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Theorem ddif 3591
 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 3075 . . . . 5
2 eldif 3441 . . . . 5
31, 2mpbiran 909 . . . 4
43con2bii 332 . . 3
51biantrur 506 . . 3
64, 5bitr2i 250 . 2
76difeqri 3579 1
 Colors of variables: wff setvar class Syntax hints:   wn 3   wa 369   wceq 1370   wcel 1758  cvv 3072   cdif 3428 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1592  ax-4 1603  ax-5 1671  ax-6 1710  ax-7 1730  ax-10 1777  ax-11 1782  ax-12 1794  ax-13 1954  ax-ext 2431 This theorem depends on definitions:  df-bi 185  df-an 371  df-tru 1373  df-ex 1588  df-nf 1591  df-sb 1703  df-clab 2438  df-cleq 2444  df-clel 2447  df-nfc 2602  df-v 3074  df-dif 3434 This theorem is referenced by:  dfun3  3691  dfin3  3692  invdif  3694  ssindif0  3835  difdifdir  3869
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