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 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
2	1	simp2bi 1015	1 ⊢ (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)

Syntax hints:   → wi 4   = wceq 1364   ∪ cun 3155   ⊆ wss 3157  ^◡ccnv 4662  dom cdm 4663   ∘ 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