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 ω⪯X and CH(X), then

X⊔X≈X×X≈X,P(X)×P(X)≈P(X).

Facts & Assumptions

[F1]

Dedekind infinitude is equivalent to a countable subset: An injected omega gives a bijection X⊔1≈X.

[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 ω⪯X, 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.

1.1F1

Fix a∈X. The two copies of X inject into P(X⊔{∗}) by x↦{x} and x↦{x,∗}. A bijection X⊔1≈X transports this to X⊔X⪯P(X). Also X⪯X⊔X.

2.1F2F3step 1.1

Local GCH makes X⊔X 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 X⊔X≈X.

3.1step 2.1

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

4.1F2F3F4step 3.1∎

Apply local GCH to X⪯X2⪯P(X). The power-set endpoint would inject P(X) into length-two sequences, again impossible; the remaining endpoint is X2≈X. Together with the product-of-powersets bijection this proves all assertions.

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