Theorem ereq2 6600

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 2206	. . 3 |- ( A = B -> ( dom R = A <-> dom R = B ) )
2	1	3anbi2d 1328	. 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 6592	. 2 |- ( R Er A <-> ( Rel R /\ dom R = A /\ ( `' R u. ( R o. R ) ) C_ R ) )
4		df-er 6592	. 2 |- ( R Er B <-> ( Rel R /\ dom R = B /\ ( `' R u. ( R o. R ) ) C_ R ) )
5	2, 3, 4	3bitr4g 223	1 |- ( A = B -> ( R Er A <-> R Er B ) )

Syntax hints:   -> wi 4   <-> wb 105   /\ w3a 980   = wceq 1364   u. cun 3155   C_ wss 3157   `'ccnv 4662   dom cdm 4663   o. ccom 4667   Rel wrel 4668   Er wer 6589

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1461  ax-gen 1463  ax-4 1524  ax-17 1540  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-3an 982  df-cleq 2189  df-er 6592

This theorem is referenced by:  iserd  6618