Theorem erdm 6300

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 6290	. 2 |- ( R Er A <-> ( Rel R /\ dom R = A /\ ( `' R u. ( R o. R ) ) C_ R ) )
2	1	simp2bi 959	1 |- ( R Er A -> dom R = A )

Syntax hints:   -> wi 4   = wceq 1289   u. cun 2997   C_ wss 2999   `'ccnv 4437   dom cdm 4438   o. ccom 4442   Rel wrel 4443   Er wer 6287

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105

This theorem depends on definitions:  df-bi 115  df-3an 926  df-er 6290

This theorem is referenced by:  ercl  6301  erref  6310  errn  6312  erssxp  6313  erexb  6315  ereldm  6333  uniqs2  6350  iinerm  6362  th3qlem1  6392  0nnq  6921  nnnq0lem1  7003  prsrlem1  7286  gt0srpr  7292  0nsr  7293