Theorem divcan5rd

Description: Cancellation of common factor in a ratio. (Contributed by Mario Carneiro, 1-Jan-2017)

	Ref	Expression
Hypotheses	div1d.1	$$⊢\phi \to A\in ℂ$$
	divcld.2	$$⊢\phi \to B\in ℂ$$
	divmuld.3	$$⊢\phi \to C\in ℂ$$
	divmuld.4	$$⊢\phi \to B\ne 0$$
	divdiv23d.5	$$⊢\phi \to C\ne 0$$
Assertion	divcan5rd	$$⊢\phi \to \frac{AC}{BC}=\frac{A}{B}$$

Proof

Step	Hyp	Ref	Expression
1		div1d.1	$$⊢\phi \to A\in ℂ$$
2		divcld.2	$$⊢\phi \to B\in ℂ$$
3		divmuld.3	$$⊢\phi \to C\in ℂ$$
4		divmuld.4	$$⊢\phi \to B\ne 0$$
5		divdiv23d.5	$$⊢\phi \to C\ne 0$$
6	1 3	mulcomd	$$⊢\phi \to AC=CA$$
7	2 3	mulcomd	$$⊢\phi \to BC=CB$$
8	6 7	oveq12d	$$⊢\phi \to \frac{AC}{BC}=\frac{CA}{CB}$$
9	1 2 3 4 5	divcan5d	$$⊢\phi \to \frac{CA}{CB}=\frac{A}{B}$$
10	8 9	eqtrd	$$⊢\phi \to \frac{AC}{BC}=\frac{A}{B}$$