Alphabeta Math
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.

Transitivity and a valuation rank bound

Statement

In ZF, for a transitive ZF ground model M and any GP, M[G] is transitive. For each name tau, rank(valG(τ))rkP(τ). No axiom satisfaction or no-new-ordinals theorem is asserted here.

Facts & Assumptions

Given: ZF; arbitrary G subset P. Subname decoding in transitive M proves transitivity, and the two recursive supremum formulas prove the rank bound even for empty G.

[F1]

Valuation of names and M[G]: Valuation selects values of subnames, and M[G] consists of values of names in M.

[F2]

Forcing names and their rank: Every descendant of a name is a name, and name rank is the supremum of predecessor name-ranks plus one.

[F3]

Membership rank under Foundation: Membership rank is the supremum of the ranks of members plus one.

Proof

1.1

If xyM[G], write y=valG(τ) for a name τM. By F1 some σ,pτ has pG and x=valG(σ). Transitivity of M, applied through the Kuratowski pair, gives σM; F2 says that sigma is a name. Thus xM[G], proving transitivity.

F1F2
2.1

Induct on the subname relation. Every member of the valuation of tau is the valuation of some sigma below tau, so F3 gives its rank as the supremum of their value-ranks plus one. By induction this is at most the supremum of rkP(σ)+1 over all subnames sigma, which F2 identifies with rkP(τ). With no selected subnames the value rank is zero and the inequality still holds. This includes both the empty name and empty G.

F1F2F3step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources