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Metamath Proof Explorer

Theorem sqrtmsqd

Description: Square root of square. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses resqrcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
resqrcld.2 ⊢ ( 𝜑 → 0 ≤ 𝐴 )
Assertion sqrtmsqd ( 𝜑 → ( √ ‘ ( 𝐴 · 𝐴 ) ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 resqrcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
2 resqrcld.2 ⊢ ( 𝜑 → 0 ≤ 𝐴 )
3 sqrtmsq ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 ≤ 𝐴 ) → ( √ ‘ ( 𝐴 · 𝐴 ) ) = 𝐴 )
4 1 2 3 syl2anc ⊢ ( 𝜑 → ( √ ‘ ( 𝐴 · 𝐴 ) ) = 𝐴 )