Theorem dmdcan2d

Description: Cancellation law for division and multiplication. (Contributed by David Moews, 28-Feb-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	dmdcan2d	$$⊢\phi \to \left(\frac{A}{B}\right)\left(\frac{B}{C}\right)=\frac{A}{C}$$

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 2 4	divcld	$$⊢\phi \to \frac{A}{B}\in ℂ$$
7	2 3 5	divcld	$$⊢\phi \to \frac{B}{C}\in ℂ$$
8	6 7	mulcomd	$$⊢\phi \to \left(\frac{A}{B}\right)\left(\frac{B}{C}\right)=\left(\frac{B}{C}\right)\left(\frac{A}{B}\right)$$
9	1 2 3 4 5	dmdcand	$$⊢\phi \to \left(\frac{B}{C}\right)\left(\frac{A}{B}\right)=\frac{A}{C}$$
10	8 9	eqtrd	$$⊢\phi \to \left(\frac{A}{B}\right)\left(\frac{B}{C}\right)=\frac{A}{C}$$