Theorem erdm 6712

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 6702	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
2	1	simp2bi 1039	1 ⊢ (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)

Syntax hints:   → wi 4   = wceq 1397   ∪ cun 3198   ⊆ wss 3200  ^◡ccnv 4724  dom cdm 4725   ∘ ccom 4729  Rel wrel 4730   Er wer 6699

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 1006  df-er 6702

This theorem is referenced by:  ercl  6713  erref  6722  errn  6724  erssxp  6725  erexb  6727  ereldm  6747  uniqs2  6764  iinerm  6776  th3qlem1  6806  0nnq  7584  nnnq0lem1  7666  prsrlem1  7962  gt0srpr  7968  0nsr  7969  divsfval  13429