Metamath Proof Explorer


Theorem mulassd

Description: Associative law for multiplication. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses addcld.1 ⊢ φ → A ∈ ℂ
addcld.2 ⊢ φ → B ∈ ℂ
addassd.3 ⊢ φ → C ∈ ℂ
Assertion mulassd ⊢ φ → A ⁢ B ⁢ C = A ⁢ B ⁢ C

Proof

Step Hyp Ref Expression
1 addcld.1 ⊢ φ → A ∈ ℂ
2 addcld.2 ⊢ φ → B ∈ ℂ
3 addassd.3 ⊢ φ → C ∈ ℂ
4 mulass ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A ⁢ B ⁢ C = A ⁢ B ⁢ C
5 1 2 3 4 syl3anc ⊢ φ → A ⁢ B ⁢ C = A ⁢ B ⁢ C