Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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.

Erdős–Rado for arbitrary infinite cardinals and finite arity

Statement

In ZFC, for every infinite cardinal κ and every n<ω,

n(κ)+(κ+)κn+1.

Here + is cardinal successor and 0(κ)=κ. In particular the theorem includes n=0.

Facts & Assumptions

Given: An infinite cardinal κ; assume AC.

[F1]

Relative beths satisfy 0(κ)=κ and m+1(κ)=2m(κ). Finite beth iteration above an infinite cardinal

[F2]

Pattern closure for μ, 1rμ, positive finite d and domain (2μ)+ gives distinct xα (α<μ+) and an outside point a whose patterns on earlier nodes agree with those of each xα. Pattern closure yields an end-homogeneous sequence

[F4]

Natural-number induction proves the assertion from zero and successor cases. The principle of mathematical induction

[F5]

The arrow means every coloring has a homogeneous subset of the target cardinality. Partition arrows and homogeneous sets

[F6]

Cantor's theorem gives ρ<2ρ for every infinite cardinal ρ. Cantor's theorem: AP(A)

[A1]

Proof

1.1

For n=0, let c:[κ+]1κ. If every color fiber had size less than κ+, each would have size at most κ by the successor-cardinal definition. AC chooses an injection of each fiber into κ, giving an injection of their union into κ×κ by recording the color and its fiber index. F3 bounds the union by κ, contrary to its size κ+. Therefore some fiber has size κ+ and is homogeneous. By F1 this is exactly the required zero case.

F1F3F5A1given
2.1

Assume the assertion at m0, and let F:[m+1(κ)+]m+2κ. Put μ=m(κ). Repeatedly applying F6 to the recurrence F1 gives μκ and μ infinite. Thus F2 applies with r=κ, d=m+11 and λ=(2μ)+=m+1(κ)+. Obtain distinct nodes xα for α<μ+ and a outside their range.

F1F2F6A1step 1.1
3.1

Define G:[μ+]m+1κ by G(v)=F({xβ:βv}{a}). The argument has size m+2 because the xβ are distinct and a is outside their range, so G is well-defined. The induction assertion at m, with μ=m(κ), gives Hμ+ of size κ+ homogeneous for G, of some color i<κ.

F1F5step 2.1
4.1

Put Y={xα:αH}; injectivity gives Y=κ+. For any m+2 nodes of Y, order their indices as α0<<αm+1 and put u={xα0,,xαm}. These are m+1 previous nodes at stage αm+1, so F2 gives F(u{xαm+1})=F(u{a})=G({α0,,αm})=i. Hence Y is homogeneous. Only the largest index was used; no ambient increase of the xα was assumed. This proves the successor step; F4 and step 1.1 prove all finite n.

F2F4F5step 1.1step 2.1step 3.1

Depends on

Used by

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources