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Theorem erdm 6023
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 6013 . 2 (𝑅 Er A ↔ (Rel 𝑅 dom 𝑅 = A (𝑅 ∪ (𝑅𝑅)) ⊆ 𝑅))
21simp2bi 906 1 (𝑅 Er A → dom 𝑅 = A)
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1226  cun 2888  wss 2890  ccnv 4267  dom cdm 4268  ccom 4272  Rel wrel 4273   Er wer 6010
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100
This theorem depends on definitions:  df-bi 110  df-3an 873  df-er 6013
This theorem is referenced by:  ercl  6024  erref  6033  errn  6035  erssxp  6036  erexb  6038  ereldm  6056  uniqs2  6073  iinerm  6085  th3qlem1  6115  0nnq  6217  nnnq0lem1  6295  prsrlem1  6486  gt0srpr  6492  0nsr  6493
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