Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge 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 no free ultrafilters on any set over ZF

Statement

If ZF is consistent, then ZF is consistent with the assertion that every ultrafilter on every set is principal.

Facts & Assumptions

Given: Con(ZF) for the fixed formal theories. This is a syntactic consistency hypothesis, not a set-model or transitive-model hypothesis.

[F1]

The Blass ultrafilter-free construction is finitely formalizable proves for every externally fixed finite target fragment that a suitable finite ZFC source proves the existence of a set model of that fragment.

[F2]

Formal consistency of ZFC plus GCH relative to ZF proves Con(ZF)Con(ZFC+GCH) by a verified proof translation and does not assume a transitive set model.

Proof

technique · contradiction from the finite support of a formal refutation
1.1

By F2, the hypothesis gives Con(ZFC+GCH). Suppose for contradiction that the target theory T=ZF+"every ultrafilter on every set is principal" is inconsistent. One formal refutation is a finite sequence and therefore uses only a finite list Δ of ZF axiom instances together with the displayed extra sentence.

F2assume-contra
2.1

Apply F1 to this exact external Δ. The finite ZFC source isolated there, and hence ZFC+GCH, proves that a set structure satisfies every sentence used in the alleged refutation. The fixed first-order soundness induction for that finite derivation would then make ZFC+GCH prove that the structure satisfies a contradiction; equality logic proves that no structure does. This contradicts step 1.1.

F1step 1.1discharge-contradiction
3.1

Consequently T is consistent. The empty-proof and zero-axiom cases cannot be refutations because no last contradiction line is present; a one-line alleged refutation is covered by the same soundness check. The argument uses only the finite support of one hypothetical proof. It invokes neither semantic completeness nor a countable transitive model of full ZF, and it concludes only conditional syntactic consistency.

step 1.1step 2.1discharge-contradiction: step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources