Metamath Proof Explorer


Theorem divcan4zi

Description: A cancellation law for division. (Contributed by NM, 12-Oct-1999)

Ref Expression
Hypotheses divclz.1 𝐴 ∈ ℂ
divclz.2 𝐵 ∈ ℂ
Assertion divcan4zi ( 𝐵 ≠ 0 → ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 divclz.1 𝐴 ∈ ℂ
2 divclz.2 𝐵 ∈ ℂ
3 divcan4 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴 )
4 1 2 3 mp3an12 ( 𝐵 ≠ 0 → ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴 )