Metamath Proof Explorer


Theorem divassd

Description: An associative law for division. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 ( 𝜑𝐴 ∈ ℂ )
divcld.2 ( 𝜑𝐵 ∈ ℂ )
divmuld.3 ( 𝜑𝐶 ∈ ℂ )
divassd.4 ( 𝜑𝐶 ≠ 0 )
Assertion divassd ( 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 divassd.4 ( 𝜑𝐶 ≠ 0 )
5 divass ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( 𝐴 · ( 𝐵 / 𝐶 ) ) )
6 1 2 3 4 5 syl112anc ( 𝜑 → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( 𝐴 · ( 𝐵 / 𝐶 ) ) )