Theorem pweq 1800

Description: Equality theorem for the power class.

Assertion

Ref	Expression
pweq	|- (A = B -> P~A = P~B)

Proof of Theorem pweq

Step	Hyp	Ref	Expression
1		sseq2 1522	. . 3 |- (A = B -> (x (_ A <-> x (_ B))
2	1	biabdv 1183	. 2 |- (A = B -> {x | x (_ A} = {x | x (_ B})
3		df-pw 1799	. 2 |- P~A = {x | x (_ A}
4		df-pw 1799	. 2 |- P~B = {x | x (_ B}
5	2, 3, 4	3eqtr4g 1147	1 |- (A = B -> P~A = P~B)

Syntax hints:   -> wi 2  {cab 1090   = wceq 1091   (_ wss 1487  P~cpw 1798

This theorem is referenced by:  pwex 1806  pwexg 1807  pwssun 1917  canth2g 3382  r1suc 3496  r1val3 3523  r1pw 3529  shsspwh 5153

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-in 1491  df-ss 1492  df-pw 1799

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