Theorem qseq1 6728

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 2729	. . 3 |- ( A = B -> ( E. x e. A y = [ x ] C <-> E. x e. B y = [ x ] C ) )
2	1	abbidv 2347	. 2 |- ( A = B -> { y | E. x e. A y = [ x ] C } = { y | E. x e. B y = [ x ] C } )
3		df-qs 6684	. 2 |- ( A /. C ) = { y | E. x e. A y = [ x ] C }
4		df-qs 6684	. 2 |- ( B /. C ) = { y | E. x e. B y = [ x ] C }
5	2, 3, 4	3eqtr4g 2287	1 |- ( A = B -> ( A /. C ) = ( B /. C ) )

Syntax hints:   -> wi 4   = wceq 1395   {cab 2215   E.wrex 2509   [cec 6676   /.cqs 6677

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-rex 2514  df-qs 6684

This theorem is referenced by: (None)