Theorem erdm 6688

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 6678	. 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 4717  dom cdm 4718   ∘ ccom 4722  Rel wrel 4723   Er wer 6675

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 6678

This theorem is referenced by:  ercl  6689  erref  6698  errn  6700  erssxp  6701  erexb  6703  ereldm  6723  uniqs2  6740  iinerm  6752  th3qlem1  6782  0nnq  7547  nnnq0lem1  7629  prsrlem1  7925  gt0srpr  7931  0nsr  7932  divsfval  13356