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Metamath Proof Explorer

Theorem rrgnz

Description: In a nonzero ring, the zero is a left zero divisor (that is, not a left-regular element). (Contributed by Thierry Arnoux, 6-May-2025)

Ref Expression
Hypotheses rrgnz.t 
|- E = ( RLReg ` R )
rrgnz.z 
|- .0. = ( 0g ` R )
Assertion rrgnz 
|- ( R e. NzRing -> -. .0. e. E )

Proof

Step Hyp Ref Expression
1 rrgnz.t 
 |-  E = ( RLReg ` R )
2 rrgnz.z 
 |-  .0. = ( 0g ` R )
3 eqid 
 |-  ( 1r ` R ) = ( 1r ` R )
4 3 2 nzrnz 
 |-  ( R e. NzRing -> ( 1r ` R ) =/= .0. )
5 4 neneqd 
 |-  ( R e. NzRing -> -. ( 1r ` R ) = .0. )
6 nzrring 
 |-  ( R e. NzRing -> R e. Ring )
7 6 adantr 
 |-  ( ( R e. NzRing /\ .0. e. E ) -> R e. Ring )
8 simpr 
 |-  ( ( R e. NzRing /\ .0. e. E ) -> .0. e. E )
9 eqid 
 |-  ( Base ` R ) = ( Base ` R )
10 9 3 ringidcl 
 |-  ( R e. Ring -> ( 1r ` R ) e. ( Base ` R ) )
11 7 10 syl 
 |-  ( ( R e. NzRing /\ .0. e. E ) -> ( 1r ` R ) e. ( Base ` R ) )
12 eqid 
 |-  ( .r ` R ) = ( .r ` R )
13 9 12 2 7 11 ringlzd 
 |-  ( ( R e. NzRing /\ .0. e. E ) -> ( .0. ( .r ` R ) ( 1r ` R ) ) = .0. )
14 1 9 12 2 rrgeq0 
 |-  ( ( R e. Ring /\ .0. e. E /\ ( 1r ` R ) e. ( Base ` R ) ) -> ( ( .0. ( .r ` R ) ( 1r ` R ) ) = .0. <-> ( 1r ` R ) = .0. ) )
15 14 biimpa 
 |-  ( ( ( R e. Ring /\ .0. e. E /\ ( 1r ` R ) e. ( Base ` R ) ) /\ ( .0. ( .r ` R ) ( 1r ` R ) ) = .0. ) -> ( 1r ` R ) = .0. )
16 7 8 11 13 15 syl31anc 
 |-  ( ( R e. NzRing /\ .0. e. E ) -> ( 1r ` R ) = .0. )
17 5 16 mtand 
 |-  ( R e. NzRing -> -. .0. e. E )