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Theorem ddif 3704
 Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)
Assertion
Ref Expression
ddif (V ∖ (V ∖ 𝐴)) = 𝐴

Proof of Theorem ddif
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 vex 3176 . . . . 5 𝑥 ∈ V
2 eldif 3550 . . . . 5 (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥𝐴))
31, 2mpbiran 955 . . . 4 (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥𝐴)
43con2bii 346 . . 3 (𝑥𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴))
51biantrur 526 . . 3 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)))
64, 5bitr2i 264 . 2 ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥𝐴)
76difeqri 3692 1 (V ∖ (V ∖ 𝐴)) = 𝐴
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ∧ wa 383   = wceq 1475   ∈ wcel 1977  Vcvv 3173   ∖ cdif 3537 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-dif 3543 This theorem is referenced by:  complss  3713  dfun3  3824  dfin3  3825  invdif  3827  ssindif0  3983  difdifdir  4008
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