CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

Relative consistency of a countable family of pairs without choice

Statement

Externally, $Con (Z F)$ implies the consistency of ZF plus a countable family of pairs with no choice function. Consequently ${A C}_{ω, 2}$ is not a theorem of ZF if ZF is consistent. The implication uses fixed finite-fragment model transfer; it does not claim a PA-verified uniform Jech–Sochor refutation transformer.

Facts & Assumptions

Given: A hypothetical finite contradiction proof from $Z F + T$ , where $T$ is the displayed socks sentence.

[F1]

Formal consistency of ZFC plus GCH relative to ZF gives $Con (Z F) \Rightarrow Con (Z F C + G C H)$ .

[F2]

Fixed finite-fragment verification for the Jech–Sochor socks transfer gives, for every externally fixed finite fragment of $Z F + T$ , a proof in a finite fragment of $Z F C + G C H$ that it has a model.

[F3]

Choice for pairs and countable finite choice identifies $T$ with failure of ${A C}_{ω, 2}$ .

Proof

1.1

A contradiction proof from $Z F + T$ uses only a finite target fragment $Δ$ . Fix $Δ$ externally. F2 supplies a proof in $Z F C + G C H$ that a set model of $Δ$ exists. The alleged contradiction proof and finite-model soundness give a proof that no such model exists. Hence target inconsistency implies inconsistency of $Z F C + G C H$ .

2.1

By F1, consistency of ZF implies consistency of $Z F C + G C H$ , so step 1.1 gives the external relative-consistency implication. F3 identifies $T$ as a countable family of pairs without a choice function. If ZF proved ${A C}_{ω, 2}$ , then $Z F + T$ would be inconsistent; therefore consistency of ZF prevents such a proof.

F1F3step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

Jech, The Axiom of Choice, Chapters 4–6 (standard reference, not scraped)