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Theorem ddif 3775
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 3234 . . . . 5 𝑥 ∈ V
2 eldif 3617 . . . . 5 (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥𝐴))
31, 2mpbiran 973 . . . 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 265 . 2 ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥𝐴)
76difeqri 3763 1 (V ∖ (V ∖ 𝐴)) = 𝐴
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wa 383   = wceq 1523  wcel 2030  Vcvv 3231  cdif 3604
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-9 2039  ax-10 2059  ax-11 2074  ax-12 2087  ax-13 2282  ax-ext 2631
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1526  df-ex 1745  df-nf 1750  df-sb 1938  df-clab 2638  df-cleq 2644  df-clel 2647  df-nfc 2782  df-v 3233  df-dif 3610
This theorem is referenced by:  complss  3784  dfun3  3898  dfin3  3899  invdif  3901  ssindif0  4064  difdifdir  4089
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