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Theorem ddif 3222
 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
StepHypRef Expression
1 vex 2730 . . . . 5
2 eldif 3088 . . . . 5
31, 2mpbiran 889 . . . 4
43con2bii 324 . . 3
51biantrur 494 . . 3
64, 5bitr2i 243 . 2
76difeqri 3213 1
 Colors of variables: wff set class Syntax hints:   wn 5   wa 360   wceq 1619   wcel 1621  cvv 2727   cdif 3075 This theorem is referenced by:  dfun3  3314  dfin3  3315  invdif  3317  ssindif0  3415  difdifdir  3447 This theorem was proved from axioms:  ax-1 7  ax-2 8  ax-3 9  ax-mp 10  ax-5 1533  ax-6 1534  ax-7 1535  ax-gen 1536  ax-8 1623  ax-11 1624  ax-17 1628  ax-12o 1664  ax-10 1678  ax-9 1684  ax-4 1692  ax-16 1926  ax-ext 2234 This theorem depends on definitions:  df-bi 179  df-or 361  df-an 362  df-tru 1315  df-ex 1538  df-nf 1540  df-sb 1883  df-clab 2240  df-cleq 2246  df-clel 2249  df-nfc 2374  df-v 2729  df-dif 3081
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