Metamath Proof Explorer
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 ) |