Metamath Proof Explorer
Description: A fraction of complex numbers is zero iff its numerator is. Deduction
form of diveq0 . (Contributed by David Moews, 28-Feb-2017)
|
|
Ref |
Expression |
|
Hypotheses |
div1d.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
|
|
divcld.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
|
|
divcld.3 |
⊢ ( 𝜑 → 𝐵 ≠ 0 ) |
|
Assertion |
diveq0ad |
⊢ ( 𝜑 → ( ( 𝐴 / 𝐵 ) = 0 ↔ 𝐴 = 0 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
div1d.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
2 |
|
divcld.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
3 |
|
divcld.3 |
⊢ ( 𝜑 → 𝐵 ≠ 0 ) |
4 |
|
diveq0 |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) = 0 ↔ 𝐴 = 0 ) ) |
5 |
1 2 3 4
|
syl3anc |
⊢ ( 𝜑 → ( ( 𝐴 / 𝐵 ) = 0 ↔ 𝐴 = 0 ) ) |