Metamath Proof Explorer


Theorem ac5

Description: An Axiom of Choice equivalent: there exists a function f (called a choice function) with domain A that maps each nonempty member of the domain to an element of that member. Axiom AC of BellMachover p. 488. Note that the assertion that f be a function is not necessary; see ac4 . (Contributed by NM, 29-Aug-1999)

Ref Expression
Hypothesis ac5.1 𝐴 ∈ V
Assertion ac5 𝑓 ( 𝑓 Fn 𝐴 ∧ ∀ 𝑥𝐴 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) )

Proof

Step Hyp Ref Expression
1 ac5.1 𝐴 ∈ V
2 fneq2 ( 𝑦 = 𝐴 → ( 𝑓 Fn 𝑦𝑓 Fn 𝐴 ) )
3 raleq ( 𝑦 = 𝐴 → ( ∀ 𝑥𝑦 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) ↔ ∀ 𝑥𝐴 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) ) )
4 2 3 anbi12d ( 𝑦 = 𝐴 → ( ( 𝑓 Fn 𝑦 ∧ ∀ 𝑥𝑦 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) ) ↔ ( 𝑓 Fn 𝐴 ∧ ∀ 𝑥𝐴 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) ) ) )
5 4 exbidv ( 𝑦 = 𝐴 → ( ∃ 𝑓 ( 𝑓 Fn 𝑦 ∧ ∀ 𝑥𝑦 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) ) ↔ ∃ 𝑓 ( 𝑓 Fn 𝐴 ∧ ∀ 𝑥𝐴 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) ) ) )
6 dfac4 ( CHOICE ↔ ∀ 𝑦𝑓 ( 𝑓 Fn 𝑦 ∧ ∀ 𝑥𝑦 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) ) )
7 6 axaci 𝑓 ( 𝑓 Fn 𝑦 ∧ ∀ 𝑥𝑦 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) )
8 1 5 7 vtocl 𝑓 ( 𝑓 Fn 𝐴 ∧ ∀ 𝑥𝐴 ( 𝑥 ≠ ∅ → ( 𝑓𝑥 ) ∈ 𝑥 ) )