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