Theorem pw0 2468

Description: Compute the power set of the empty set. Theorem 89 of [Suppes] p. 47. (The proof was shortened by Eric Schmidt, 4-Apr-2007.)

Assertion

Ref	Expression
pw0	|- P~(/) = {(/)}

Proof of Theorem pw0

Step	Hyp	Ref	Expression
1		df-pw 2402	. . . . 5 |- P~(/) = {x | x (_ (/)}
2	1	abeq2i 1570	. . . 4 |- (x e. P~(/) <-> x (_ (/))
3		ss0b 2302	. . . 4 |- (x (_ (/) <-> x = (/))
4	2, 3	bitr 173	. . 3 |- (x e. P~(/) <-> x = (/))
5	4	abbi2i 1574	. 2 |- P~(/) = {x | x = (/)}
6		df-sn 2412	. 2 |- {(/)} = {x | x = (/)}
7	5, 6	eqtr4 1498	1 |- P~(/) = {(/)}

Syntax hints:   = wceq 956   e. wcel 958  {cab 1463   (_ wss 2047  (/)c0 2280  P~cpw 2401  {csn 2409

This theorem is referenced by:  pwfiOLD 4571

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-v 1812  df-dif 2049  df-in 2051  df-ss 2053  df-nul 2281  df-pw 2402  df-sn 2412

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