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

Step	Hyp	Ref	Expression
1		eqeq2 2097	. . 3 |- ( A = B -> ( dom R = A <-> dom R = B ) )
2	1	3anbi2d 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 ) )
5	2, 3, 4	3bitr4g 221	1 |- ( A = B -> ( R Er A <-> R Er B ) )

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