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Theorem ereq2 6405
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 2127 . . 3  |-  ( A  =  B  ->  ( dom  R  =  A  <->  dom  R  =  B ) )
213anbi2d 1280 . 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 6397 . 2  |-  ( R  Er  A  <->  ( Rel  R  /\  dom  R  =  A  /\  ( `' R  u.  ( R  o.  R ) ) 
C_  R ) )
4 df-er 6397 . 2  |-  ( R  Er  B  <->  ( Rel  R  /\  dom  R  =  B  /\  ( `' R  u.  ( R  o.  R ) ) 
C_  R ) )
52, 3, 43bitr4g 222 1  |-  ( A  =  B  ->  ( R  Er  A  <->  R  Er  B ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 104    /\ w3a 947    = wceq 1316    u. cun 3039    C_ wss 3041   `'ccnv 4508   dom cdm 4509    o. ccom 4513   Rel wrel 4514    Er wer 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-5 1408  ax-gen 1410  ax-4 1472  ax-17 1491  ax-ext 2099
This theorem depends on definitions:  df-bi 116  df-3an 949  df-cleq 2110  df-er 6397
This theorem is referenced by:  iserd  6423
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