Metamath Proof Explorer


Theorem divcan4i

Description: A cancellation law for division. (Contributed by NM, 18-May-1999)

Ref Expression
Hypotheses divclz.1 𝐴 ∈ ℂ
divclz.2 𝐵 ∈ ℂ
divcl.3 𝐵 ≠ 0
Assertion divcan4i ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴

Proof

Step Hyp Ref Expression
1 divclz.1 𝐴 ∈ ℂ
2 divclz.2 𝐵 ∈ ℂ
3 divcl.3 𝐵 ≠ 0
4 1 2 divcan4zi ( 𝐵 ≠ 0 → ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴 )
5 3 4 ax-mp ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴