TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)

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.

Check-name evaluation and reconstruction of G

Statement

In ZF, for every nonempty $G \subseteq P$ , ${val}_{G} (\overset{ˇ}{x}) = x$ and ${val}_{G} (\dot{G}) = G$ . Hence for a transitive ZF model M containing P, $M \subseteq M [G]$ and $G \in M [G]$ . The inclusion is not asserted to be elementary.

Facts & Assumptions

Given: ZF; G nonempty. Direct valuation calculations establish check recovery and dot G recovery, then ground membership of the names gives M subset M[G] and G in M[G].

[F1]

Check names without a largest condition: Check names have every condition as a coefficient; dot G uses the name of p with coefficient p, and these names belong to the ground model.

Proof

1.1

By membership induction suppose the assertion holds for each $y \in x$ . The valuation equation for check x gives exactly ${{val}_{G} (\overset{ˇ}{y}) : y \in x, \exists p \in G} = {y : y \in x} = x$ . The nonemptiness of G supplies the existential coefficient for every y. For x empty both sides are empty.

F1given

2.1

Apply step 1.1 to every p in P. The valuation equation gives ${val}_{G} (\dot{G}) = {{val}_{G} (\overset{ˇ}{p}) : p \in G} = G$ . For each $x \in M$ , F1 puts check x in M, so step 1.1 puts x in M[G]. F1 also puts dot G in M, so its value G lies in M[G]. No genericity or directedness was needed for these identities.

F1step 1.1∎

Depends on

Check names without a largest condition

Used by

Dependency tree · two levels

3 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 Exercises 2.6–2.7 and Proposition 2.9 pp6–7; Marks Lemma 24.3 p98 (standard reference, not scraped)