Theorem erdm 6630

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 6620	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
2	1	simp2bi 1016	1 ⊢ (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)

Syntax hints:   → wi 4   = wceq 1373   ∪ cun 3164   ⊆ wss 3166  ^◡ccnv 4674  dom cdm 4675   ∘ ccom 4679  Rel wrel 4680   Er wer 6617

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 983  df-er 6620

This theorem is referenced by:  ercl  6631  erref  6640  errn  6642  erssxp  6643  erexb  6645  ereldm  6665  uniqs2  6682  iinerm  6694  th3qlem1  6724  0nnq  7477  nnnq0lem1  7559  prsrlem1  7855  gt0srpr  7861  0nsr  7862  divsfval  13160