Metamath Proof Explorer


Theorem dmdcand

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

Ref Expression
Hypotheses div1d.1 ( 𝜑𝐴 ∈ ℂ )
divcld.2 ( 𝜑𝐵 ∈ ℂ )
divmuld.3 ( 𝜑𝐶 ∈ ℂ )
divmuld.4 ( 𝜑𝐵 ≠ 0 )
divdiv23d.5 ( 𝜑𝐶 ≠ 0 )
Assertion dmdcand ( 𝜑 → ( ( 𝐵 / 𝐶 ) · ( 𝐴 / 𝐵 ) ) = ( 𝐴 / 𝐶 ) )

Proof

Step Hyp Ref Expression
1 div1d.1 ( 𝜑𝐴 ∈ ℂ )
2 divcld.2 ( 𝜑𝐵 ∈ ℂ )
3 divmuld.3 ( 𝜑𝐶 ∈ ℂ )
4 divmuld.4 ( 𝜑𝐵 ≠ 0 )
5 divdiv23d.5 ( 𝜑𝐶 ≠ 0 )
6 dmdcan ( ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ∧ 𝐴 ∈ ℂ ) → ( ( 𝐵 / 𝐶 ) · ( 𝐴 / 𝐵 ) ) = ( 𝐴 / 𝐶 ) )
7 2 4 3 5 1 6 syl221anc ( 𝜑 → ( ( 𝐵 / 𝐶 ) · ( 𝐴 / 𝐵 ) ) = ( 𝐴 / 𝐶 ) )