Alphabeta Math
Session-authored (Fable 5 assisted)
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.

3 results · all verified · 0 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 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Classical and Log-Log Erdős–Hajnal Bounds

1 · Prerequisites

2 · Summary

Homogeneous sets, sparse induced subgraphs, greedy colouring, the product bound Vχα, complementation, and base-2 logarithms are the ingredients behind the quantitative Erdős–Hajnal estimates. The page uses them in one fixed order: a density theorem isolates a large induced subgraph with few edges or few nonedges, trimming turns low density into bounded degree, colouring extracts a large stable set, and the complement turns the same argument into a clique bound.

The page then records two source-cited quantitative density inputs, one on the classical log2n scale and one on the improved log2nlog2log2n scale, and proves from them the corresponding homogeneous-set lower bounds for H-free graphs. A final corollary compares the two exponents and shows that the log-log scale eventually dominates every fixed classical scale.

3 · Logical flowchart

4 · Definitions, theorems and proofs

RemarkRemark: AI-adaptedProof: Not supplied sources checked 2026-08-26 not proved hereOpen item page →
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Fox–Sudakov: a quantitative density form of Rödl's theorem

Statement

For every finite graph H there exists a constant CH>0 such that for every real x with 0<x<1/2 and every finite H-free graph G, there is a vertex set SV(G) with S2CH(log2(1/x))2V(G) such that one of the induced graphs G[S] and G[S] has at most x(S2) edges.

Remarks

This page uses only that H-free specialization and keeps the source's base-2 logarithm convention. A local proof belongs to the quantitative induced-density track rather than to this page, so the result is recorded here and cited by Every H-free graph has a homogeneous set of size at least 2clog2n.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26 rests on unproved materialOpen item page →
Rests on 1 statement not proved in this library. Every dependency marked below is recorded with a citation but is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Every H-free graph has a homogeneous set of size at least 2clog2n

Statement

Let H be a finite graph. Then there exists a constant cH>0 such that every nonnull finite H-free graph G with n:=V(G)2 satisfies hom(G)2cHlog2n. Equivalently, G has a clique or a stable set of size at least 2cHlog2n.

Facts & Assumptions

Given: A finite graph H, a nonnull finite H-free graph G, and n:=V(G)2.

[L1]

The homogeneous number is hom(G)=max{ω(G),α(G)} (Homogeneous vertex sets and the homogeneous number hom(G)=max{ω(G),α(G)}).

[L2]

Fox-Sudakov quantitative density: there exists CH>0 such that for every real x with 0<x<1/2 there is SV(G) with S2CH(log2(1/x))2n and one of G[S] and G[S] has at most x(S2) edges (Fox–Sudakov: a quantitative density form of Rödl's theorem ).

[L3]

If a nonempty vertex set X satisfies dG(X,X)c, then some XX has XX/2 and is 4c-sparse (A set of self-density at most c has a subset of at least half its size that is 4c-sparse).

[L4]

A nonempty set X is c-sparse exactly when every vertex of G[X] has degree at most cX (A set is c-sparse exactly when the maximum degree of the graph it induces is at most c times its size).

[L5]

Every nonnull finite graph F satisfies χ(F)Δ(F)+1 (The greedy colouring bound χ(G)Δ(G)+1 for every nonnull finite graph).

[L6]

Every finite graph F satisfies V(F)χ(F)α(F) (The bounds ω(G)χ(G) and V(G)χ(G)α(G)).

[L7]

A vertex set is a clique in G if and only if it is a stable set in G (Complementation swaps cliques with stable sets, so ω(G)=α(G)).

[L8]

For nonempty X, the self-density is dG(X,X)=2E(G[X])/X2 (Edge counts and densities between nonempty vertex sets).

Proof

technique · direct
1.1

By [L2], choose a constant CH>0. Write L:=log2n and set x:=2L/(2CH). Since n2, one has L>0, so 0<x<1/2.

L2choose
2.1

Because log2(1/x)=L/(2CH), [L2] gives a set SV(G) with S2CH(log2(1/x))2n=2L/2n=n, and one of G[S] and G[S] has at most x(S2) edges. For that chosen graph F on vertex set S, [L8] gives dF(S,S)=2E(F)/S22x(S2)/S2=x(S1)/Sx.

step 1.1 L2L8algebra
3.1

If F=G[S], then [L3] gives XS with XS/2 and X 4x-sparse in G. By [L4] every vertex of G[X] has degree at most 4xX, so [L5] gives χ(G[X])4xX+1, and then [L6] yields α(G[X])X/(4xX+1).

step 2.1L3L4L5L6
3.2

If F=G[S], then [L3] gives XS with XS/2 and X 4x-sparse in G. Applying [L4], [L5], and [L6] inside the complement shows that G[X] has a stable set of size at least X/(4xX+1), and [L7] turns that stable set into a clique of the same size in G[X].

step 2.1L3L4L5L6L7
4.1

Steps 3.1 and 3.2 show that G has a homogeneous set Y with YX/(4xX+1) for some XS satisfying XS/2n/2.

step 3.1step 3.2step 2.1L1
5.1

Because xn=2L/2L/(2CH), choose NH2 so that xn1 whenever nNH. For such n, step 4.1 gives 4xX2, hence 4xX+18xX, so Y1/(8x)=2L/(2CH)3.

step 4.1step 1.1choosealgebra
6.1

Set cH:=1/(42CH). Choose NHNH so that L/(2CH)3cHL whenever nNH. Then step 5.1 gives Y2cHL for all nNH. Shrinking cH if necessary handles the finitely many integers 2n<NH, because every nonnull graph has hom(G)1.

step 5.1L1choosealgebra
7.1

Therefore every nonnull finite H-free graph G with V(G)=n2 satisfies hom(G)2cHlog2n.

step 6.1L1
RemarkRemark: AI-adaptedProof: Not supplied sources checked 2026-08-26 not proved hereOpen item page →
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Bucić–Nguyen–Scott–Seymour: a log-log quantitative density theorem

Statement

For every finite graph H there exists a constant CH>0 such that for every real x with 0<x<1/2 and every finite H-free graph G, there is a vertex set SV(G) with S2CH(log2(1/x))2/log2log2(1/x)V(G) such that one of the induced graphs G[S] and G[S] has at most x(S2) edges.

Remarks

Again the statement is recorded exactly in the paper's base-2 convention. The page uses it only as a source-cited input for Every H-free graph has a homogeneous set of size at least 2clog2nlog2log2n.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26 rests on unproved materialOpen item page →
Rests on 1 statement not proved in this library. Every dependency marked below is recorded with a citation but is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Every H-free graph has a homogeneous set of size at least 2clog2nlog2log2n

Statement

Let H be a finite graph. Then there exists a constant cH>0 such that every nonnull finite H-free graph G with n:=V(G)2 satisfies hom(G)2cHlog2nlog2log2n.

Facts & Assumptions

Given: A finite graph H, a nonnull finite H-free graph G, and n:=V(G)2.

[L1]

The homogeneous number is hom(G)=max{ω(G),α(G)} (Homogeneous vertex sets and the homogeneous number hom(G)=max{ω(G),α(G)}).

[L2]

Bucić-Nguyen-Scott-Seymour quantitative density: there exists CH>0 such that for every real x with 0<x<1/2 there is SV(G) with S2CH(log2(1/x))2/log2log2(1/x)n and one of G[S] and G[S] has at most x(S2) edges (Bucić–Nguyen–Scott–Seymour: a log-log quantitative density theorem ).

[L3]

If a nonempty vertex set X satisfies dG(X,X)c, then some XX has XX/2 and is 4c-sparse (A set of self-density at most c has a subset of at least half its size that is 4c-sparse).

[L4]

A nonempty set X is c-sparse exactly when every vertex of G[X] has degree at most cX (A set is c-sparse exactly when the maximum degree of the graph it induces is at most c times its size).

[L5]

Every nonnull finite graph F satisfies χ(F)Δ(F)+1 (The greedy colouring bound χ(G)Δ(G)+1 for every nonnull finite graph).

[L6]

Every finite graph F satisfies V(F)χ(F)α(F) (The bounds ω(G)χ(G) and V(G)χ(G)α(G)).

[L7]

A vertex set is a clique in G if and only if it is a stable set in G (Complementation swaps cliques with stable sets, so ω(G)=α(G)).

[L8]

For nonempty X, the self-density is dG(X,X)=2E(G[X])/X2 (Edge counts and densities between nonempty vertex sets).

Proof

technique · direct
1.1

By [L2], choose a constant CH>0. Because every nonnull graph has hom(G)1, it is enough to prove the bound for all sufficiently large n; assume from now on that n is large enough that L:=log2n4. Set β:=1/(4CH) and x:=2βLlog2L. Then 0<x<1/2.

L1 L2choose
2.1

For large enough L, the inequality βLlog2LL holds, so log2log2(1/x)=log2(βLlog2L)12log2L. Therefore CH(log2(1/x))2/log2log2(1/x)2CHβ2L=L/8. Using [L2], obtain SV(G) with S2L/8n=27L/8n, and one of G[S] and G[S] has at most x(S2) edges. For that chosen graph F on vertex set S, [L8] gives dF(S,S)2x(S2)/S2=x(S1)/Sx.

step 1.1 L2L8algebra
3.1

If F=G[S], then [L3] gives XS with XS/2 and X 4x-sparse in G. By [L4], [L5], and [L6], α(G[X])X/(4xX+1).

step 2.1L3L4L5L6
3.2

If F=G[S], then the same argument inside the complement produces a stable set of G[X] of size at least X/(4xX+1) for some XS with XS/2, and [L7] turns it into a clique of G[X].

step 2.1L3L4L5L6L7
4.1

Steps 3.1 and 3.2 show that G has a homogeneous set Y with YX/(4xX+1) for some XS satisfying XS/2n/2.

step 3.1step 3.2step 2.1L1
5.1

Because xn=2L/2βLlog2L, choose a threshold NH2 so that xn1 whenever nNH. For those n, step 4.1 gives 4xX2, hence 4xX+18xX, and therefore Y1/(8x)=2βLlog2L3.

step 4.1step 1.1choosealgebra
6.1

Set cH:=β/2. For all sufficiently large n, the inequality βLlog2L3cHLlog2L holds, so step 5.1 gives Y2cHLlog2L. Shrinking cH if necessary handles the finitely many smaller values of n.

step 5.1choosealgebra
7.1

Hence every nonnull finite H-free graph G with V(G)=n2 satisfies hom(G)2cHlog2nlog2log2n.

step 6.1L1
CorollaryStatement: AI-generatedProof: AI-generatedprecheck passaudited 2026-08-26 rests on unproved material (inherited)Open item page →
Rests on 2 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Fox–Sudakov: a quantitative density form of Rödl's theorem and Bucić–Nguyen–Scott–Seymour: a log-log quantitative density theorem. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

For fixed H, the log-log scale eventually exceeds every classical scale 2clog2n

Statement

Let a,b>0. Then there exists N2 such that for every integer nN, 2blog2nlog2log2n2alog2n. In particular, for each fixed finite graph H, the log-log lower bound of Every H-free graph has a homogeneous set of size at least 2clog2nlog2log2n eventually exceeds every classical scale 2alog2n.

Facts & Assumptions

Given: Positive reals a,b.

[L1]

For n>1, log2n is defined (The logarithm to a positive base other than one).

Proof

technique · direct
1.1

For every integer n>2, one has log2nlog2log2n=log2log2nlog2n.

L1algebra
2.1

Choose N4 so that blog2log2na whenever nN. This is possible because log2log2n tends to + with n.

step 1.1choose
3.1

For every nN, step 2.1 gives blog2nlog2log2nalog2n, and exponentiating base 2 preserves the inequality.

step 1.1step 2.1algebra
4.1

This proves the displayed eventual inequality, and the final sentence is its application with the constant supplied by Every H-free graph has a homogeneous set of size at least 2clog2nlog2log2n.

step 3.1

5 · Examples, counterexamples and false statements

None yet.

Sources