Metamath Proof Explorer


Theorem sqrtsqi

Description: Square root of square. (Contributed by NM, 11-Aug-1999)

Ref Expression
Hypothesis sqrtthi.1 𝐴 ∈ ℝ
Assertion sqrtsqi ( 0 ≤ 𝐴 → ( √ ‘ ( 𝐴 ↑ 2 ) ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 sqrtthi.1 𝐴 ∈ ℝ
2 sqrtsq ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 ↑ 2 ) ) = 𝐴 )
3 1 2 mpan ( 0 ≤ 𝐴 → ( √ ‘ ( 𝐴 ↑ 2 ) ) = 𝐴 )