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 ˙ = 0 R
Assertion rrgnz ⊢ R ∈ NzRing → ¬ 0 ˙ ∈ E

Proof

Step Hyp Ref Expression
1 rrgnz.t ⊢ E = RLReg ⁡ R
2 rrgnz.z ⊢ 0 ˙ = 0 R
3 eqid ⊢ 1 R = 1 R
4 3 2 nzrnz ⊢ R ∈ NzRing → 1 R ≠ 0 ˙
5 4 neneqd ⊢ R ∈ NzRing → ¬ 1 R = 0 ˙
6 nzrring ⊢ R ∈ NzRing → R ∈ Ring
7 6 adantr ⊢ R ∈ NzRing ∧ 0 ˙ ∈ E → R ∈ Ring
8 simpr ⊢ R ∈ NzRing ∧ 0 ˙ ∈ E → 0 ˙ ∈ E
9 eqid ⊢ Base R = Base R
10 9 3 ringidcl ⊢ R ∈ Ring → 1 R ∈ Base R
11 7 10 syl ⊢ R ∈ NzRing ∧ 0 ˙ ∈ E → 1 R ∈ Base R
12 eqid ⊢ ⋅ R = ⋅ R
13 9 12 2 7 11 ringlzd ⊢ R ∈ NzRing ∧ 0 ˙ ∈ E → 0 ˙ ⋅ R 1 R = 0 ˙
14 1 9 12 2 rrgeq0 ⊢ R ∈ Ring ∧ 0 ˙ ∈ E ∧ 1 R ∈ Base R → 0 ˙ ⋅ R 1 R = 0 ˙ ↔ 1 R = 0 ˙
15 14 biimpa ⊢ R ∈ Ring ∧ 0 ˙ ∈ E ∧ 1 R ∈ Base R ∧ 0 ˙ ⋅ R 1 R = 0 ˙ → 1 R = 0 ˙
16 7 8 11 13 15 syl31anc ⊢ R ∈ NzRing ∧ 0 ˙ ∈ E → 1 R = 0 ˙
17 5 16 mtand ⊢ R ∈ NzRing → ¬ 0 ˙ ∈ E