Theorem erdm 6698

Description: The domain of an equivalence relation. (Contributed by Mario Carneiro, 12-Aug-2015.)

Assertion

Ref	Expression
erdm	|- ( R Er A -> dom R = A )

Proof of Theorem erdm

Step	Hyp	Ref	Expression
1		df-er 6688	. 2 |- ( R Er A <-> ( Rel R /\ dom R = A /\ ( `' R u. ( R o. R ) ) C_ R ) )
2	1	simp2bi 1037	1 |- ( R Er A -> dom R = A )

Syntax hints:   -> wi 4   = wceq 1395   u. cun 3195   C_ wss 3197   `'ccnv 4718   dom cdm 4719   o. ccom 4723   Rel wrel 4724   Er wer 6685

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-er 6688

This theorem is referenced by:  ercl  6699  erref  6708  errn  6710  erssxp  6711  erexb  6713  ereldm  6733  uniqs2  6750  iinerm  6762  th3qlem1  6792  0nnq  7562  nnnq0lem1  7644  prsrlem1  7940  gt0srpr  7946  0nsr  7947  divsfval  13377