Metamath Proof Explorer


Theorem acnrcl

Description: Reverse closure for the choice set predicate. (Contributed by Mario Carneiro, 31-Aug-2015)

Ref Expression
Assertion acnrcl ( 𝑋AC 𝐴𝐴 ∈ V )

Proof

Step Hyp Ref Expression
1 ne0i ( 𝑋 ∈ { 𝑥 ∣ ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) } → { 𝑥 ∣ ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) } ≠ ∅ )
2 abn0 ( { 𝑥 ∣ ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) } ≠ ∅ ↔ ∃ 𝑥 ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) )
3 simpl ( ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) → 𝐴 ∈ V )
4 3 exlimiv ( ∃ 𝑥 ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) → 𝐴 ∈ V )
5 2 4 sylbi ( { 𝑥 ∣ ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) } ≠ ∅ → 𝐴 ∈ V )
6 1 5 syl ( 𝑋 ∈ { 𝑥 ∣ ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) } → 𝐴 ∈ V )
7 df-acn AC 𝐴 = { 𝑥 ∣ ( 𝐴 ∈ V ∧ ∀ 𝑓 ∈ ( ( 𝒫 𝑥 ∖ { ∅ } ) ↑m 𝐴 ) ∃ 𝑔𝑦𝐴 ( 𝑔𝑦 ) ∈ ( 𝑓𝑦 ) ) }
8 6 7 eleq2s ( 𝑋AC 𝐴𝐴 ∈ V )