Theorem ereq2 6509

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 2175	. . 3 |- ( A = B -> ( dom R = A <-> dom R = B ) )
2	1	3anbi2d 1307	. 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 6501	. 2 |- ( R Er A <-> ( Rel R /\ dom R = A /\ ( `' R u. ( R o. R ) ) C_ R ) )
4		df-er 6501	. 2 |- ( R Er B <-> ( Rel R /\ dom R = B /\ ( `' R u. ( R o. R ) ) C_ R ) )
5	2, 3, 4	3bitr4g 222	1 |- ( A = B -> ( R Er A <-> R Er B ) )

Syntax hints:   -> wi 4   <-> wb 104   /\ w3a 968   = wceq 1343   u. cun 3114   C_ wss 3116   `'ccnv 4603   dom cdm 4604   o. ccom 4608   Rel wrel 4609   Er wer 6498

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 1435  ax-gen 1437  ax-4 1498  ax-17 1514  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-3an 970  df-cleq 2158  df-er 6501

This theorem is referenced by:  iserd  6527