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Theorem pw0 3736
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 3459 . . 3  |-  ( x 
C_  (/)  <->  x  =  (/) )
21abbii 2370 . 2  |-  { x  |  x  C_  (/) }  =  { x  |  x  =  (/) }
3 df-pw 3601 . 2  |-  ~P (/)  =  {
x  |  x  C_  (/)
}
4 df-sn 3620 . 2  |-  { (/) }  =  { x  |  x  =  (/) }
52, 3, 43eqtr4i 2288 1  |-  ~P (/)  =  { (/)
}
Colors of variables: wff set class
Syntax hints:    = wceq 1619   {cab 2244    C_ wss 3127   (/)c0 3430   ~Pcpw 3599   {csn 3614
This theorem is referenced by:  p0ex  4169  pwfi  7119  ackbij1lem14  7827  fin1a2lem12  8005  0tsk  8345  hashbc  11356  sn0topon  16697  sn0cld  16789  rankeq1o  24176  ssoninhaus  24262
This theorem was proved from axioms:  ax-1 7  ax-2 8  ax-3 9  ax-mp 10  ax-5 1533  ax-6 1534  ax-7 1535  ax-gen 1536  ax-8 1623  ax-11 1624  ax-17 1628  ax-12o 1664  ax-10 1678  ax-9 1684  ax-4 1692  ax-16 1927  ax-ext 2239
This theorem depends on definitions:  df-bi 179  df-or 361  df-an 362  df-tru 1315  df-ex 1538  df-nf 1540  df-sb 1884  df-clab 2245  df-cleq 2251  df-clel 2254  df-nfc 2383  df-v 2765  df-dif 3130  df-in 3134  df-ss 3141  df-nul 3431  df-pw 3601  df-sn 3620
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