Metamath Proof Explorer


Theorem ac6s

Description: Equivalent of Axiom of Choice. Using the Boundedness Axiom bnd2 , we derive this strong version of ac6 that doesn't require B to be a set. (Contributed by NM, 4-Feb-2004)

Ref Expression
Hypotheses ac6s.1 𝐴 ∈ V
ac6s.2 ( 𝑦 = ( 𝑓𝑥 ) → ( 𝜑𝜓 ) )
Assertion ac6s ( ∀ 𝑥𝐴𝑦𝐵 𝜑 → ∃ 𝑓 ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜓 ) )

Proof

Step Hyp Ref Expression
1 ac6s.1 𝐴 ∈ V
2 ac6s.2 ( 𝑦 = ( 𝑓𝑥 ) → ( 𝜑𝜓 ) )
3 1 bnd2 ( ∀ 𝑥𝐴𝑦𝐵 𝜑 → ∃ 𝑧 ( 𝑧𝐵 ∧ ∀ 𝑥𝐴𝑦𝑧 𝜑 ) )
4 vex 𝑧 ∈ V
5 1 4 2 ac6 ( ∀ 𝑥𝐴𝑦𝑧 𝜑 → ∃ 𝑓 ( 𝑓 : 𝐴𝑧 ∧ ∀ 𝑥𝐴 𝜓 ) )
6 5 anim2i ( ( 𝑧𝐵 ∧ ∀ 𝑥𝐴𝑦𝑧 𝜑 ) → ( 𝑧𝐵 ∧ ∃ 𝑓 ( 𝑓 : 𝐴𝑧 ∧ ∀ 𝑥𝐴 𝜓 ) ) )
7 6 eximi ( ∃ 𝑧 ( 𝑧𝐵 ∧ ∀ 𝑥𝐴𝑦𝑧 𝜑 ) → ∃ 𝑧 ( 𝑧𝐵 ∧ ∃ 𝑓 ( 𝑓 : 𝐴𝑧 ∧ ∀ 𝑥𝐴 𝜓 ) ) )
8 fss ( ( 𝑓 : 𝐴𝑧𝑧𝐵 ) → 𝑓 : 𝐴𝐵 )
9 8 expcom ( 𝑧𝐵 → ( 𝑓 : 𝐴𝑧𝑓 : 𝐴𝐵 ) )
10 9 anim1d ( 𝑧𝐵 → ( ( 𝑓 : 𝐴𝑧 ∧ ∀ 𝑥𝐴 𝜓 ) → ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜓 ) ) )
11 10 eximdv ( 𝑧𝐵 → ( ∃ 𝑓 ( 𝑓 : 𝐴𝑧 ∧ ∀ 𝑥𝐴 𝜓 ) → ∃ 𝑓 ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜓 ) ) )
12 11 imp ( ( 𝑧𝐵 ∧ ∃ 𝑓 ( 𝑓 : 𝐴𝑧 ∧ ∀ 𝑥𝐴 𝜓 ) ) → ∃ 𝑓 ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜓 ) )
13 12 exlimiv ( ∃ 𝑧 ( 𝑧𝐵 ∧ ∃ 𝑓 ( 𝑓 : 𝐴𝑧 ∧ ∀ 𝑥𝐴 𝜓 ) ) → ∃ 𝑓 ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜓 ) )
14 3 7 13 3syl ( ∀ 𝑥𝐴𝑦𝐵 𝜑 → ∃ 𝑓 ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜓 ) )