Metamath Proof Explorer


Theorem isdrng

Description: The predicate "is a division ring". (Contributed by NM, 18-Oct-2012) (Revised by Mario Carneiro, 2-Dec-2014)

Ref Expression
Hypotheses isdrng.b B=BaseR
isdrng.u U=UnitR
isdrng.z 0˙=0R
Assertion isdrng RDivRingRRingU=B0˙

Proof

Step Hyp Ref Expression
1 isdrng.b B=BaseR
2 isdrng.u U=UnitR
3 isdrng.z 0˙=0R
4 fveq2 r=RUnitr=UnitR
5 4 2 eqtr4di r=RUnitr=U
6 fveq2 r=RBaser=BaseR
7 6 1 eqtr4di r=RBaser=B
8 fveq2 r=R0r=0R
9 8 3 eqtr4di r=R0r=0˙
10 9 sneqd r=R0r=0˙
11 7 10 difeq12d r=RBaser0r=B0˙
12 5 11 eqeq12d r=RUnitr=Baser0rU=B0˙
13 df-drng DivRing=rRing|Unitr=Baser0r
14 12 13 elrab2 RDivRingRRingU=B0˙