Metamath Proof Explorer


Theorem fldidom

Description: A field is an integral domain. (Contributed by Mario Carneiro, 29-Mar-2015)

Ref Expression
Assertion fldidom ( 𝑅 ∈ Field → 𝑅 ∈ IDomn )

Proof

Step Hyp Ref Expression
1 isfld ( 𝑅 ∈ Field ↔ ( 𝑅 ∈ DivRing ∧ 𝑅 ∈ CRing ) )
2 1 simprbi ( 𝑅 ∈ Field → 𝑅 ∈ CRing )
3 1 simplbi ( 𝑅 ∈ Field → 𝑅 ∈ DivRing )
4 drngdomn ( 𝑅 ∈ DivRing → 𝑅 ∈ Domn )
5 3 4 syl ( 𝑅 ∈ Field → 𝑅 ∈ Domn )
6 isidom ( 𝑅 ∈ IDomn ↔ ( 𝑅 ∈ CRing ∧ 𝑅 ∈ Domn ) )
7 2 5 6 sylanbrc ( 𝑅 ∈ Field → 𝑅 ∈ IDomn )