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Theorem pw0 3662
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 3391 . . 3  |-  ( x 
C_  (/)  <->  x  =  (/) )
21abbii 2361 . 2  |-  { x  |  x  C_  (/) }  =  { x  |  x  =  (/) }
3 df-pw 3532 . 2  |-  ~P (/)  =  {
x  |  x  C_  (/)
}
4 df-sn 3550 . 2  |-  { (/) }  =  { x  |  x  =  (/) }
52, 3, 43eqtr4i 2283 1  |-  ~P (/)  =  { (/)
}
Colors of variables: wff set class
Syntax hints:    = wceq 1619   {cab 2239    C_ wss 3078   (/)c0 3362   ~Pcpw 3530   {csn 3544
This theorem is referenced by:  p0ex  4091  pwfi  7035  ackbij1lem14  7743  fin1a2lem12  7921  0tsk  8257  hashbc  11268  sn0topon  16567  sn0cld  16659  rankeq1o  23975  ssoninhaus  24061
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 1926  ax-ext 2234
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 1883  df-clab 2240  df-cleq 2246  df-clel 2249  df-nfc 2374  df-v 2729  df-dif 3081  df-in 3085  df-ss 3089  df-nul 3363  df-pw 3532  df-sn 3550
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