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Theorem pw0 3762
Description: Compute the power set of the empty set. Theorem 89 of [Suppes] p. 47. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)
Assertion
Ref Expression
pw0  |-  ~P (/)  =  { (/)
}

Proof of Theorem pw0
StepHypRef Expression
1 ss0b 3484 . . 3  |-  ( x 
C_  (/)  <->  x  =  (/) )
21abbii 2395 . 2  |-  { x  |  x  C_  (/) }  =  { x  |  x  =  (/) }
3 df-pw 3627 . 2  |-  ~P (/)  =  {
x  |  x  C_  (/)
}
4 df-sn 3646 . 2  |-  { (/) }  =  { x  |  x  =  (/) }
52, 3, 43eqtr4i 2313 1  |-  ~P (/)  =  { (/)
}
Colors of variables: wff set class
Syntax hints:    = wceq 1623   {cab 2269    C_ wss 3152   (/)c0 3455   ~Pcpw 3625   {csn 3640
This theorem is referenced by:  p0ex  4197  pwfi  7151  ackbij1lem14  7859  fin1a2lem12  8037  0tsk  8377  hashbc  11391  incexclem  12295  sn0topon  16735  sn0cld  16827  rankeq1o  24212  ssoninhaus  24298  usgra0v  27516
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 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866  ax-ext 2264
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 1630  df-clab 2270  df-cleq 2276  df-clel 2279  df-nfc 2408  df-v 2790  df-dif 3155  df-in 3159  df-ss 3166  df-nul 3456  df-pw 3627  df-sn 3646
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