Metamath Proof Explorer


Theorem resqrtthlem

Description: Lemma for resqrtth . (Contributed by Mario Carneiro, 9-Jul-2013)

Ref Expression
Assertion resqrtthlem ⊢ A ∈ ℝ ∧ 0 ≤ A → A 2 = A ∧ 0 ≤ ℜ ⁡ A ∧ i ⁢ A ∉ ℝ +

Proof

Step Hyp Ref Expression
1 recn ⊢ A ∈ ℝ → A ∈ ℂ
2 sqrtval ⊢ A ∈ ℂ → A = ι x ∈ ℂ | x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ +
3 2 eqcomd ⊢ A ∈ ℂ → ι x ∈ ℂ | x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ + = A
4 1 3 syl ⊢ A ∈ ℝ → ι x ∈ ℂ | x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ + = A
5 4 adantr ⊢ A ∈ ℝ ∧ 0 ≤ A → ι x ∈ ℂ | x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ + = A
6 resqrtcl ⊢ A ∈ ℝ ∧ 0 ≤ A → A ∈ ℝ
7 6 recnd ⊢ A ∈ ℝ ∧ 0 ≤ A → A ∈ ℂ
8 resqreu ⊢ A ∈ ℝ ∧ 0 ≤ A → ∃! x ∈ ℂ x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ +
9 oveq1 ⊢ x = A → x 2 = A 2
10 9 eqeq1d ⊢ x = A → x 2 = A ↔ A 2 = A
11 fveq2 ⊢ x = A → ℜ ⁡ x = ℜ ⁡ A
12 11 breq2d ⊢ x = A → 0 ≤ ℜ ⁡ x ↔ 0 ≤ ℜ ⁡ A
13 oveq2 ⊢ x = A → i ⁢ x = i ⁢ A
14 neleq1 ⊢ i ⁢ x = i ⁢ A → i ⁢ x ∉ ℝ + ↔ i ⁢ A ∉ ℝ +
15 13 14 syl ⊢ x = A → i ⁢ x ∉ ℝ + ↔ i ⁢ A ∉ ℝ +
16 10 12 15 3anbi123d ⊢ x = A → x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ + ↔ A 2 = A ∧ 0 ≤ ℜ ⁡ A ∧ i ⁢ A ∉ ℝ +
17 16 riota2 ⊢ A ∈ ℂ ∧ ∃! x ∈ ℂ x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ + → A 2 = A ∧ 0 ≤ ℜ ⁡ A ∧ i ⁢ A ∉ ℝ + ↔ ι x ∈ ℂ | x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ + = A
18 7 8 17 syl2anc ⊢ A ∈ ℝ ∧ 0 ≤ A → A 2 = A ∧ 0 ≤ ℜ ⁡ A ∧ i ⁢ A ∉ ℝ + ↔ ι x ∈ ℂ | x 2 = A ∧ 0 ≤ ℜ ⁡ x ∧ i ⁢ x ∉ ℝ + = A
19 5 18 mpbird ⊢ A ∈ ℝ ∧ 0 ≤ A → A 2 = A ∧ 0 ≤ ℜ ⁡ A ∧ i ⁢ A ∉ ℝ +