Theorem qseq1 4288

Description: Equality theorem for quotient set.

Assertion

Ref	Expression
qseq1	|- (A = B -> (A/.C) = (B/.C))

Proof of Theorem qseq1

Step	Hyp	Ref	Expression
1		rexeq1 1787	. . 3 |- (A = B -> (E.x e. A y = [x]C <-> E.x e. B y = [x]C))
2	1	abbidv 1577	. 2 |- (A = B -> {y | E.x e. A y = [x]C} = {y | E.x e. B y = [x]C})
3		df-qs 4266	. 2 |- (A/.C) = {y | E.x e. A y = [x]C}
4		df-qs 4266	. 2 |- (B/.C) = {y | E.x e. B y = [x]C}
5	2, 3, 4	3eqtr4g 1531	1 |- (A = B -> (A/.C) = (B/.C))

Syntax hints:   -> wi 3   = wceq 956  {cab 1463  E.wrex 1646  [cec 4259  /.cqs 4260

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-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-rex 1650  df-qs 4266

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