Metamath Proof Explorer


Theorem divmuld

Description: Relationship between division and multiplication. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 ⊢ φ → A ∈ ℂ
divcld.2 ⊢ φ → B ∈ ℂ
divmuld.3 ⊢ φ → C ∈ ℂ
divmuld.4 ⊢ φ → B ≠ 0
Assertion divmuld ⊢ φ → A B = C ↔ B ⁢ C = A

Proof

Step Hyp Ref Expression
1 div1d.1 ⊢ φ → A ∈ ℂ
2 divcld.2 ⊢ φ → B ∈ ℂ
3 divmuld.3 ⊢ φ → C ∈ ℂ
4 divmuld.4 ⊢ φ → B ≠ 0
5 divmul ⊢ A ∈ ℂ ∧ C ∈ ℂ ∧ B ∈ ℂ ∧ B ≠ 0 → A B = C ↔ B ⁢ C = A
6 1 3 2 4 5 syl112anc ⊢ φ → A B = C ↔ B ⁢ C = A