Metamath Proof Explorer


Theorem divdiv1d

Description: Division into a fraction. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 ( 𝜑𝐴 ∈ ℂ )
divcld.2 ( 𝜑𝐵 ∈ ℂ )
divmuld.3 ( 𝜑𝐶 ∈ ℂ )
divmuld.4 ( 𝜑𝐵 ≠ 0 )
divdiv23d.5 ( 𝜑𝐶 ≠ 0 )
Assertion divdiv1d ( 𝜑 → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 div1d.1 ( 𝜑𝐴 ∈ ℂ )
2 divcld.2 ( 𝜑𝐵 ∈ ℂ )
3 divmuld.3 ( 𝜑𝐶 ∈ ℂ )
4 divmuld.4 ( 𝜑𝐵 ≠ 0 )
5 divdiv23d.5 ( 𝜑𝐶 ≠ 0 )
6 divdiv1 ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )
7 1 2 4 3 5 6 syl122anc ( 𝜑 → ( ( 𝐴 / 𝐵 ) / 𝐶 ) = ( 𝐴 / ( 𝐵 · 𝐶 ) ) )