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 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
2	1	simp2bi 1037	1 ⊢ (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)

Syntax hints:   → wi 4   = wceq 1395   ∪ cun 3195   ⊆ wss 3197  ^◡ccnv 4718  dom cdm 4719   ∘ 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  13376