Metamath Proof Explorer


Theorem isdmn

Description: The predicate "is a domain". (Contributed by Jeff Madsen, 10-Jun-2010)

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 ) )