Theorem divcan5d

Description: Cancellation of common factor in a ratio. (Contributed by Mario Carneiro, 27-May-2016)

	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	divcan5d	$$⊢\phi \to \frac{CA}{CB}=\frac{A}{B}$$

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		divcan5	$$⊢\left(A\in ℂ\wedge \left(B\in ℂ\wedge B\ne 0\right)\wedge \left(C\in ℂ\wedge C\ne 0\right)\right)\to \frac{CA}{CB}=\frac{A}{B}$$
7	1 2 4 3 5 6	syl122anc	$$⊢\phi \to \frac{CA}{CB}=\frac{A}{B}$$