Metamath Proof Explorer

Theorem fldidom

Description: A field is an integral domain. (Contributed by Mario Carneiro, 29-Mar-2015) (Proof shortened by SN, 11-Nov-2024)

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

Proof

Step Hyp Ref Expression
1 drngdomn 
 |-  ( R e. DivRing -> R e. Domn )
2 1 anim1ci 
 |-  ( ( R e. DivRing /\ R e. CRing ) -> ( R e. CRing /\ R e. Domn ) )
3 isfld 
 |-  ( R e. Field <-> ( R e. DivRing /\ R e. CRing ) )
4 isidom 
 |-  ( R e. IDomn <-> ( R e. CRing /\ R e. Domn ) )
5 2 3 4 3imtr4i 
 |-  ( R e. Field -> R e. IDomn )