Metamath Proof Explorer


Theorem divcan5rd

Description: Cancellation of common factor in a ratio. (Contributed by Mario Carneiro, 1-Jan-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 divcan5rd ( 𝜑 → ( ( 𝐴 · 𝐶 ) / ( 𝐵 · 𝐶 ) ) = ( 𝐴 / 𝐵 ) )

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 3 mulcomd ⊢ ( 𝜑 → ( 𝐴 · 𝐶 ) = ( 𝐶 · 𝐴 ) )
7 2 3 mulcomd ⊢ ( 𝜑 → ( 𝐵 · 𝐶 ) = ( 𝐶 · 𝐵 ) )
8 6 7 oveq12d ⊢ ( 𝜑 → ( ( 𝐴 · 𝐶 ) / ( 𝐵 · 𝐶 ) ) = ( ( 𝐶 · 𝐴 ) / ( 𝐶 · 𝐵 ) ) )
9 1 2 3 4 5 divcan5d ⊢ ( 𝜑 → ( ( 𝐶 · 𝐴 ) / ( 𝐶 · 𝐵 ) ) = ( 𝐴 / 𝐵 ) )
10 8 9 eqtrd ⊢ ( 𝜑 → ( ( 𝐴 · 𝐶 ) / ( 𝐵 · 𝐶 ) ) = ( 𝐴 / 𝐵 ) )