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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 ) )