Metamath Proof Explorer


Theorem divmul2

Description: Relationship between division and multiplication. (Contributed by NM, 7-Feb-2006)

Ref Expression
Assertion divmul2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) = 𝐵𝐴 = ( 𝐶 · 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 divmul ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) = 𝐵 ↔ ( 𝐶 · 𝐵 ) = 𝐴 ) )
2 eqcom ( ( 𝐶 · 𝐵 ) = 𝐴𝐴 = ( 𝐶 · 𝐵 ) )
3 1 2 bitrdi ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) = 𝐵𝐴 = ( 𝐶 · 𝐵 ) ) )