Metamath Proof Explorer


Theorem ac6mapd

Description: Axiom of choice equivalent, deduction form. (Contributed by Thierry Arnoux, 13-Oct-2025)

Ref Expression
Hypotheses ac6mapd.1 ( 𝑦 = ( 𝑓𝑥 ) → ( 𝜓𝜒 ) )
ac6mapd.2 ( 𝜑𝐴𝑉 )
ac6mapd.3 ( 𝜑𝐵𝑊 )
ac6mapd.4 ( ( 𝜑𝑥𝐴 ) → ∃ 𝑦𝐵 𝜓 )
Assertion ac6mapd ( 𝜑 → ∃ 𝑓 ∈ ( 𝐵m 𝐴 ) ∀ 𝑥𝐴 𝜒 )

Proof

Step Hyp Ref Expression
1 ac6mapd.1 ( 𝑦 = ( 𝑓𝑥 ) → ( 𝜓𝜒 ) )
2 ac6mapd.2 ( 𝜑𝐴𝑉 )
3 ac6mapd.3 ( 𝜑𝐵𝑊 )
4 ac6mapd.4 ( ( 𝜑𝑥𝐴 ) → ∃ 𝑦𝐵 𝜓 )
5 4 ralrimiva ( 𝜑 → ∀ 𝑥𝐴𝑦𝐵 𝜓 )
6 1 ac6sg ( 𝐴𝑉 → ( ∀ 𝑥𝐴𝑦𝐵 𝜓 → ∃ 𝑓 ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜒 ) ) )
7 2 5 6 sylc ( 𝜑 → ∃ 𝑓 ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜒 ) )
8 3 2 elmapd ( 𝜑 → ( 𝑓 ∈ ( 𝐵m 𝐴 ) ↔ 𝑓 : 𝐴𝐵 ) )
9 8 biimprd ( 𝜑 → ( 𝑓 : 𝐴𝐵𝑓 ∈ ( 𝐵m 𝐴 ) ) )
10 9 anim1d ( 𝜑 → ( ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜒 ) → ( 𝑓 ∈ ( 𝐵m 𝐴 ) ∧ ∀ 𝑥𝐴 𝜒 ) ) )
11 10 eximdv ( 𝜑 → ( ∃ 𝑓 ( 𝑓 : 𝐴𝐵 ∧ ∀ 𝑥𝐴 𝜒 ) → ∃ 𝑓 ( 𝑓 ∈ ( 𝐵m 𝐴 ) ∧ ∀ 𝑥𝐴 𝜒 ) ) )
12 7 11 mpd ( 𝜑 → ∃ 𝑓 ( 𝑓 ∈ ( 𝐵m 𝐴 ) ∧ ∀ 𝑥𝐴 𝜒 ) )
13 df-rex ( ∃ 𝑓 ∈ ( 𝐵m 𝐴 ) ∀ 𝑥𝐴 𝜒 ↔ ∃ 𝑓 ( 𝑓 ∈ ( 𝐵m 𝐴 ) ∧ ∀ 𝑥𝐴 𝜒 ) )
14 12 13 sylibr ( 𝜑 → ∃ 𝑓 ∈ ( 𝐵m 𝐴 ) ∀ 𝑥𝐴 𝜒 )