Theorem divmul24d

Description: Swap the numerators in the product of two ratios. (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	$$⊢\phi \to C\in ℂ$$
	divmuldivd.4	$$⊢\phi \to D\in ℂ$$
	divmuldivd.5	$$⊢\phi \to B\ne 0$$
	divmuldivd.6	$$⊢\phi \to D\ne 0$$
Assertion	divmul24d	$$⊢\phi \to \left(\frac{A}{B}\right)\left(\frac{C}{D}\right)=\left(\frac{A}{D}\right)\left(\frac{C}{B}\right)$$

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		divmuldivd.4	$$⊢\phi \to D\in ℂ$$
5		divmuldivd.5	$$⊢\phi \to B\ne 0$$
6		divmuldivd.6	$$⊢\phi \to D\ne 0$$
7	2 5	jca	$$⊢\phi \to \left(B\in ℂ\wedge B\ne 0\right)$$
8	4 6	jca	$$⊢\phi \to \left(D\in ℂ\wedge D\ne 0\right)$$
9		divmul24	$$⊢\left(\left(A\in ℂ\wedge C\in ℂ\right)\wedge \left(\left(B\in ℂ\wedge B\ne 0\right)\wedge \left(D\in ℂ\wedge D\ne 0\right)\right)\right)\to \left(\frac{A}{B}\right)\left(\frac{C}{D}\right)=\left(\frac{A}{D}\right)\left(\frac{C}{B}\right)$$
10	1 3 7 8 9	syl22anc	$$⊢\phi \to \left(\frac{A}{B}\right)\left(\frac{C}{D}\right)=\left(\frac{A}{D}\right)\left(\frac{C}{B}\right)$$