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Metamath Proof Explorer

Theorem mulcomd

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

Ref Expression
Hypotheses addcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
addcld.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
Assertion mulcomd ( 𝜑 → ( 𝐴 · 𝐵 ) = ( 𝐵 · 𝐴 ) )

Proof

Step Hyp Ref Expression
1 addcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 addcld.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 mulcom ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 · 𝐵 ) = ( 𝐵 · 𝐴 ) )
4 1 2 3 syl2anc ⊢ ( 𝜑 → ( 𝐴 · 𝐵 ) = ( 𝐵 · 𝐴 ) )