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Theorem qseq1 6443
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.
StepHypRef Expression
1 rexeq 2602 . . 3  |-  ( A  =  B  ->  ( E. x  e.  A  y  =  [ x ] C  <->  E. x  e.  B  y  =  [ x ] C ) )
21abbidv 2233 . 2  |-  ( A  =  B  ->  { y  |  E. x  e.  A  y  =  [
x ] C }  =  { y  |  E. x  e.  B  y  =  [ x ] C } )
3 df-qs 6401 . 2  |-  ( A /. C )  =  { y  |  E. x  e.  A  y  =  [ x ] C }
4 df-qs 6401 . 2  |-  ( B /. C )  =  { y  |  E. x  e.  B  y  =  [ x ] C }
52, 3, 43eqtr4g 2173 1  |-  ( A  =  B  ->  ( A /. C )  =  ( B /. C
) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    = wceq 1314   {cab 2101   E.wrex 2392   [cec 6393   /.cqs 6394
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 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-bndl 1469  ax-4 1470  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498  ax-ext 2097
This theorem depends on definitions:  df-bi 116  df-tru 1317  df-nf 1420  df-sb 1719  df-clab 2102  df-cleq 2108  df-clel 2111  df-nfc 2245  df-rex 2397  df-qs 6401
This theorem is referenced by: (None)
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