Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Local GCH absorbs sums and squares

Statement

In ZF, if ωX and CH(X), then

XXX×XX,P(X)×P(X)P(X).

Facts & Assumptions

[F1]

Dedekind infinitude is equivalent to a countable subset: An injected omega gives a bijection X1X.

[F2]

Local GCH for arbitrary sets: An intermediate size between X and its power set equals one endpoint under local GCH.

[F3]

No injection of a power set into finite sequences: For ωX, its power set does not inject into finite sequences.

[F4]

The Schröder-Bernstein theorem: Opposite injections yield a bijection in ZF.

Proof

Given: The objects and hypotheses in the statement.

1.1

Fix aX. The two copies of X inject into P(X{}) by x{x} and x{x,}. A bijection X1X transports this to XXP(X). Also XXX.

F1
2.1

Local GCH makes XX equinumerous either with X or with P(X). The latter would inject the power set into finite sequences: use (x) for the first copy and (a,x) for the second. This is impossible. Thus XXX.

F2F3step 1.1
3.1

The map taking a subset of a tagged disjoint union to its two component subsets is a bijection P(XX)P(X)2. Transport along the previous bijection gives P(X)2P(X). Singleton coordinates inject X2 into this product, while x(x,a) injects X into X2.

step 2.1
4.1

Apply local GCH to XX2P(X). The power-set endpoint would inject P(X) into length-two sequences, again impossible; the remaining endpoint is X2X. Together with the product-of-powersets bijection this proves all assertions.

F2F3F4step 3.1

Depends on

Used by

Dependency tree · two levels

14 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