Metamath Proof Explorer


Theorem dmdcan2d

Description: Cancellation law for division and multiplication. (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses div1d.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
divcld.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
divmuld.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
divmuld.4 ⊢ ( 𝜑 → 𝐵 ≠ 0 )
divdiv23d.5 ⊢ ( 𝜑 → 𝐶 ≠ 0 )
Assertion dmdcan2d ( 𝜑 → ( ( 𝐴 / 𝐵 ) · ( 𝐵 / 𝐶 ) ) = ( 𝐴 / 𝐶 ) )

Proof

Step Hyp Ref Expression
1 div1d.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 divcld.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 divmuld.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4 divmuld.4 ⊢ ( 𝜑 → 𝐵 ≠ 0 )
5 divdiv23d.5 ⊢ ( 𝜑 → 𝐶 ≠ 0 )
6 1 2 4 divcld ⊢ ( 𝜑 → ( 𝐴 / 𝐵 ) ∈ ℂ )
7 2 3 5 divcld ⊢ ( 𝜑 → ( 𝐵 / 𝐶 ) ∈ ℂ )
8 6 7 mulcomd ⊢ ( 𝜑 → ( ( 𝐴 / 𝐵 ) · ( 𝐵 / 𝐶 ) ) = ( ( 𝐵 / 𝐶 ) · ( 𝐴 / 𝐵 ) ) )
9 1 2 3 4 5 dmdcand ⊢ ( 𝜑 → ( ( 𝐵 / 𝐶 ) · ( 𝐴 / 𝐵 ) ) = ( 𝐴 / 𝐶 ) )
10 8 9 eqtrd ⊢ ( 𝜑 → ( ( 𝐴 / 𝐵 ) · ( 𝐵 / 𝐶 ) ) = ( 𝐴 / 𝐶 ) )