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Theorem acneq 9965
Description: Equality theorem for the choice set function. (Contributed by Mario Carneiro, 31-Aug-2015.)
Assertion
Ref Expression
acneq (𝐴 = 𝐶AC 𝐴 = AC 𝐶)

Proof of Theorem acneq
Dummy variables 𝑓 𝑔 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eleq1 2824 . . . 4 (𝐴 = 𝐶 → (𝐴 ∈ V ↔ 𝐶 ∈ V))
2 oveq2 7375 . . . . 5 (𝐴 = 𝐶 → ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴) = ((𝒫 𝑥 ∖ {∅}) ↑m 𝐶))
3 raleq 3292 . . . . . 6 (𝐴 = 𝐶 → (∀𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦) ↔ ∀𝑦𝐶 (𝑔𝑦) ∈ (𝑓𝑦)))
43exbidv 1923 . . . . 5 (𝐴 = 𝐶 → (∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦) ↔ ∃𝑔𝑦𝐶 (𝑔𝑦) ∈ (𝑓𝑦)))
52, 4raleqbidv 3311 . . . 4 (𝐴 = 𝐶 → (∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦) ↔ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐶)∃𝑔𝑦𝐶 (𝑔𝑦) ∈ (𝑓𝑦)))
61, 5anbi12d 633 . . 3 (𝐴 = 𝐶 → ((𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦)) ↔ (𝐶 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐶)∃𝑔𝑦𝐶 (𝑔𝑦) ∈ (𝑓𝑦))))
76abbidv 2802 . 2 (𝐴 = 𝐶 → {𝑥 ∣ (𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦))} = {𝑥 ∣ (𝐶 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐶)∃𝑔𝑦𝐶 (𝑔𝑦) ∈ (𝑓𝑦))})
8 df-acn 9866 . 2 AC 𝐴 = {𝑥 ∣ (𝐴 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐴)∃𝑔𝑦𝐴 (𝑔𝑦) ∈ (𝑓𝑦))}
9 df-acn 9866 . 2 AC 𝐶 = {𝑥 ∣ (𝐶 ∈ V ∧ ∀𝑓 ∈ ((𝒫 𝑥 ∖ {∅}) ↑m 𝐶)∃𝑔𝑦𝐶 (𝑔𝑦) ∈ (𝑓𝑦))}
107, 8, 93eqtr4g 2796 1 (𝐴 = 𝐶AC 𝐴 = AC 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1542  wex 1781  wcel 2114  {cab 2714  wral 3051  Vcvv 3429  cdif 3886  c0 4273  𝒫 cpw 4541  {csn 4567  cfv 6498  (class class class)co 7367  m cmap 8773  AC wacn 9862
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-ral 3052  df-rex 3062  df-rab 3390  df-v 3431  df-dif 3892  df-un 3894  df-ss 3906  df-nul 4274  df-if 4467  df-sn 4568  df-pr 4570  df-op 4574  df-uni 4851  df-br 5086  df-iota 6454  df-fv 6506  df-ov 7370  df-acn 9866
This theorem is referenced by:  acndom  9973  dfacacn  10064  dfac13  10065
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