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Theorem ddif 3635
 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 3112 . . . . 5
2 eldif 3485 . . . . 5
31, 2mpbiran 918 . . . 4
43con2bii 332 . . 3
51biantrur 506 . . 3
64, 5bitr2i 250 . 2
76difeqri 3623 1
 Colors of variables: wff setvar class Syntax hints:  -.wn 3  /\wa 369  =wceq 1395  e.wcel 1818   cvv 3109  \cdif 3472 This theorem is referenced by:  dfun3  3735  dfin3  3736  invdif  3738  ssindif0  3880  difdifdir  3915 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1618  ax-4 1631  ax-5 1704  ax-6 1747  ax-7 1790  ax-10 1837  ax-11 1842  ax-12 1854  ax-13 1999  ax-ext 2435 This theorem depends on definitions:  df-bi 185  df-an 371  df-tru 1398  df-ex 1613  df-nf 1617  df-sb 1740  df-clab 2443  df-cleq 2449  df-clel 2452  df-nfc 2607  df-v 3111  df-dif 3478
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