Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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.

AC supplies the dependent-choice instances used in vector-bundle constructions

Statement

In ZF, the Axiom of Choice implies the prescribed-initial-point form of the Axiom of Dependent Choice. Explicitly, if X is nonempty, R is an entire relation on X, and a∈X, then AC supplies a sequence x:N→X such that

x0=aandxnRxn+1

for every n∈N. Consequently, a vector-bundle theorem stated under AC may discharge a separately declared DC hypothesis without assuming another choice principle.

Facts & Assumptions

Given: ZF, AC, a nonempty set X, a relation R entire on X, and a prescribed point a∈X.

[F1]

AC says that every family of nonempty sets has a choice function (The Axiom of Choice).

[F2]

DC with prescribed initial point asks for a function x:N→X with x0=a and xnRxn+1 for every n (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain).

[F3]

Given a set X, a point a∈X, and a function s:X→X, recursion on N supplies a unique function x:N→X with x0=a and xn+1=s(xn) (The recursion theorem).

Proof

technique · direct
1.1F1givenconstruct

For each u∈X, let Su={v∈X:uRv}. Every Su is nonempty because R is entire. Apply [F1] to the set S={Su:u∈X} and let c be its choice function. Define s:X→X by s(u)=c(Su). This is well-defined even when two successor sets coincide, and uRs(u) for every u∈X.

2.1F3step 1.1

Apply [F3] to s and the prescribed a. It gives x:N→X with x0=a and xn+1=s(xn). Step 1.1 then gives xnRxn+1 for every n.

3.1F2step 1.1step 2.1∎

Since X, the entire relation R, and a were arbitrary, the sequence in step 2.1 satisfies exactly the prescribed-initial-point formulation in [F2]. Thus AC implies DC, and every later use of this lemma spends AC only in the simultaneous choice made in step 1.1.

Depends on

Dependency tree · 2 levels

Sources