Theorem qseq1 6693

Description: Equality theorem for quotient set. (Contributed by NM, 23-Jul-1995.)

Assertion

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

Proof of Theorem qseq1

Dummy variables x y are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		rexeq 2706	. . 3 |- ( A = B -> ( E. x e. A y = [ x ] C <-> E. x e. B y = [ x ] C ) )
2	1	abbidv 2325	. 2 |- ( A = B -> { y | E. x e. A y = [ x ] C } = { y | E. x e. B y = [ x ] C } )
3		df-qs 6649	. 2 |- ( A /. C ) = { y | E. x e. A y = [ x ] C }
4		df-qs 6649	. 2 |- ( B /. C ) = { y | E. x e. B y = [ x ] C }
5	2, 3, 4	3eqtr4g 2265	1 |- ( A = B -> ( A /. C ) = ( B /. C ) )

Syntax hints:   -> wi 4   = wceq 1373   {cab 2193   E.wrex 2487   [cec 6641   /.cqs 6642

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-nfc 2339  df-rex 2492  df-qs 6649

This theorem is referenced by: (None)