Metamath Proof Explorer


Theorem bndatandm

Description: A point in the open unit disk is in the domain of the arctangent. (Contributed by Mario Carneiro, 5-Apr-2015)

Ref Expression
Assertion bndatandm ⊢ A ∈ ℂ ∧ A < 1 → A ∈ dom ⁡ arctan

Proof

Step Hyp Ref Expression
1 simpl ⊢ A ∈ ℂ ∧ A < 1 → A ∈ ℂ
2 sqcl ⊢ A ∈ ℂ → A 2 ∈ ℂ
3 2 adantr ⊢ A ∈ ℂ ∧ A < 1 → A 2 ∈ ℂ
4 3 abscld ⊢ A ∈ ℂ ∧ A < 1 → A 2 ∈ ℝ
5 2nn0 ⊢ 2 ∈ ℕ 0
6 absexp ⊢ A ∈ ℂ ∧ 2 ∈ ℕ 0 → A 2 = A 2
7 1 5 6 sylancl ⊢ A ∈ ℂ ∧ A < 1 → A 2 = A 2
8 simpr ⊢ A ∈ ℂ ∧ A < 1 → A < 1
9 abscl ⊢ A ∈ ℂ → A ∈ ℝ
10 9 adantr ⊢ A ∈ ℂ ∧ A < 1 → A ∈ ℝ
11 1red ⊢ A ∈ ℂ ∧ A < 1 → 1 ∈ ℝ
12 absge0 ⊢ A ∈ ℂ → 0 ≤ A
13 12 adantr ⊢ A ∈ ℂ ∧ A < 1 → 0 ≤ A
14 0le1 ⊢ 0 ≤ 1
15 14 a1i ⊢ A ∈ ℂ ∧ A < 1 → 0 ≤ 1
16 10 11 13 15 lt2sqd ⊢ A ∈ ℂ ∧ A < 1 → A < 1 ↔ A 2 < 1 2
17 8 16 mpbid ⊢ A ∈ ℂ ∧ A < 1 → A 2 < 1 2
18 sq1 ⊢ 1 2 = 1
19 17 18 breqtrdi ⊢ A ∈ ℂ ∧ A < 1 → A 2 < 1
20 7 19 eqbrtrd ⊢ A ∈ ℂ ∧ A < 1 → A 2 < 1
21 4 20 ltned ⊢ A ∈ ℂ ∧ A < 1 → A 2 ≠ 1
22 fveq2 ⊢ A 2 = − 1 → A 2 = − 1
23 ax-1cn ⊢ 1 ∈ ℂ
24 23 absnegi ⊢ − 1 = 1
25 abs1 ⊢ 1 = 1
26 24 25 eqtri ⊢ − 1 = 1
27 22 26 eqtrdi ⊢ A 2 = − 1 → A 2 = 1
28 27 necon3i ⊢ A 2 ≠ 1 → A 2 ≠ − 1
29 21 28 syl ⊢ A ∈ ℂ ∧ A < 1 → A 2 ≠ − 1
30 atandm3 ⊢ A ∈ dom ⁡ arctan ↔ A ∈ ℂ ∧ A 2 ≠ − 1
31 1 29 30 sylanbrc ⊢ A ∈ ℂ ∧ A < 1 → A ∈ dom ⁡ arctan