Theorem erdm 6599

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 6589	. 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 3152   ⊆ wss 3154  ^◡ccnv 4659  dom cdm 4660   ∘ ccom 4664  Rel wrel 4665   Er wer 6586

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 6589

This theorem is referenced by:  ercl  6600  erref  6609  errn  6611  erssxp  6612  erexb  6614  ereldm  6634  uniqs2  6651  iinerm  6663  th3qlem1  6693  0nnq  7426  nnnq0lem1  7508  prsrlem1  7804  gt0srpr  7810  0nsr  7811  divsfval  12914