Theorem divcan6d

Description: Cancellation of inverted fractions. (Contributed by Mario Carneiro, 27-May-2016)

	Ref	Expression
Hypotheses	div1d.1	$$⊢\phi \to A\in ℂ$$
	divcld.2	$$⊢\phi \to B\in ℂ$$
	divne0d.3	$$⊢\phi \to A\ne 0$$
	divne0d.4	$$⊢\phi \to B\ne 0$$
Assertion	divcan6d	$$⊢\phi \to \left(\frac{A}{B}\right)\left(\frac{B}{A}\right)=1$$

Proof

Step	Hyp	Ref	Expression
1		div1d.1	$$⊢\phi \to A\in ℂ$$
2		divcld.2	$$⊢\phi \to B\in ℂ$$
3		divne0d.3	$$⊢\phi \to A\ne 0$$
4		divne0d.4	$$⊢\phi \to B\ne 0$$
5		divcan6	$$⊢\left(\left(A\in ℂ\wedge A\ne 0\right)\wedge \left(B\in ℂ\wedge B\ne 0\right)\right)\to \left(\frac{A}{B}\right)\left(\frac{B}{A}\right)=1$$
6	1 3 2 4 5	syl22anc	$$⊢\phi \to \left(\frac{A}{B}\right)\left(\frac{B}{A}\right)=1$$