Metamath Proof Explorer


Theorem divcan2d

Description: A cancellation law for division. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 ( 𝜑𝐴 ∈ ℂ )
divcld.2 ( 𝜑𝐵 ∈ ℂ )
divcld.3 ( 𝜑𝐵 ≠ 0 )
Assertion divcan2d ( 𝜑 → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 div1d.1 ( 𝜑𝐴 ∈ ℂ )
2 divcld.2 ( 𝜑𝐵 ∈ ℂ )
3 divcld.3 ( 𝜑𝐵 ≠ 0 )
4 divcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )
5 1 2 3 4 syl3anc ( 𝜑 → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )