Metamath Proof Explorer


Theorem divdiv3d

Description: Division into a fraction. (Contributed by Glauco Siliprandi, 24-Dec-2020)

Ref Expression
Hypotheses divdiv3d.1 ( 𝜑𝐴 ∈ ℂ )
divdiv3d.2 ( 𝜑𝐵 ∈ ℂ )
divdiv3d.3 ( 𝜑𝐶 ∈ ℂ )
divdiv3d.4 ( 𝜑𝐵 ≠ 0 )
divdiv3d.5 ( 𝜑𝐶 ≠ 0 )
Assertion divdiv3d ( 𝜑 → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐶 · 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 divdiv3d.1 ( 𝜑𝐴 ∈ ℂ )
2 divdiv3d.2 ( 𝜑𝐵 ∈ ℂ )
3 divdiv3d.3 ( 𝜑𝐶 ∈ ℂ )
4 divdiv3d.4 ( 𝜑𝐵 ≠ 0 )
5 divdiv3d.5 ( 𝜑𝐶 ≠ 0 )
6 1 2 3 4 5 divdiv1d ( 𝜑 → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )
7 2 3 mulcomd ( 𝜑 → ( 𝐵 · 𝐶 ) = ( 𝐶 · 𝐵 ) )
8 7 oveq2d ( 𝜑 → ( 𝐴 / ( 𝐵 · 𝐶 ) ) = ( 𝐴 / ( 𝐶 · 𝐵 ) ) )
9 6 8 eqtrd ( 𝜑 → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐶 · 𝐵 ) ) )