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 is nonempty, is an entire relation on , and , then AC supplies a sequence such that
for every . 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 , a relation entire on , and a prescribed point .
AC says that every family of nonempty sets has a choice function (The Axiom of Choice).
DC with prescribed initial point asks for a function with and for every (The axiom of dependent choice: a relation in which every element is related to something admits an -indexed chain).
Given a set , a point , and a function , recursion on supplies a unique function with and (The recursion theorem).
Proof
For each , let . Every is nonempty because is entire. Apply [F1] to the set and let be its choice function. Define by . This is well-defined even when two successor sets coincide, and for every .
Apply [F3] to and the prescribed . It gives with and . Step 1.1 then gives for every .
Since , the entire relation , and 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.