TheoremStatement: AI-adaptedProof: AI-generatedPipeline-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.

Induction on well-founded setlike relations

Statement

Let $R$ be well-founded and setlike on a definable class $X$ . If a definable property $P$ is progressive, meaning that for every $x \in X$ , $[\forall y R x P (y)] \Rightarrow P (x)$ , then $P (x)$ holds for all $x \in X$ . Set parameters in $P$ are allowed. This holds without Foundation.

Facts & Assumptions

Given: Work in ZF unless the statement explicitly weakens or supplements it; fix the objects and hypotheses of the statement.

[F1]

For every setlike relation $R$ on a definable class $X$ and $x \in X$ , there is a least predecessor-closed set $C (x) \subseteq X$ containing $x$ . It consists exactly of nodes reachable from $x$ by a finite sequence of predecessor steps. Well-foundedness is not needed. (Finite predecessor closures are sets)

Proof

1.1F1given

If there is a counterexample $x$ , take its predecessor-closed set $C (x)$ and separate the nonempty set $B = {y \in C (x) : \neg P (y)}$ . Well-foundedness gives a minimal $b \in B$ .

2.1step 1.1given∎

Every predecessor of $b$ belongs to $C (x)$ , and none belongs to $B$ by minimality. Hence all satisfy $P$ . Progressiveness gives $P (b)$ , contrary to $b \in B$ . Therefore no counterexample exists.

Depends on

Finite predecessor closures are sets

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

Marks, Set Theory, Berkeley edition — Lemma 6.2 and Theorem 6.5 p.30. (standard reference, not scraped)