Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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.

The first constructible levels

Example

In ZF,

L0=∅,L1={∅},L2={∅,{∅}},L3=P(L2).

For every n<ω, Ln=Vn, and therefore Lω=Vω.

Facts & Assumptions

Given: ZF. Calculated the first four levels and explicitly defined every subset of a finite carrier; finite induction and the omega union give the promised V-level identities.

[F1]

The constructible hierarchy and constructible rank: The L hierarchy starts from empty, uses Def at successors, and takes unions at nonzero limits.

[F2]

The cumulative hierarchy: The V hierarchy starts from empty, uses power sets at successors, and takes unions at nonzero limits.

[F3]

Definable subsets of a membership structure: Def includes each subset definable with finitely many parameters and has the designated empty case.

Verification

1.1F3

For a finite set A every subset B is finite. If B={b1,…,bk} is nonempty, the formula x=b1∨⋯∨x=bk defines it over A using those parameters. The formula x≠x defines the empty subset when A is nonempty; the designated clause treats A empty. Hence Def⁡(A)=P(A) for finite A.

2.1F1F2step 1.1∎

Starting with empty gives L1={∅} and L2={∅,{∅}}. The four elements of L3 are ∅, {∅}, {{∅}} and {∅,{∅}}. Induction gives finiteness at each stage (a set with k elements has subsets encoded by k binary choices) and Ln+1=P(Ln)=P(Vn)=Vn+1. The zero stages coincide. Taking the unions over n yields Lω=Vω.

Depends on

Used by

Dependency tree · two levels

7 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