Metamath Proof Explorer


Theorem sqrtsq2d

Description: Relationship between square root and squares. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses resqrcld.1 ⊢ φ → A ∈ ℝ
resqrcld.2 ⊢ φ → 0 ≤ A
sqr11d.3 ⊢ φ → B ∈ ℝ
sqr11d.4 ⊢ φ → 0 ≤ B
Assertion sqrtsq2d ⊢ φ → A = B ↔ A = B 2

Proof

Step Hyp Ref Expression
1 resqrcld.1 ⊢ φ → A ∈ ℝ
2 resqrcld.2 ⊢ φ → 0 ≤ A
3 sqr11d.3 ⊢ φ → B ∈ ℝ
4 sqr11d.4 ⊢ φ → 0 ≤ B
5 sqrtsq2 ⊢ A ∈ ℝ ∧ 0 ≤ A ∧ B ∈ ℝ ∧ 0 ≤ B → A = B ↔ A = B 2
6 1 2 3 4 5 syl22anc ⊢ φ → A = B ↔ A = B 2