Metamath Proof Explorer


Theorem divasszi

Description: An associative law for division. (Contributed by NM, 12-Aug-1999)

Ref Expression
Hypotheses divclz.1 𝐴 ∈ ℂ
divclz.2 𝐵 ∈ ℂ
divmulz.3 𝐶 ∈ ℂ
Assertion divasszi ( 𝐶 ≠ 0 → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( 𝐴 · ( 𝐵 / 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 divclz.1 𝐴 ∈ ℂ
2 divclz.2 𝐵 ∈ ℂ
3 divmulz.3 𝐶 ∈ ℂ
4 divass ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( 𝐴 · ( 𝐵 / 𝐶 ) ) )
5 1 2 4 mp3an12 ( ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( 𝐴 · ( 𝐵 / 𝐶 ) ) )
6 3 5 mpan ( 𝐶 ≠ 0 → ( ( 𝐴 · 𝐵 ) / 𝐶 ) = ( 𝐴 · ( 𝐵 / 𝐶 ) ) )