Metamath Proof Explorer


Theorem sqrtthlem

Description: Lemma for sqrtth . (Contributed by Mario Carneiro, 10-Jul-2013)

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

Proof

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