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Theorem erdm 6711
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
StepHypRef Expression
1 df-er 6701 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
21simp2bi 1039 1 (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1397  cun 3198  wss 3200  ccnv 4724  dom cdm 4725  ccom 4729  Rel wrel 4730   Er wer 6698
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 6701
This theorem is referenced by:  ercl  6712  erref  6721  errn  6723  erssxp  6724  erexb  6726  ereldm  6746  uniqs2  6763  iinerm  6775  th3qlem1  6805  0nnq  7583  nnnq0lem1  7665  prsrlem1  7961  gt0srpr  7967  0nsr  7968  divsfval  13410
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