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Theorem ddif 3550
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 3037 . . . . 5  |-  x  e. 
_V
2 eldif 3399 . . . . 5  |-  ( x  e.  ( _V  \  A )  <->  ( x  e.  _V  /\  -.  x  e.  A ) )
31, 2mpbiran 916 . . . 4  |-  ( x  e.  ( _V  \  A )  <->  -.  x  e.  A )
43con2bii 330 . . 3  |-  ( x  e.  A  <->  -.  x  e.  ( _V  \  A
) )
51biantrur 504 . . 3  |-  ( -.  x  e.  ( _V 
\  A )  <->  ( x  e.  _V  /\  -.  x  e.  ( _V  \  A
) ) )
64, 5bitr2i 250 . 2  |-  ( ( x  e.  _V  /\  -.  x  e.  ( _V  \  A ) )  <-> 
x  e.  A )
76difeqri 3538 1  |-  ( _V 
\  ( _V  \  A ) )  =  A
Colors of variables: wff setvar class
Syntax hints:   -. wn 3    /\ wa 367    = wceq 1399    e. wcel 1826   _Vcvv 3034    \ cdif 3386
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1626  ax-4 1639  ax-5 1712  ax-6 1755  ax-7 1798  ax-10 1845  ax-11 1850  ax-12 1862  ax-13 2006  ax-ext 2360
This theorem depends on definitions:  df-bi 185  df-an 369  df-tru 1402  df-ex 1621  df-nf 1625  df-sb 1748  df-clab 2368  df-cleq 2374  df-clel 2377  df-nfc 2532  df-v 3036  df-dif 3392
This theorem is referenced by:  dfun3  3661  dfin3  3662  invdif  3664  ssindif0  3796  difdifdir  3831
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