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Theorem ddif 3422
 Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)
Assertion
Ref Expression
ddif

Proof of Theorem ddif
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 vex 2902 . . . . 5
2 eldif 3273 . . . . 5
31, 2mpbiran 885 . . . 4
43con2bii 323 . . 3
51biantrur 493 . . 3
64, 5bitr2i 242 . 2
76difeqri 3410 1
 Colors of variables: wff set class Syntax hints:   wn 3   wa 359   wceq 1649   wcel 1717  cvv 2899   cdif 3260 This theorem is referenced by:  dfun3  3522  dfin3  3523  invdif  3525  ssindif0  3624  difdifdir  3658 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1552  ax-5 1563  ax-17 1623  ax-9 1661  ax-8 1682  ax-6 1736  ax-7 1741  ax-11 1753  ax-12 1939  ax-ext 2368 This theorem depends on definitions:  df-bi 178  df-or 360  df-an 361  df-tru 1325  df-ex 1548  df-nf 1551  df-sb 1656  df-clab 2374  df-cleq 2380  df-clel 2383  df-nfc 2512  df-v 2901  df-dif 3266
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