Metamath Proof Explorer


Theorem fldidom

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

Ref Expression
Assertion fldidom
|- ( R e. Field -> R e. IDomn )

Proof

Step Hyp Ref Expression
1 isfld
 |-  ( R e. Field <-> ( R e. DivRing /\ R e. CRing ) )
2 1 simprbi
 |-  ( R e. Field -> R e. CRing )
3 1 simplbi
 |-  ( R e. Field -> R e. DivRing )
4 drngdomn
 |-  ( R e. DivRing -> R e. Domn )
5 3 4 syl
 |-  ( R e. Field -> R e. Domn )
6 isidom
 |-  ( R e. IDomn <-> ( R e. CRing /\ R e. Domn ) )
7 2 5 6 sylanbrc
 |-  ( R e. Field -> R e. IDomn )