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 ) ) )