Theorem ereq2 6114

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 2049	. . 3 |- ( A = B -> ( dom R = A <-> dom R = B ) )
2	1	3anbi2d 1212	. 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 6106	. 2 |- ( R Er A <-> ( Rel R /\ dom R = A /\ ( `' R u. ( R o. R ) ) C_ R ) )
4		df-er 6106	. 2 |- ( R Er B <-> ( Rel R /\ dom R = B /\ ( `' R u. ( R o. R ) ) C_ R ) )
5	2, 3, 4	3bitr4g 212	1 |- ( A = B -> ( R Er A <-> R Er B ) )

Syntax hints:   -> wi 4   <-> wb 98   /\ w3a 885   = wceq 1243   u. cun 2915   C_ wss 2917   `'ccnv 4344   dom cdm 4345   o. ccom 4349   Rel wrel 4350   Er wer 6103

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-4 1400  ax-17 1419  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-3an 887  df-cleq 2033  df-er 6106

This theorem is referenced by:  iserd  6132