Theorem qseq1 6585

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 2674	. . 3 |- ( A = B -> ( E. x e. A y = [ x ] C <-> E. x e. B y = [ x ] C ) )
2	1	abbidv 2295	. 2 |- ( A = B -> { y | E. x e. A y = [ x ] C } = { y | E. x e. B y = [ x ] C } )
3		df-qs 6543	. 2 |- ( A /. C ) = { y | E. x e. A y = [ x ] C }
4		df-qs 6543	. 2 |- ( B /. C ) = { y | E. x e. B y = [ x ] C }
5	2, 3, 4	3eqtr4g 2235	1 |- ( A = B -> ( A /. C ) = ( B /. C ) )

Syntax hints:   -> wi 4   = wceq 1353   {cab 2163   E.wrex 2456   [cec 6535   /.cqs 6536

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-rex 2461  df-qs 6543

This theorem is referenced by: (None)