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Theorem pw0 3945
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 3657 . . 3  |-  ( x 
C_  (/)  <->  x  =  (/) )
21abbii 2548 . 2  |-  { x  |  x  C_  (/) }  =  { x  |  x  =  (/) }
3 df-pw 3801 . 2  |-  ~P (/)  =  {
x  |  x  C_  (/)
}
4 df-sn 3820 . 2  |-  { (/) }  =  { x  |  x  =  (/) }
52, 3, 43eqtr4i 2466 1  |-  ~P (/)  =  { (/)
}
Colors of variables: wff set class
Syntax hints:    = wceq 1652   {cab 2422    C_ wss 3320   (/)c0 3628   ~Pcpw 3799   {csn 3814
This theorem is referenced by:  p0ex  4386  pwfi  7402  ackbij1lem14  8113  fin1a2lem12  8291  0tsk  8630  hashbc  11702  incexclem  12616  sn0topon  17062  sn0cld  17154  ust0  18249  uhgra0v  21345  usgra0v  21391  esumnul  24443  rankeq1o  26112  ssoninhaus  26198
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1555  ax-5 1566  ax-17 1626  ax-9 1666  ax-8 1687  ax-6 1744  ax-7 1749  ax-11 1761  ax-12 1950  ax-ext 2417
This theorem depends on definitions:  df-bi 178  df-or 360  df-an 361  df-tru 1328  df-ex 1551  df-nf 1554  df-sb 1659  df-clab 2423  df-cleq 2429  df-clel 2432  df-nfc 2561  df-v 2958  df-dif 3323  df-in 3327  df-ss 3334  df-nul 3629  df-pw 3801  df-sn 3820
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