Theorem pw0 2459

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 2392	. . . . 5 |- P~(/) = {x | x (_ (/)}
2	1	abeq2i 1562	. . . 4 |- (x e. P~(/) <-> x (_ (/))
3		ss0b 2292	. . . 4 |- (x (_ (/) <-> x = (/))
4	2, 3	bitr 173	. . 3 |- (x e. P~(/) <-> x = (/))
5	4	abbi2i 1566	. 2 |- P~(/) = {x | x = (/)}
6		df-sn 2402	. 2 |- {(/)} = {x | x = (/)}
7	5, 6	eqtr4 1490	1 |- P~(/) = {(/)}

Syntax hints:   = wceq 953   e. wcel 955  {cab 1456   (_ wss 2037  (/)c0 2270  P~cpw 2391  {csn 2399

This theorem is referenced by:  pwfi 4545

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-8 961  ax-10 963  ax-12 965  ax-17 968  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-16 1206  ax-11o 1213  ax-ext 1452

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 978  df-sb 1168  df-clab 1457  df-cleq 1462  df-clel 1465  df-v 1803  df-dif 2039  df-in 2041  df-ss 2043  df-nul 2271  df-pw 2392  df-sn 2402

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