Theorem sqrtmuld

Description: Square root distributes over multiplication. (Contributed by Mario Carneiro, 29-May-2016)

	Ref	Expression
Hypotheses	resqrcld.1	$$⊢\phi \to A\in ℝ$$
	resqrcld.2	$$⊢\phi \to 0\le A$$
	sqr11d.3	$$⊢\phi \to B\in ℝ$$
	sqr11d.4	$$⊢\phi \to 0\le B$$
Assertion	sqrtmuld	$$⊢\phi \to \sqrt{AB}=\sqrt{A}\sqrt{B}$$

Proof

Step	Hyp	Ref	Expression
1		resqrcld.1	$$⊢\phi \to A\in ℝ$$
2		resqrcld.2	$$⊢\phi \to 0\le A$$
3		sqr11d.3	$$⊢\phi \to B\in ℝ$$
4		sqr11d.4	$$⊢\phi \to 0\le B$$
5		sqrtmul	$$⊢\left(\left(A\in ℝ\wedge 0\le A\right)\wedge \left(B\in ℝ\wedge 0\le B\right)\right)\to \sqrt{AB}=\sqrt{A}\sqrt{B}$$
6	1 2 3 4 5	syl22anc	$$⊢\phi \to \sqrt{AB}=\sqrt{A}\sqrt{B}$$