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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  esumnul  23427  rankeq1o  24801  ssoninhaus  24887  usgra0v  28117
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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