Metamath Proof Explorer


Theorem isdmn

Description: Obsolete theorem, use isidom2 instead. The predicate "is a domain". (Contributed by Jeff Madsen, 10-Jun-2010) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion isdmn ⊢ R ∈ Dmn ↔ R ∈ PrRing ∧ R ∈ Com2

Proof

Step Hyp Ref Expression
1 df-dmn ⊢ Dmn = PrRing ∩ Com2
2 1 elin2 ⊢ R ∈ Dmn ↔ R ∈ PrRing ∧ R ∈ Com2