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Theorem erdm 6448
 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 6438 . 2 (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
21simp2bi 998 1 (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)
 Colors of variables: wff set class Syntax hints:   → wi 4   = wceq 1332   ∪ cun 3075   ⊆ wss 3077  ◡ccnv 4547  dom cdm 4548   ∘ ccom 4552  Rel wrel 4553   Er wer 6435 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106 This theorem depends on definitions:  df-bi 116  df-3an 965  df-er 6438 This theorem is referenced by:  ercl  6449  erref  6458  errn  6460  erssxp  6461  erexb  6463  ereldm  6481  uniqs2  6498  iinerm  6510  th3qlem1  6540  0nnq  7216  nnnq0lem1  7298  prsrlem1  7594  gt0srpr  7600  0nsr  7601
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