TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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.

Rasiowa–Sikorski with its choice use exposed

Statement

In ZFC, given a nonempty forcing preorder P, a sequence $(D_{n})_{n < ω}$ of dense subsets and $p \in P$ , a filter G contains p and meets every $D_{n}$ . If a surjection $e : ω \to P$ is supplied, the conclusion has a ZF proof without AC.

Facts & Assumptions

Given: General branch assumes AC for the omega cross P refinement family; supplied-enumeration branch is ZF. Explicit descending sequence and upward closure verify nonemptiness, direction and all dense-set meetings.

[F1]

Dense open sets and generic filters over a model: Density supplies refinements; filters are upward closed and internally downward directed.

[F2]

The Axiom of Choice: AC selects an element from each set in a set-indexed family of nonempty sets.

[F3]

Transfinite recursion: The definable-rule recursion schema applies on omega and uses no Choice.

Proof

1.1F1F2F3construct

For $(n, q) \in ω \times P$ , let $E_{n, q} = {r \in D_{n} : r \leq q}$ . These sets are nonempty by density. AC selects $h (n, q) \in E_{n, q}$ ; this is the sole AC use. Recursively set $p_{0} = p$ and $p_{n + 1} = h (n, p_{n})$ . Thus $p_{n + 1} \leq p_{n}$ and $p_{n + 1} \in D_{n}$ .

2.1F1step 1.1

Define $G = {q \in P : \exists n (p_{n} \leq q)}$ . It contains p and is upward closed. If q,r have witnesses n,m, then $p_{\max (n, m)} \in G$ strengthens both. Hence G is a filter and $p_{n + 1} \in G \cap D_{n}$ for every n.

3.1F1F3step 1.1step 2.1∎

If e is supplied, instead define h(n,q) as $e (k)$ for the least k with $e (k) \in D_{n}$ and $e (k) \leq q$ . Such k exists by density and surjectivity, and is unique by leastness. This rule and the recursion and verification above require only ZF.

Depends on

Used by

Strong compactness and the product-measure extension interface Theorem

Dependency tree · two levels

9 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 Theorem 1.14 and Corollary 1.15 p4; Marks Lemma 24.6 p99 (standard reference, not scraped)