be a y3-sparse P5-free graph with V(G)y4. Suppose that G is not 2y4-sparse, and that there do not exist disjoint sets X,YV(G) such that

Xy4V(G),Y(14y)V(G),

and Y is x-sparse to X. Then there exist disjoint sets A,BV(G) such that:

  1. A(13y)V(G) and By4V(G);
  2. A is y2-sparse to B; and
  3. every vertex of B has at least x2A neighbours in A.

Facts & Assumptions

Given: Parameters x,y and a graph G satisfying the displayed hypotheses.

[L1]

Claim 5.2.1 of the cited source proves exactly the displayed conclusion under these hypotheses after translating notation.

Proof

technique · translate the cited source claim
1.1

The cited source claim proves exactly this large nearly covered sparse-pair conclusion after translating notation.

L1given
2.1

Therefore the present statement follows.

step 1.1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-31Open item page →

Anticonnected block contraction turns an upside-down comb into a pure blockade

Statement

Let G be a P5-free graph, let k4 be an integer, and let BV(G). Suppose

((ai,Bi):i[k])

is a (k,B/k8)-comb in G with BiB and aiB for every i[k]. Suppose also that there is a vertex

vV(G)(B{a1,,ak})

that is complete to B and anticomplete to {a1,,ak}. Then G[B] contains a pure (k,B/k10)-blockade.

Facts & Assumptions

Given: The graph G, the integer k, the set B, the displayed comb, and the vertex v satisfying the hypotheses above.

[F1]

In the displayed comb, ai is complete to Bi and anticomplete to Bj for ij, while the blocks are pairwise disjoint and satisfy BiB/k8 (Combs in a graph).

[F2]

A set D is an anticonnected component of G[X] precisely when it is the vertex set of a connected component of G[X] (Anticonnected graphs and anticonnected components).

[L1]

Distinct anticonnected components of a graph are complete to one another (Distinct connected components are anticomplete, and distinct anticonnected components are complete).

[F3]

A blockade is pure when every pair of distinct blocks is either complete or anticomplete (Complete, anticomplete, pure, weakly sparse, and x-sparse blockades).

[F5]

Complementation exchanges edges and nonedges (Graph isomorphisms, automorphisms and graph complements).

Proof

technique · contract each comb block to a large anticomponent
1.1

Fix i[k] and put ni:=Bi. Suppose first that every anticonnected component of G[Bi] has size less than ni/k. Partition the anticonnected components into a minimum number of unions S0,,Sr, each of size less than ni/k, and order them by nondecreasing size. Since the parts cover Bi, one has r+1>k. Minimality gives Sj1+Sjni/k for j1, and hence Sjni/(2k)ni/k2 because k4. By [L1], the sets S1,,Sk are pairwise complete. They therefore form a complete, hence pure, (k,ni/k2)-blockade in G[B]. By [F1], ni/k2B/k10, which proves the result in this case.

F1L1F3givenchoosealgebra
2.1

We may consequently assume that, for every i[k], the graph G[Bi] has an anticonnected component Di with DiBi/k. Choose one such Di. Then DiBi/kB/k9B/k10, giving the required width bound for every chosen component.

F1F2step 1.1choosealgebra
3.1

Let ij, and suppose that some uDj is mixed on Di. The sets of neighbours and nonneighbours of u in Di are both nonempty. Since G[Di] is connected, some edge of that complement crosses these two sets. Thus there are w,zDi such that uwE(G), uzE(G), and wzE(G).

F2step 2.1given
4.1

Among the five vertices v,ai,u,z,w, the nonedges are exactly vai,aiu,uz,zw. Indeed, the hypotheses on v determine its four incidences; [F1] determines the incidences from ai to DiBi and to uDjBj; and step 3.1 determines the three remaining incidences. Hence those four nonedges form the path vaiuzw, so these vertices induce P5 by [F4] and [F5], contrary to the hypothesis on G. Therefore no vertex of Dj is mixed on Di.

F1F4F5step 3.1given
5.1

Applying step 4.1 with both orders of i,j shows that no vertex of either set is mixed on the other. If the pair (Di,Dj) had both an edge and a nonedge, then the vertices of Dj would include one complete to Di and one anticomplete to Di; every endpoint in Di of the cross-edge would then be mixed on Dj, a contradiction. Thus (Di,Dj) is pure.

step 4.1cases
6.1

The sets D1,,Dk are pairwise disjoint subsets of B, have size at least B/k10 by step 2.1, and every cross-pair is pure by step 5.1. Hence they form the required pure (k,B/k10)-blockade in G[B].

F1F3step 2.1step 5.1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A sparse P5-free graph either sparsifies further or yields a pure blockade or a large sparse pair

Statement

Let x,y>0 with xy28, and let G be a y3-sparse P5-free graph with V(G)y4. Then at least one of the following holds:

  1. G is 2y4-sparse;
  2. there exists an integer k[y1/4,1/x] and a pure (k,V(G)/k26)-blockade in G; or
  3. there are disjoint sets X,YV(G) such that Xy4V(G), Y(14y)V(G), and Y is x-sparse to X.

Facts & Assumptions

Given: Parameters x,y and a graph G satisfying the displayed hypotheses.

[L1]

Lemma 5.2 of the cited source proves exactly the displayed trichotomy under these hypotheses.

Proof

technique · translate the cited source lemma
1.1

The cited source lemma proves exactly the three displayed alternatives under these hypotheses.

L1given
2.1

Therefore the present trichotomy holds.

step 1.1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

An x-sparse blockade iteration yields further sparsification or a pure blockade

Statement

Let x(0,12). There exist constants c>0 and d1 such that the following holds. If G is a sparse P5-free graph and B=(B0,,Bn) is a maximal x-sparse blockade whose last block still has linear size, then either:

  1. some induced subgraph of Bn is substantially sparser than G; or
  2. G has a pure blockade of polynomial width.

Facts & Assumptions

Given: A sparse P5-free graph G and a maximal x-sparse blockade B=(B0,,Bn) with large last block Bn.

[L1]

Lemma 5.3 of the cited source proves the stated maximal-blockade alternative with explicit constants, sizes, and exponents.

Proof

technique · translate the cited source lemma
1.1

The cited source lemma applies the preceding sparse-pair trichotomy to the final block and uses the correctly oriented sparse relations to extend the blockade whenever the sparse-pair outcome occurs.

L1given
2.1

Maximality excludes that extension, leaving exactly a substantially sparser induced subgraph or a polynomial-width pure blockade.

step 1.1L1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-31Open item page →

An iterative sparsification step for sparse P5-free graphs

Statement

Put c:=28. Let x(0,c5), and let G be a c16-sparse P5-free graph with V(G)x7. Then at least one of the following holds:

  1. for some k[c1,x1], there is a pure (k,V(G)/k34)-blockade in G; or
  2. for some y[x,c5], there is an x-sparse (y1,y7V(G))-blockade in G.

Facts & Assumptions

Given: The constant c=28, a parameter x(0,c5), and a c16-sparse P5-free graph G with V(G)x7.

[F1]

Lemma 5.4 of Nguyen, Scott, and Seymour's cited paper gives the displayed two outcomes with these constants and exponents. Its statement prints Gy7 before y is bound; the proof shows that the intended hypothesis is Gx7 by using it to deduce cxGx2Gx5.

[F2]

The source proof chooses a minimal threshold y[cx,c5], applies its preceding three-outcome sparse-blockade lemma, and rules out the deeper-sparsification branch by minimality. The remaining branches give the pure blockade in outcome 1 or the x-sparse blockade in outcome 2.

Proof

technique · translate the cited source lemma
1.1

Apply the corrected, well-formed reading of the cited source lemma recorded in [F1]. Its two alternatives are exactly outcomes 1 and 2, and [F2] records the minimal-threshold argument establishing them.

F1F2givencases
2.1

Therefore the present statement follows.

step 1.1
TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-31Open item page →

P5-free graphs admit a pure or x-sparse polynomial blockade

Statement

There exists d40 such that for every x(0,2d) and every P5-free graph G with V(G)xd, there exists an integer k[2,x1] and either

  1. a pure (k,V(G)/kd)-blockade in G; or
  2. an x-sparse (k,V(G)/kd)-blockade in G.

Facts & Assumptions

Given: After the exponent d is chosen below, a parameter x(0,2d) and a P5-free graph G with V(G)xd.

[L1]

Lemma 5.5 of the cited source supplies an exponent D40 such that, under its convention allowing a real blockade-length threshold, there is some r[2,x1] and a pure or x-sparse (r,V(G)/rD)-blockade whenever x(0,2D) and V(G)xD.

[F2]

In this library, the first parameter of an (,w)-blockade must be a natural number, and the actual length is at least (Blockades, their length, their width, and their support).

Proof

technique · translate the cited source theorem
1.1

Let D40 be supplied by [L1], and set d:=2D. Fix x and G as in the Statement. Since dD, one has x<2D and V(G)xD. Thus [L1] gives a real r[2,x1] and a pure or x-sparse blockade whose actual length is at least r and whose width is at least V(G)/rD.

L1givenalgebra
2.1

Put k:=r. Then k is an integer and 2krx1. The blockade's integral actual length, being at least r, is in particular at least k, as required by [F2].

step 1.1F2algebra
3.1

Since k2 and kr<k+1, one has rD<(k+1)D(3k/2)Dk2D=kd. Consequently V(G)/rDV(G)/kd. The blockade from step 1.1 is therefore a pure or x-sparse (k,V(G)/kd)-blockade in the library's sense.

step 1.1step 2.1algebra
4.1

The chosen d=2D satisfies d40, and steps 1.1--3.1 prove the stated conclusion for every admissible x and G.

step 1.1step 3.1
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A maximal layout has at most ϵ1 blocks

Statement

Let ϵ(0,12) and d1. In the counterexample layout construction used for Theorem 6.1 of the cited source, the chosen maximal layout has fewer than ϵ1 blocks; otherwise its blocks already contain the blockade required by that theorem.

Facts & Assumptions

Given: The maximal layout and counterexample hypotheses in the proof of Theorem 6.1 of the cited source.

[L1]

Claim 6.1.1 of the cited source proves that a chosen layout with at least ϵ1 blocks already satisfies the target blockade conclusion.

Proof

technique · direct from the cited source claim
1.1

If the chosen layout had at least ϵ1 blocks, [L1] would make its blocks a blockade satisfying the conclusion of the surrounding theorem.

L1given
2.1

The enclosing counterexample excludes that conclusion, so the layout has fewer than ϵ1 blocks.

step 1.1
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Refining the largest layout block forces local blockade length at least ϵ1

Statement

In the setting of the previous lemma, let A be the largest block of the maximal layout. If A is refined by a pure or x-sparse polynomial blockade, then that local blockade has length at least ϵ1.

Facts & Assumptions

Given: A maximal layout, its largest block A, and a pure or x-sparse polynomial blockade inside A.

[L1]

The cited source claim proves that substituting a local blockade of length below ϵ1 into the largest layout block preserves the three defining layout bounds while strictly increasing the number of blocks.

Proof

technique · contradiction by substitution
1.1

Suppose the local blockade inside A had length k<ϵ1. By [L1], substituting its pattern for the layout vertex corresponding to A produces another admissible layout with strictly more blocks.

L1assume-contra
2.1

This contradicts the maximal choice of the original layout. Hence the local blockade has length at least ϵ1.

step 1.1discharge-contradiction
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Local pure or x-sparse blockades yield a nice blockade

Statement

Let ϵ(0,12) and d1, and put x:=ϵ5d. Let G be a graph with V(G)ϵ10d2. Assume that every induced subgraph F of G with V(F)ϵdV(G) has a pure or x-sparse (k,V(F)/kd)-blockade for some integer k[2,x1]. Then G has a (ϵ1,x2dV(G))-blockade whose distinct block pairs are either complete or weakly ϵd-sparse.

Facts & Assumptions

Given: The hypotheses in the statement.

[L1]

Theorem 6.1 of the cited source proves exactly the displayed local-to-global blockade conclusion, with its layout carrying both the block-size power-sum condition and the wrong-pair bound.

Proof

technique · translate the cited source theorem
1.1

The cited source theorem applies to the hypotheses above and produces a blockade of length at least ϵ1, width at least x2dV(G), and complete-or-weakly-ϵd-sparse cross-pairs.

L1given
2.1

Since blockade length is integral, length at least ϵ1 is equivalent to length at least ϵ1. This is exactly the stated conclusion.

step 1.1algebra
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

The five-vertex path is nice

Statement

The graph P5 is nice.

Facts & Assumptions

Given: The five-vertex path P5.

[L1]

Every sufficiently large P5-free graph admits a pure or x-sparse polynomial blockade when x is below the source threshold (P5-free graphs admit a pure or x-sparse polynomial blockade).

[L2]

Such local pure or sparse blockades force a nice blockade (Local pure or x-sparse blockades yield a nice blockade).

[L3]

A graph is nice exactly when some exponent d makes the conclusion of step 2.1 hold for every sufficiently large H-free graph (A nice graph).

Proof

technique · direct
1.1

Let d be a common exponent large enough to dominate the polynomial width bound in [L1] and the layout theorem [L2]. Fix ϵ(0,12) and a P5-free graph G with V(G)ϵ10d2, and set x:=ϵ5d. Then x<2d. If F is an induced subgraph of G with V(F)ϵdV(G), then V(F)ϵdϵ10d2=ϵ10d2+dϵ5d2=xd. Therefore [L1] applies to F and gives a pure or x-sparse (k,V(F)/kd)-blockade for some k[2,ϵ5d2].

L1choosegivenalgebra
2.1

Applying [L2] to G yields an (ϵ1,ϵ10d2V(G))-blockade whose distinct block pairs are either complete or weakly ϵd-sparse. By [L3], this is exactly the niceness condition for H=P5.

step 1.1L2L3
3.1

Therefore P5 is nice.

step 2.1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A semisparse blockade can be sampled to anticonnected blocks with nearly pure relations

Statement

There exist constants c>0 and d1 such that every sufficiently large sparse P5-free graph has at least one of the following:

  1. a complete blockade of linear width; or
  2. a blockade (B1,,Bt) with tc1, every block of size at least cV(G), each Bi anticonnected, and every pair (Bi,Bj) with ij either complete or weakly c-sparse.

Facts & Assumptions

Given: A sufficiently large sparse P5-free graph G.

[L1]

Claim 7.1.1 of the cited source yields exactly the displayed two-outcome alternative after translating exponents into constants.

Proof

technique · translate the cited source claim
1.1

The cited source claim yields exactly this semisparse-blockade or complete-blockade alternative after translating exponents into constants.

L1given
2.1

Therefore one of the two displayed outcomes holds.

step 1.1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

No vertex is mixed on many blocks of a semisparse blockade

Statement

There exist constants c>0 and d1 such that the following holds. Let G be a sufficiently large c-sparse P5-free graph, and let (B1,,Bt) be a blockade from outcome 2 of A semisparse blockade can be sampled to anticonnected blocks with nearly pure relations. Then at least one of the following holds:

  1. G has a cd-sparse induced subgraph of linear size; or
  2. every vertex outside the blockade is mixed on fewer than ct blocks, where a vertex v is mixed on Bi when the pair ({v},Bi) is mixed in the sense of Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs.

Facts & Assumptions

Given: A graph G and a blockade (B1,,Bt) as in the statement.

[L1]

Claim 7.1.2 of the cited source proves exactly the displayed dichotomy for such semisparse blockades after translating exponents into constants.

Proof

technique · translate the cited source claim
1.1

The cited source claim proves exactly this mixed-block dichotomy after translating exponents into constants.

L1given
2.1

Therefore one of the two displayed outcomes holds.

step 1.1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-31Open item page →

A sparse P5-free graph yields deeper sparsification or a complete blockade or a large anticomplete set

Statement

There exists a constant d40 such that, for every y(0,12) and every y-sparse P5-free graph G, at least one of the following holds:

  1. there is a set SV(G) with Sy30d3V(G) such that G[S] is y2d-sparse;
  2. there is a complete (y1,y33d3V(G))-blockade in G; or
  3. there are disjoint sets X,YV(G) such that Xy33d3V(G),Y(13y)V(G), and X is anticomplete to Y.

Facts & Assumptions

Given: A parameter y(0,12) and a y-sparse P5-free graph G.

[F1]

Lemma 7.1 of Nguyen, Scott, and Seymour's cited paper states the displayed trichotomy, with the same constant d40 and the same exponents and blockade parameters.

[F2]

In the proof of that lemma, Claim 7.1.1 constructs either the complete blockade in outcome 2 or a long semisparse blockade with anticonnected blocks. Claim 7.1.2 shows that a vertex mixed on many of those blocks yields outcome 1; otherwise averaging over the blocks gives a block anticomplete to a set of size at least (13y)V(G), which is outcome 3.

Proof

technique · translate the cited source lemma
1.1

Apply [F1] to the graph in the Given. Its three alternatives are exactly outcomes 1, 2, and 3 in the statement; [F2] records how the semisparse and mixed-block cases in the source proof produce those alternatives.

F1F2givencases
2.1

Therefore the present trichotomy holds.

step 1.1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

A sparse P5-free graph yields a complete or anticomplete blockade or a sparser subgraph

Statement

There exist constants c>0 and d1 such that every sufficiently large c-sparse P5-free graph has at least one of the following:

  1. a complete blockade of polynomial width;
  2. an anticomplete blockade of polynomial width; or
  3. a cd-sparse induced subgraph of linear size.

Facts & Assumptions

Given: A sufficiently large c-sparse P5-free graph G.

[L1]

Lemma 7.2 of the cited source proves the displayed complete-or-anticomplete blockade versus deeper-sparsification alternative with explicit parameters.

Proof

technique · translate the cited source lemma
1.1

The cited source lemma builds a maximal anticomplete blockade and verifies the size normalization needed before applying the preceding sparse trichotomy inside its last block.

L1given
2.1

Its three resulting cases are precisely a complete blockade, an anticomplete blockade, or a linearly large induced subgraph with strictly deeper sparsity.

step 1.1L1
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

The minimal sparsity parameter drops below the target

Statement

Let x(0,12). In the minimal-threshold setup of the source, the least sparsity parameter y for which a linear-sized y-sparse induced subgraph exists satisfies y<x.

Facts & Assumptions

Given: A parameter x(0,12) and a counterexample graph G in which y is minimal with the stated property.

[L1]

Claim 7.3.1 of the cited source proves exactly that the minimal parameter in its fully quantified threshold setup satisfies y<x.

Proof

technique · direct translation
1.1

The cited source claim applies the exact sparsity, size, and exponent bounds to the minimal witness and excludes its blockade outcome by the enclosing counterexample hypothesis.

L1given
2.1

The remaining outcome contradicts minimality unless y<x, which proves the statement.

step 1.1L1
TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

P5-free graphs yield a polynomial restricted set or a complete or anticomplete blockade

Statement

There exists a1 such that for every x(0,12) and every P5-free graph G, at least one of the following holds:

  1. G has an x-restricted induced subgraph with at least xaV(G) vertices; or
  2. G has a complete or anticomplete (k,V(G)/ka)-blockade for some integer k[2,x1].

Facts & Assumptions

Given: A parameter x(0,12) and a P5-free graph G.

[L1]

Lemma 7.3 of the cited source proves exactly the displayed restricted-set or complete/anticomplete-blockade alternative, with explicit constants.

Proof

technique · direct from the minimal-threshold alternative
1.1

The cited source lemma runs the minimal-threshold argument with all size and exponent bounds and proves the first alternative whenever the blockade alternative is absent.

L1given
2.1

Consequently either the stated polynomial restricted set exists or the source's complete/anticomplete polynomial blockade exists, exactly as claimed.

step 1.1L1
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

The five-vertex path has the polynomial Rödl property

Statement

The graph P5 has the polynomial Rödl property.

Facts & Assumptions

Given: The graph P5.

[L1]

There exists a1 such that every P5-free graph either has an x-restricted induced subgraph of size at least xaV(G) or has a complete or anticomplete (k,V(G)/ka)-blockade for some k[2,x1] (P5-free graphs yield a polynomial restricted set or a complete or anticomplete blockade).

[L2]

The blockade alternative alone already forces an x-restricted induced subgraph of size at least x3aV(G) (Complete or anticomplete blockade hypotheses force an ϵ-restricted induced subgraph).

Proof

technique · direct
1.1

Let a be as in [L1]. Fix x(0,12) and a P5-free graph G. Then G is P5-free. Apply [L1] to every induced subgraph F of G with V(F)x2aV(G). If any such F has an x-restricted set of size at least xaV(F), that set has size at least x3aV(G) and is also x-restricted in G.

L1givencasesalgebra
2.1

Otherwise every such F has the complete-or-anticomplete blockade supplied by [L1], so the hypothesis of [L2] holds for G. Applying [L2] yields an x-restricted induced subgraph on at least x3aV(G) vertices; the same vertex set is x-restricted in G.

step 1.1L2cases
3.1

Therefore P5 has the polynomial Rödl property.

step 2.1
CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

The five-vertex path and its complement have the Erdős-Hajnal property

Statement

Both P5 and P5 have the Erdős-Hajnal property.

Facts & Assumptions

Given: The graph P5.

[L1]

The graph P5 has the polynomial Rödl property (The five-vertex path has the polynomial Rödl property).

[L2]

Every finite family with the polynomial Rödl property has the Erdős-Hajnal property (The polynomial Rödl property implies the Erdős–Hajnal property).

[F1]

The polynomial Rödl property and the Erdős-Hajnal property are both invariant under complementation of the forbidden graph.

Proof

technique · direct
1.1

Applying [L2] to the singleton family {P5} and using [L1], we conclude that P5 has the Erdős-Hajnal property.

L1L2
2.1

By [F1], the same holds for P5.

step 1.1
3.1

Therefore both P5 and P5 have the Erdős-Hajnal property.

step 1.1step 2.1

5 · Examples, counterexamples and false statements

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Sources