Theorem qseq1 6549

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 2662	. . 3 |- ( A = B -> ( E. x e. A y = [ x ] C <-> E. x e. B y = [ x ] C ) )
2	1	abbidv 2284	. 2 |- ( A = B -> { y | E. x e. A y = [ x ] C } = { y | E. x e. B y = [ x ] C } )
3		df-qs 6507	. 2 |- ( A /. C ) = { y | E. x e. A y = [ x ] C }
4		df-qs 6507	. 2 |- ( B /. C ) = { y | E. x e. B y = [ x ] C }
5	2, 3, 4	3eqtr4g 2224	1 |- ( A = B -> ( A /. C ) = ( B /. C ) )

Syntax hints:   -> wi 4   = wceq 1343   {cab 2151   E.wrex 2445   [cec 6499   /.cqs 6500

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-rex 2450  df-qs 6507

This theorem is referenced by: (None)