Metamath Proof Explorer


Theorem diveq0ad

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 ) )