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Theorem ereq2 6290
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 2097 . . 3  |-  ( A  =  B  ->  ( dom  R  =  A  <->  dom  R  =  B ) )
213anbi2d 1253 . 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 6282 . 2  |-  ( R  Er  A  <->  ( Rel  R  /\  dom  R  =  A  /\  ( `' R  u.  ( R  o.  R ) ) 
C_  R ) )
4 df-er 6282 . 2  |-  ( R  Er  B  <->  ( Rel  R  /\  dom  R  =  B  /\  ( `' R  u.  ( R  o.  R ) ) 
C_  R ) )
52, 3, 43bitr4g 221 1  |-  ( A  =  B  ->  ( R  Er  A  <->  R  Er  B ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 103    /\ w3a 924    = wceq 1289    u. cun 2997    C_ wss 2999   `'ccnv 4435   dom cdm 4436    o. ccom 4440   Rel wrel 4441    Er wer 6279
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1381  ax-gen 1383  ax-4 1445  ax-17 1464  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-3an 926  df-cleq 2081  df-er 6282
This theorem is referenced by:  iserd  6308
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