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Theorem qseq1 6709
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 2737 . . 3  |-  ( A  =  B  ->  ( E. x  e.  A  y  =  [ x ] C  <->  E. x  e.  B  y  =  [ x ] C ) )
21abbidv 2397 . 2  |-  ( A  =  B  ->  { y  |  E. x  e.  A  y  =  [
x ] C }  =  { y  |  E. x  e.  B  y  =  [ x ] C } )
3 df-qs 6666 . 2  |-  ( A /. C )  =  { y  |  E. x  e.  A  y  =  [ x ] C }
4 df-qs 6666 . 2  |-  ( B /. C )  =  { y  |  E. x  e.  B  y  =  [ x ] C }
52, 3, 43eqtr4g 2340 1  |-  ( A  =  B  ->  ( A /. C )  =  ( B /. C
) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    = wceq 1623   {cab 2269   E.wrex 2544   [cec 6658   /.cqs 6659
This theorem is referenced by:  pi1bas  18536  aishp  26172
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866  ax-ext 2264
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1630  df-clab 2270  df-cleq 2276  df-clel 2279  df-nfc 2408  df-rex 2549  df-qs 6666
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