Theorem sqrtmsqd

Description: Square root of square. (Contributed by Mario Carneiro, 29-May-2016)

	Ref	Expression
Hypotheses	resqrcld.1	⊢ ( 𝜑 → 𝐴 ∈ ℝ )
	resqrcld.2	⊢ ( 𝜑 → 0 ≤ 𝐴 )
Assertion	sqrtmsqd	⊢ ( 𝜑 → ( √ ‘ ( 𝐴 · 𝐴 ) ) = 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		resqrcld.1	⊢ ( 𝜑 → 𝐴 ∈ ℝ )
2		resqrcld.2	⊢ ( 𝜑 → 0 ≤ 𝐴 )
3		sqrtmsq	⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 · 𝐴 ) ) = 𝐴 )
4	1 2 3	syl2anc	⊢ ( 𝜑 → ( √ ‘ ( 𝐴 · 𝐴 ) ) = 𝐴 )