Metamath Proof Explorer


Theorem divcan2i

Description: A cancellation law for division. (Contributed by NM, 9-Feb-1995)

Ref Expression
Hypotheses divclz.1 𝐴 ∈ ℂ
divclz.2 𝐵 ∈ ℂ
divcl.3 𝐵 ≠ 0
Assertion divcan2i ( 𝐵 · ( 𝐴 / 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 divcan2zi ( 𝐵 ≠ 0 → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )
5 3 4 ax-mp ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴