Theorem erdm 6116

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 6106	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
2	1	simp2bi 920	1 ⊢ (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)

Syntax hints:   → wi 4   = wceq 1243   ∪ cun 2915   ⊆ wss 2917  ^◡ccnv 4344  dom cdm 4345   ∘ ccom 4349  Rel wrel 4350   Er wer 6103

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 887  df-er 6106

This theorem is referenced by:  ercl  6117  erref  6126  errn  6128  erssxp  6129  erexb  6131  ereldm  6149  uniqs2  6166  iinerm  6178  th3qlem1  6208  0nnq  6462  nnnq0lem1  6544  prsrlem1  6827  gt0srpr  6833  0nsr  6834