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Theorem ddif 3309
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 2792 . . . . 5  |-  x  e. 
_V
2 eldif 3163 . . . . 5  |-  ( x  e.  ( _V  \  A )  <->  ( x  e.  _V  /\  -.  x  e.  A ) )
31, 2mpbiran 884 . . . 4  |-  ( x  e.  ( _V  \  A )  <->  -.  x  e.  A )
43con2bii 322 . . 3  |-  ( x  e.  A  <->  -.  x  e.  ( _V  \  A
) )
51biantrur 492 . . 3  |-  ( -.  x  e.  ( _V 
\  A )  <->  ( x  e.  _V  /\  -.  x  e.  ( _V  \  A
) ) )
64, 5bitr2i 241 . 2  |-  ( ( x  e.  _V  /\  -.  x  e.  ( _V  \  A ) )  <-> 
x  e.  A )
76difeqri 3297 1  |-  ( _V 
\  ( _V  \  A ) )  =  A
Colors of variables: wff set class
Syntax hints:   -. wn 3    /\ wa 358    = wceq 1623    e. wcel 1685   _Vcvv 2789    \ cdif 3150
This theorem is referenced by:  dfun3  3408  dfin3  3409  invdif  3411  ssindif0  3509  difdifdir  3542
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1636  ax-8 1644  ax-6 1704  ax-7 1709  ax-11 1716  ax-12 1868  ax-ext 2265
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1631  df-clab 2271  df-cleq 2277  df-clel 2280  df-nfc 2409  df-v 2791  df-dif 3156
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