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Theorem pw0 3937
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 3649 . . 3  |-  ( x 
C_  (/)  <->  x  =  (/) )
21abbii 2547 . 2  |-  { x  |  x  C_  (/) }  =  { x  |  x  =  (/) }
3 df-pw 3793 . 2  |-  ~P (/)  =  {
x  |  x  C_  (/)
}
4 df-sn 3812 . 2  |-  { (/) }  =  { x  |  x  =  (/) }
52, 3, 43eqtr4i 2465 1  |-  ~P (/)  =  { (/)
}
Colors of variables: wff set class
Syntax hints:    = wceq 1652   {cab 2421    C_ wss 3312   (/)c0 3620   ~Pcpw 3791   {csn 3806
This theorem is referenced by:  p0ex  4378  pwfi  7393  ackbij1lem14  8102  fin1a2lem12  8280  0tsk  8619  hashbc  11690  incexclem  12604  sn0topon  17050  sn0cld  17142  ust0  18237  uhgra0v  21333  usgra0v  21379  esumnul  24431  rankeq1o  26060  ssoninhaus  26146
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 2416
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 2422  df-cleq 2428  df-clel 2431  df-nfc 2560  df-v 2950  df-dif 3315  df-in 3319  df-ss 3326  df-nul 3621  df-pw 3793  df-sn 3812
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