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Theorem ereq2 6668
Description: Equality theorem for equivalence predicate. (Contributed by Mario Carneiro, 12-Aug-2015.)
Assertion
Ref Expression
ereq2  |-  ( A  =  B  ->  ( R  Er  A  <->  R  Er  B ) )

Proof of Theorem ereq2
StepHypRef Expression
1 eqeq2 2292 . . 3  |-  ( A  =  B  ->  ( dom  R  =  A  <->  dom  R  =  B ) )
213anbi2d 1257 . 2  |-  ( A  =  B  ->  (
( Rel  R  /\  dom  R  =  A  /\  ( `' R  u.  ( R  o.  R )
)  C_  R )  <->  ( Rel  R  /\  dom  R  =  B  /\  ( `' R  u.  ( R  o.  R )
)  C_  R )
) )
3 df-er 6660 . 2  |-  ( R  Er  A  <->  ( Rel  R  /\  dom  R  =  A  /\  ( `' R  u.  ( R  o.  R ) ) 
C_  R ) )
4 df-er 6660 . 2  |-  ( R  Er  B  <->  ( Rel  R  /\  dom  R  =  B  /\  ( `' R  u.  ( R  o.  R ) ) 
C_  R ) )
52, 3, 43bitr4g 279 1  |-  ( A  =  B  ->  ( R  Er  A  <->  R  Er  B ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176    /\ w3a 934    = wceq 1623    u. cun 3150    C_ wss 3152   `'ccnv 4688   dom cdm 4689    o. ccom 4693   Rel wrel 4694    Er wer 6657
This theorem is referenced by:  iserd  6686  efgval  15026  frgp0  15069  frgpmhm  15074  isibcg  26191
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-11 1715  ax-ext 2264
This theorem depends on definitions:  df-bi 177  df-an 360  df-3an 936  df-cleq 2276  df-er 6660
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