Theorem erdm 6139

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 6129	. 2 ⊢ (𝑅 Er 𝐴 ↔ (Rel 𝑅 ∧ dom 𝑅 = 𝐴 ∧ (◡𝑅 ∪ (𝑅 ∘ 𝑅)) ⊆ 𝑅))
2	1	simp2bi 954	1 ⊢ (𝑅 Er 𝐴 → dom 𝑅 = 𝐴)

Syntax hints:   → wi 4   = wceq 1284   ∪ cun 2971   ⊆ wss 2973  ^◡ccnv 4362  dom cdm 4363   ∘ ccom 4367  Rel wrel 4368   Er wer 6126

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105

This theorem depends on definitions:  df-bi 115  df-3an 921  df-er 6129

This theorem is referenced by:  ercl  6140  erref  6149  errn  6151  erssxp  6152  erexb  6154  ereldm  6172  uniqs2  6189  iinerm  6201  th3qlem1  6231  0nnq  6554  nnnq0lem1  6636  prsrlem1  6919  gt0srpr  6925  0nsr  6926