Theorem erdm 6629

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 6619	. 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 1372   ∪ cun 3163   ⊆ wss 3165  ^◡ccnv 4673  dom cdm 4674   ∘ ccom 4678  Rel wrel 4679   Er wer 6616

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 6619

This theorem is referenced by:  ercl  6630  erref  6639  errn  6641  erssxp  6642  erexb  6644  ereldm  6664  uniqs2  6681  iinerm  6693  th3qlem1  6723  0nnq  7476  nnnq0lem1  7558  prsrlem1  7854  gt0srpr  7860  0nsr  7861  divsfval  13131