Theorem qseq1 6751

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 2731	. . 3 |- ( A = B -> ( E. x e. A y = [ x ] C <-> E. x e. B y = [ x ] C ) )
2	1	abbidv 2349	. 2 |- ( A = B -> { y | E. x e. A y = [ x ] C } = { y | E. x e. B y = [ x ] C } )
3		df-qs 6707	. 2 |- ( A /. C ) = { y | E. x e. A y = [ x ] C }
4		df-qs 6707	. 2 |- ( B /. C ) = { y | E. x e. B y = [ x ] C }
5	2, 3, 4	3eqtr4g 2289	1 |- ( A = B -> ( A /. C ) = ( B /. C ) )

Syntax hints:   -> wi 4   = wceq 1397   {cab 2217   E.wrex 2511   [cec 6699   /.cqs 6700

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-rex 2516  df-qs 6707

This theorem is referenced by: (None)