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Theorem pw0 3905
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 3617 . . 3  |-  ( x 
C_  (/)  <->  x  =  (/) )
21abbii 2516 . 2  |-  { x  |  x  C_  (/) }  =  { x  |  x  =  (/) }
3 df-pw 3761 . 2  |-  ~P (/)  =  {
x  |  x  C_  (/)
}
4 df-sn 3780 . 2  |-  { (/) }  =  { x  |  x  =  (/) }
52, 3, 43eqtr4i 2434 1  |-  ~P (/)  =  { (/)
}
Colors of variables: wff set class
Syntax hints:    = wceq 1649   {cab 2390    C_ wss 3280   (/)c0 3588   ~Pcpw 3759   {csn 3774
This theorem is referenced by:  p0ex  4346  pwfi  7360  ackbij1lem14  8069  fin1a2lem12  8247  0tsk  8586  hashbc  11657  incexclem  12571  sn0topon  17017  sn0cld  17109  ust0  18202  uhgra0v  21298  usgra0v  21344  esumnul  24396  rankeq1o  26016  ssoninhaus  26102
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1552  ax-5 1563  ax-17 1623  ax-9 1662  ax-8 1683  ax-6 1740  ax-7 1745  ax-11 1757  ax-12 1946  ax-ext 2385
This theorem depends on definitions:  df-bi 178  df-an 361  df-tru 1325  df-ex 1548  df-nf 1551  df-sb 1656  df-clab 2391  df-cleq 2397  df-clel 2400  df-nfc 2529  df-v 2918  df-dif 3283  df-in 3287  df-ss 3294  df-nul 3589  df-pw 3761  df-sn 3780
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