Theorem erdm 6602

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 6592	. 2 |- ( R Er A <-> ( Rel R /\ dom R = A /\ ( `' R u. ( R o. R ) ) C_ R ) )
2	1	simp2bi 1015	1 |- ( R Er A -> dom R = A )

Syntax hints:   -> wi 4   = 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

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

This theorem is referenced by:  ercl  6603  erref  6612  errn  6614  erssxp  6615  erexb  6617  ereldm  6637  uniqs2  6654  iinerm  6666  th3qlem1  6696  0nnq  7431  nnnq0lem1  7513  prsrlem1  7809  gt0srpr  7815  0nsr  7816  divsfval  12971