Metamath Proof Explorer


Theorem sn-remul0ord

Description: A product is zero iff one of its factors are zero. (Contributed by SN, 24-Nov-2025)

Ref Expression
Hypotheses sn-remul0ord.a ⊢ φ → A ∈ ℝ
sn-remul0ord.b ⊢ φ → B ∈ ℝ
Assertion sn-remul0ord ⊢ φ → A ⁢ B = 0 ↔ A = 0 ∨ B = 0

Proof

Step Hyp Ref Expression
1 sn-remul0ord.a ⊢ φ → A ∈ ℝ
2 sn-remul0ord.b ⊢ φ → B ∈ ℝ
3 remul02 ⊢ B ∈ ℝ → 0 ⋅ B = 0
4 2 3 syl ⊢ φ → 0 ⋅ B = 0
5 4 adantr ⊢ φ ∧ B ≠ 0 → 0 ⋅ B = 0
6 5 eqeq2d ⊢ φ ∧ B ≠ 0 → A ⁢ B = 0 ⋅ B ↔ A ⁢ B = 0
7 1 adantr ⊢ φ ∧ B ≠ 0 → A ∈ ℝ
8 0red ⊢ φ ∧ B ≠ 0 → 0 ∈ ℝ
9 2 adantr ⊢ φ ∧ B ≠ 0 → B ∈ ℝ
10 simpr ⊢ φ ∧ B ≠ 0 → B ≠ 0
11 7 8 9 10 remulcan2d ⊢ φ ∧ B ≠ 0 → A ⁢ B = 0 ⋅ B ↔ A = 0
12 6 11 bitr3d ⊢ φ ∧ B ≠ 0 → A ⁢ B = 0 ↔ A = 0
13 12 biimpd ⊢ φ ∧ B ≠ 0 → A ⁢ B = 0 → A = 0
14 13 impancom ⊢ φ ∧ A ⁢ B = 0 → B ≠ 0 → A = 0
15 14 necon1bd ⊢ φ ∧ A ⁢ B = 0 → ¬ A = 0 → B = 0
16 15 orrd ⊢ φ ∧ A ⁢ B = 0 → A = 0 ∨ B = 0
17 16 ex ⊢ φ → A ⁢ B = 0 → A = 0 ∨ B = 0
18 oveq1 ⊢ A = 0 → A ⁢ B = 0 ⋅ B
19 18 eqeq1d ⊢ A = 0 → A ⁢ B = 0 ↔ 0 ⋅ B = 0
20 4 19 syl5ibrcom ⊢ φ → A = 0 → A ⁢ B = 0
21 remul01 ⊢ A ∈ ℝ → A ⋅ 0 = 0
22 1 21 syl ⊢ φ → A ⋅ 0 = 0
23 oveq2 ⊢ B = 0 → A ⁢ B = A ⋅ 0
24 23 eqeq1d ⊢ B = 0 → A ⁢ B = 0 ↔ A ⋅ 0 = 0
25 22 24 syl5ibrcom ⊢ φ → B = 0 → A ⁢ B = 0
26 20 25 jaod ⊢ φ → A = 0 ∨ B = 0 → A ⁢ B = 0
27 17 26 impbid ⊢ φ → A ⁢ B = 0 ↔ A = 0 ∨ B = 0