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Theorem pw0 3778
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 3497 . . 3  |-  ( x 
C_  (/)  <->  x  =  (/) )
21abbii 2408 . 2  |-  { x  |  x  C_  (/) }  =  { x  |  x  =  (/) }
3 df-pw 3640 . 2  |-  ~P (/)  =  {
x  |  x  C_  (/)
}
4 df-sn 3659 . 2  |-  { (/) }  =  { x  |  x  =  (/) }
52, 3, 43eqtr4i 2326 1  |-  ~P (/)  =  { (/)
}
Colors of variables: wff set class
Syntax hints:    = wceq 1632   {cab 2282    C_ wss 3165   (/)c0 3468   ~Pcpw 3638   {csn 3653
This theorem is referenced by:  p0ex  4213  pwfi  7167  ackbij1lem14  7875  fin1a2lem12  8053  0tsk  8393  hashbc  11407  incexclem  12311  sn0topon  16751  sn0cld  16843  esumnul  23442  rankeq1o  24873  ssoninhaus  24959  usgra0v  28251
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547  ax-17 1606  ax-9 1644  ax-8 1661  ax-6 1715  ax-7 1720  ax-11 1727  ax-12 1878  ax-ext 2277
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639  df-clab 2283  df-cleq 2289  df-clel 2292  df-nfc 2421  df-v 2803  df-dif 3168  df-in 3172  df-ss 3179  df-nul 3469  df-pw 3640  df-sn 3659
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