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.

Rank characterizes hierarchy membership

Statement

In ZF, for every set $x$ and ordinal $α$ ,

$x \in V_{α} ⟺ rank (x) < α, x \subseteq V_{α} ⟺ rank (x) \leq α .$

Thus $rank (x)$ is the least $α$ with $x \subseteq V_{α}$ , and $rank (x) = α$ iff $x \in V_{α + 1} ∖ V_{α}$ .

Facts & Assumptions

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

[F1]

Now assume ZF, including Foundation. Membership on the universe is well-founded and setlike, so its ordinal rank is defined for every set. Write $rank (x) = \sup {rank (y) + 1 : y \in x} .$ The empty supremum is $0$ , so $rank (\emptyset) = 0$ . If $y \in x$ , then $rank (y) < rank (x)$ . This is the Foundation-dependent special case of relation rank. The earlier construction of $V_{α}$ did not require Foundation. Conventions and prerequisites: def-rank-of-a-well-founded-relation, thm-foundation-equivalent-to-hierarchy-exhaustion. (Membership rank under Foundation)

[F2]

In ZF without Foundation, every $V_{α}$ is transitive and $α \leq β$ implies $V_{α} \subseteq V_{β}$ . Also $V_{α} \cap Ord = α$ , and both $α$ and $V_{α}$ belong to $V_{α + 1} ∖ V_{α}$ . (Transitivity and growth of hierarchy stages)

Proof

1.1

Induct on $α$ for the membership equivalence, for all sets $x$ at once. At zero neither membership in the empty stage nor rank below zero holds. At a successor $β + 1$ , $x \in V_{β + 1}$ iff every $y \in x$ lies in $V_{β}$ , iff every $rank (y) < β$ , iff $\sup_{y \in x} (rank (y) + 1) \leq β$ , iff $rank (x) < β + 1$ . All equivalences include the empty supremum case.

F1F2

2.1

At a nonzero limit $λ$ , membership means membership in some $V_{β}$ with $β < λ$ . By induction this implies rank below $λ$ . Conversely if $ρ = rank (x) < λ$ , then $ρ + 1 < λ$ and induction puts $x$ in $V_{ρ + 1}$ , hence in $V_{λ}$ .

F1F2step 1.1

3.1

Now $x \subseteq V_{α}$ iff each $y \in x$ has rank below $α$ , iff the supremum of their successor ranks is at most $α$ . This proves the subset equivalence. Taking $α = rank (x)$ gives the least-stage assertion; applying membership at $α + 1$ and at $α$ gives the successor-shell assertion.

F1step 1.1step 2.1∎

Depends on

Used by

The universe is the class union of its stages Corollary
Minimum-rank selection and Collection Proposition
Ranks of ordinals and hierarchy stages Proposition
Hₖappa is a transitive subset of Vₖappa Theorem

Dependency tree · two levels

6 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 — 7.3 p.34; Weiss Exercise 33 p.100. (standard reference, not scraped)