Metamath Proof Explorer


Theorem ringn0

Description: Rings exist. (Contributed by AV, 29-Apr-2019)

Ref Expression
Assertion ringn0
|- Ring =/= (/)

Proof

Step Hyp Ref Expression
1 vex
 |-  z e. _V
2 eqid
 |-  { <. ( Base ` ndx ) , { z } >. , <. ( +g ` ndx ) , { <. <. z , z >. , z >. } >. , <. ( .r ` ndx ) , { <. <. z , z >. , z >. } >. } = { <. ( Base ` ndx ) , { z } >. , <. ( +g ` ndx ) , { <. <. z , z >. , z >. } >. , <. ( .r ` ndx ) , { <. <. z , z >. , z >. } >. }
3 2 ring1
 |-  ( z e. _V -> { <. ( Base ` ndx ) , { z } >. , <. ( +g ` ndx ) , { <. <. z , z >. , z >. } >. , <. ( .r ` ndx ) , { <. <. z , z >. , z >. } >. } e. Ring )
4 ne0i
 |-  ( { <. ( Base ` ndx ) , { z } >. , <. ( +g ` ndx ) , { <. <. z , z >. , z >. } >. , <. ( .r ` ndx ) , { <. <. z , z >. , z >. } >. } e. Ring -> Ring =/= (/) )
5 1 3 4 mp2b
 |-  Ring =/= (/)