Metamath Proof Explorer


Theorem ringunitnzdiv

Description: In a unitary ring, a unit is not a zero divisor. (Contributed by AV, 7-Mar-2025)

Ref Expression
Hypotheses ringunitnzdiv.b ⊢ B = Base R
ringunitnzdiv.z ⊢ 0 ˙ = 0 R
ringunitnzdiv.t ⊢ · ˙ = ⋅ R
ringunitnzdiv.r ⊢ φ → R ∈ Ring
ringunitnzdiv.y ⊢ φ → Y ∈ B
ringunitnzdiv.x ⊢ φ → X ∈ Unit ⁡ R
Assertion ringunitnzdiv ⊢ φ → X · ˙ Y = 0 ˙ ↔ Y = 0 ˙

Proof

Step Hyp Ref Expression
1 ringunitnzdiv.b ⊢ B = Base R
2 ringunitnzdiv.z ⊢ 0 ˙ = 0 R
3 ringunitnzdiv.t ⊢ · ˙ = ⋅ R
4 ringunitnzdiv.r ⊢ φ → R ∈ Ring
5 ringunitnzdiv.y ⊢ φ → Y ∈ B
6 ringunitnzdiv.x ⊢ φ → X ∈ Unit ⁡ R
7 eqid ⊢ 1 R = 1 R
8 eqid ⊢ Unit ⁡ R = Unit ⁡ R
9 1 8 unitcl ⊢ X ∈ Unit ⁡ R → X ∈ B
10 6 9 syl ⊢ φ → X ∈ B
11 eqid ⊢ inv r ⁡ R = inv r ⁡ R
12 8 11 1 ringinvcl ⊢ R ∈ Ring ∧ X ∈ Unit ⁡ R → inv r ⁡ R ⁡ X ∈ B
13 4 6 12 syl2anc ⊢ φ → inv r ⁡ R ⁡ X ∈ B
14 oveq1 ⊢ e = inv r ⁡ R ⁡ X → e · ˙ X = inv r ⁡ R ⁡ X · ˙ X
15 14 eqeq1d ⊢ e = inv r ⁡ R ⁡ X → e · ˙ X = 1 R ↔ inv r ⁡ R ⁡ X · ˙ X = 1 R
16 15 adantl ⊢ φ ∧ e = inv r ⁡ R ⁡ X → e · ˙ X = 1 R ↔ inv r ⁡ R ⁡ X · ˙ X = 1 R
17 8 11 3 7 unitlinv ⊢ R ∈ Ring ∧ X ∈ Unit ⁡ R → inv r ⁡ R ⁡ X · ˙ X = 1 R
18 4 6 17 syl2anc ⊢ φ → inv r ⁡ R ⁡ X · ˙ X = 1 R
19 13 16 18 rspcedvd ⊢ φ → ∃ e ∈ B e · ˙ X = 1 R
20 1 3 7 2 4 10 19 5 ringinvnzdiv ⊢ φ → X · ˙ Y = 0 ˙ ↔ Y = 0 ˙