Theorem erdm 6523

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 6513	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
2	1	simp2bi 1008	1 ⊢ (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)

Syntax hints:   → wi 4   = wceq 1348   ∪ cun 3119   ⊆ wss 3121  ^◡ccnv 4610  dom cdm 4611   ∘ ccom 4615  Rel wrel 4616   Er wer 6510

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106

This theorem depends on definitions:  df-bi 116  df-3an 975  df-er 6513

This theorem is referenced by:  ercl  6524  erref  6533  errn  6535  erssxp  6536  erexb  6538  ereldm  6556  uniqs2  6573  iinerm  6585  th3qlem1  6615  0nnq  7326  nnnq0lem1  7408  prsrlem1  7704  gt0srpr  7710  0nsr  7711