Theorem absmulrposd

Description: Specialization of absmuld with absidd . (Contributed by Stanislas Polu, 9-Mar-2020)

	Ref	Expression
Hypotheses	absmulrposd.1	$$⊢\phi \to 0\le A$$
	absmulrposd.2	$$⊢\phi \to A\in ℝ$$
	absmulrposd.3	$$⊢\phi \to B\in ℝ$$
Assertion	absmulrposd	$$⊢\phi \to \left|AB\right|=A\left|B\right|$$

Proof

Step	Hyp	Ref	Expression
1		absmulrposd.1	$$⊢\phi \to 0\le A$$
2		absmulrposd.2	$$⊢\phi \to A\in ℝ$$
3		absmulrposd.3	$$⊢\phi \to B\in ℝ$$
4	2	recnd	$$⊢\phi \to A\in ℂ$$
5	3	recnd	$$⊢\phi \to B\in ℂ$$
6	4 5	absmuld	$$⊢\phi \to \left|AB\right|=\left|A\right|\left|B\right|$$
7	2 1	absidd	$$⊢\phi \to \left|A\right|=A$$
8	7	oveq1d	$$⊢\phi \to \left|A\right|\left|B\right|=A\left|B\right|$$
9	6 8	eqtrd	$$⊢\phi \to \left|AB\right|=A\left|B\right|$$