Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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.

For every t1, the class of Kt-free graphs has the Erdős–Hajnal property

Statement

For every positive integer t, the hereditary class of Kt-free finite graphs has the Erdős–Hajnal property.

Facts & Assumptions

Given: A positive integer t and the class Ct of Kt-free finite graphs.

[L1]

For a graph G, hom(G)=max{ω(G),α(G)} (Homogeneous vertex sets and the homogeneous number hom(G)=max{ω(G),α(G)}).

[L2]

A hereditary class has the Erdős–Hajnal property when some ϵ>0 satisfies hom(G)V(G)ϵ for every nonempty member G (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).

[L3]

A graph is Kt-free when it has no induced copy of Kt (H-free and F-free graphs under the induced-subgraph convention), and the class of graphs free of any fixed family is hereditary (Every class defined by forbidden induced subgraphs is hereditary).

[L4]

The graph Kt has every pair of its t vertices as an edge, while an empty graph has no edges (Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices).

[L5]

For positive a,b, every graph on at least (a+b2a1) vertices contains an a-clique or a b-vertex stable set (Finite graph Ramsey theorem: (s+t2s1)(s,t)2 for all positive s,t).

[L6]

The binomial coefficient (mr) counts the r-subsets of an m-set (The set [A]k of k-element subsets and the binomial coefficient (nk):=[n]k).

[L7]

For x>0 and real u, xu=exp(ulogx) (Real powers for positive bases, with the zero-base positive-exponent convention).

[L8]

The logarithm is strictly increasing, maps 1 to 0, and obeys the product and quotient laws (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L9]

The exponential is strictly increasing (The exponential function is strictly increasing).

Proof

technique · direct
1.1

By [L3], Ct is hereditary. If t=1, it has no nonempty member, so any positive exponent works in [L2]; if t=2, every member is empty by [L3] and [L4], so hom(G)=V(G) and exponent 1 works.

L1L2L3L4
1.2

Assume t3. For all sufficiently large integers n, the integer s=12n1/(t1) satisfies st2, sn1/t, and s1; these assertions follow from [L7], [L8], and [L9] because n1/(t1)/n1/t=n1/(t(t1)) tends to infinity.

givenL7L8L9algebra
2.1

For such n, t+s22s, and hence (t+s2t1)(t+s2)t1(2s)t1n; the first inequality counts ordered choices containing every (t1)-subset.

step 1.2L6algebra
3.1

If GCt has sufficiently large order n, [L5] with parameters (t,s) and step 2.1 give a t-clique or an s-vertex stable set; the first is forbidden, so hom(G)α(G)sn1/t.

step 1.2step 2.1L1L3L5
4.1

Choose an integer threshold N2 beyond which step 3.1 applies, and choose 0<ϵ1/t so small that Nϵ2. Such an ϵ exists by [L7], [L8], and [L9].

step 3.1L7L8L9choose
5.1

If GCt has nN, then hom(G)n1/tnϵ; if 2n<N, an edge gives a two-vertex clique and a nonedge gives a two-vertex stable set, so hom(G)2nϵ; and if n=1, both sides equal 1. Thus ϵ is an Erdős–Hajnal constant for Ct.

step 3.1step 4.1L1L2L4algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 106 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources