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Theorem ereq2 6915
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 2447 . . 3  |-  ( A  =  B  ->  ( dom  R  =  A  <->  dom  R  =  B ) )
213anbi2d 1260 . 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 6907 . 2  |-  ( R  Er  A  <->  ( Rel  R  /\  dom  R  =  A  /\  ( `' R  u.  ( R  o.  R ) ) 
C_  R ) )
4 df-er 6907 . 2  |-  ( R  Er  B  <->  ( Rel  R  /\  dom  R  =  B  /\  ( `' R  u.  ( R  o.  R ) ) 
C_  R ) )
52, 3, 43bitr4g 281 1  |-  ( A  =  B  ->  ( R  Er  A  <->  R  Er  B ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 178    /\ w3a 937    = wceq 1653    u. cun 3320    C_ wss 3322   `'ccnv 4879   dom cdm 4880    o. ccom 4884   Rel wrel 4885    Er wer 6904
This theorem is referenced by:  iserd  6933  efgval  15351  frgp0  15394  frgpmhm  15399
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1556  ax-5 1567  ax-17 1627  ax-9 1667  ax-8 1688  ax-11 1762  ax-ext 2419
This theorem depends on definitions:  df-bi 179  df-an 362  df-3an 939  df-ex 1552  df-cleq 2431  df-er 6907
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