ExampleConstruction: Literature-sourcedVerification: 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.

Where countable-union proofs spend choice

Example

For an omega family $(A_{n})$ of at most countable sets, a supplied family of enumerations of the nonempty $A_{n}$ suffices in ZF to enumerate $⋃_{n} A_{n}$ . Countable choice supplies those enumerations if only individual countability is given. In ZF plus countable choice, the real line cannot be a countable union of countable sets.

Facts & Assumptions

[F1]

$N \times N \approx N$ : Omega squared is explicitly countable.

[F2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): One member of each omega-indexed nonempty set can be selected.

[F3]

Countable unions of at most countable sets, assuming ${A C}_{ω}$ : Under countable choice the union is at most countable.

[F4]

$R$ is uncountable (Cantor's nested intervals, 1874): The real line is not at most countable.

Verification

Given: The objects and hypotheses in the statement.

1.1

If all $A_{n}$ are empty, the union is empty. Otherwise let $I = {n : A_{n} \neq \emptyset}$ . Given surjections $e_{n} : ω \to A_{n}$ for $n \in I$ , enumerate $(n, k) \in I \times ω$ by a fixed pairing enumeration and output $e_{n} (k)$ , skipping other pairs. This lists every union member; first occurrences give an injection of the union into omega.

2.1

Without the supplied $e_{n}$ , the set of surjections onto each nonempty $A_{n}$ is nonempty. For empty $A_{n}$ use a singleton dummy set instead. Countable choice selects these data simultaneously. The pairing operation itself costs no choice.

F2step 1.1

3.1

If the real line were such a countable union under countable choice, the union theorem would make it at most countable, contradicting its uncountability.

F3F4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

45 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, §2.4.2 and Corollary 1, pp.20–21 (standard reference, not scraped)