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Metamath Proof Explorer

Theorem sqrtsq

Description: Square root of square. (Contributed by NM, 14-Jan-2006) (Revised by Mario Carneiro, 29-May-2016)

Ref Expression
Assertion sqrtsq ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 ↑ 2 ) ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 eqid ⊢ ( 𝐴 ↑ 2 ) = ( 𝐴 ↑ 2 )
2 resqcl ⊢ ( 𝐴 ∈ ℝ → ( 𝐴 ↑ 2 ) ∈ ℝ )
3 sqge0 ⊢ ( 𝐴 ∈ ℝ → 0 ≤ ( 𝐴 ↑ 2 ) )
4 2 3 jca ⊢ ( 𝐴 ∈ ℝ → ( ( 𝐴 ↑ 2 ) ∈ ℝ ∧ 0 ≤ ( 𝐴 ↑ 2 ) ) )
5 4 adantr ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( ( 𝐴 ↑ 2 ) ∈ ℝ ∧ 0 ≤ ( 𝐴 ↑ 2 ) ) )
6 sqrtsq2 ⊢ ( ( ( ( 𝐴 ↑ 2 ) ∈ ℝ ∧ 0 ≤ ( 𝐴 ↑ 2 ) ) ∧ ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) ) → ( ( √ ‘ ( 𝐴 ↑ 2 ) ) = 𝐴 ↔ ( 𝐴 ↑ 2 ) = ( 𝐴 ↑ 2 ) ) )
7 5 6 mpancom ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( ( √ ‘ ( 𝐴 ↑ 2 ) ) = 𝐴 ↔ ( 𝐴 ↑ 2 ) = ( 𝐴 ↑ 2 ) ) )
8 1 7 mpbiri ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 ↑ 2 ) ) = 𝐴 )