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Theorem acnrcl 10025
Description: Reverse closure for the choice set predicate. (Contributed by Mario Carneiro, 31-Aug-2015.)
Assertion
Ref Expression
acnrcl (𝑋AC 𝐴𝐴 ∈ V)

Proof of Theorem acnrcl
Dummy variables 𝑓 𝑔 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 ne0i 4302 . . 3 (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 4348 . . . 4 ({𝑥 ∣ (𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦))} ≠ ∅ ↔ ∃𝑥(𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦)))
3 simpl 487 . . . . 5 ((𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦)) → 𝐴 ∈ V)
43exlimiv 1957 . . . 4 (∃𝑥(𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦)) → 𝐴 ∈ V)
52, 4sylbi 220 . . 3 ({𝑥 ∣ (𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦))} ≠ ∅ → 𝐴 ∈ V)
61, 5syl 18 . 2 (𝑋 ∈ {𝑥 ∣ (𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦))} → 𝐴 ∈ V)
7 df-acn 9927 . 2 AC 𝐴 = {𝑥 ∣ (𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦))}
86, 7eleq2s 2887 1 (𝑋AC 𝐴𝐴 ∈ V)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 400  wex 1806  wcel 2149  {cab 2747  wne 2964  wral 3085  Vcvv 3463  cdif 3910  c0 4294  𝒫 cpw 4567  {csn 4594  cfv 6537  (class class class)co 7411  m cmap 8823  AC wacn 9923
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-10 2182  ax-11 2198  ax-12 2219  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1570  df-fal 1580  df-ex 1807  df-nf 1811  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ne 2965  df-dif 3916  df-nul 4295  df-acn 9927
This theorem is referenced by:  acni  10028  acni2  10029  acndom2  10037  fodomacn  10039  iundom2g  10523
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