Theorem qseq1 4294

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 1790	. . 3 |- (A = B -> (E.x e. A y = [x]C <-> E.x e. B y = [x]C))
2	1	abbidv 1580	. 2 |- (A = B -> {y | E.x e. A y = [x]C} = {y | E.x e. B y = [x]C})
3		df-qs 4272	. 2 |- (A/.C) = {y | E.x e. A y = [x]C}
4		df-qs 4272	. 2 |- (B/.C) = {y | E.x e. B y = [x]C}
5	2, 3, 4	3eqtr4g 1534	1 |- (A = B -> (A/.C) = (B/.C))

Syntax hints:   -> wi 3   = wceq 958  {cab 1466  E.wrex 1649  [cec 4265  /.cqs 4266

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-rex 1653  df-qs 4272

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