TheoremStatement: Literature-sourcedProof: 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.

MA(aleph_0) and the implication from CH to MA

Statement

In ZFC, $M A (ℵ_{0})$ holds for every nonempty preorder, without ccc. Consequently CH implies MA, because under CH every infinite cardinal below the continuum is $ℵ_{0}$ .

Facts & Assumptions

Given: AC, a nonempty preorder, and a countable family of dense sets.

[F1]

Martin's Axiom at a cardinal and Martin's Axiom fixes the desired filter and the strict continuum range.

[F2]

Transfinite recursion constructs the descending sequence.

Proof

1.1

Enumerate the dense family as $D_{0}, D_{1}, \dots$ (repetitions allowed, and for an empty family choose any $p_{0}$ ). Choose recursively $p_{n + 1} \leq p_{n}$ in $D_{n}$ . The upward closure $G = {q : \exists n p_{n} \leq q}$ is directed and upward closed, and it meets each $D_{n}$ . No chain condition was used. AC provides the enumeration and choices.

F1F2

2.1

Under CH, $2^{ℵ_{0}} = ℵ_{1}$ . The only infinite cardinal strictly below the continuum is $ℵ_{0}$ , so the MA scheme of F1 is exactly the instance proved in step 1.1.

F1step 1.1∎

Depends on

Used by

Martin's Axiom implies CH False statement

Dependency tree · two levels

10 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

Karagila, Forcing & Symmetric Extensions, Exercise 7.2 and Theorem 1.14 (Rasiowa–Sikorski) (standard reference, not scraped)