Metamath Proof Explorer


Theorem efopnlem1

Description: Lemma for efopn . (Contributed by Mario Carneiro, 23-Apr-2015) (Revised by Mario Carneiro, 8-Sep-2015)

Ref Expression
Assertion efopnlem1 ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → ℑ ⁡ A < π

Proof

Step Hyp Ref Expression
1 simpr ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → A ∈ 0 ball ⁡ abs ∘ − R
2 rpxr ⊢ R ∈ ℝ + → R ∈ ℝ *
3 2 ad2antrr ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → R ∈ ℝ *
4 eqid ⊢ abs ∘ − = abs ∘ −
5 4 cnbl0 ⊢ R ∈ ℝ * → abs -1 0 R = 0 ball ⁡ abs ∘ − R
6 3 5 syl ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → abs -1 0 R = 0 ball ⁡ abs ∘ − R
7 1 6 eleqtrrd ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → A ∈ abs -1 0 R
8 absf ⊢ abs : ℂ ⟶ ℝ
9 ffn ⊢ abs : ℂ ⟶ ℝ → abs Fn ℂ
10 elpreima ⊢ abs Fn ℂ → A ∈ abs -1 0 R ↔ A ∈ ℂ ∧ A ∈ 0 R
11 8 9 10 mp2b ⊢ A ∈ abs -1 0 R ↔ A ∈ ℂ ∧ A ∈ 0 R
12 11 simplbi ⊢ A ∈ abs -1 0 R → A ∈ ℂ
13 7 12 syl ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → A ∈ ℂ
14 13 imcld ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → ℑ ⁡ A ∈ ℝ
15 14 recnd ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → ℑ ⁡ A ∈ ℂ
16 15 abscld ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → ℑ ⁡ A ∈ ℝ
17 rpre ⊢ R ∈ ℝ + → R ∈ ℝ
18 17 ad2antrr ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → R ∈ ℝ
19 pire ⊢ π ∈ ℝ
20 19 a1i ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → π ∈ ℝ
21 13 abscld ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → A ∈ ℝ
22 absimle ⊢ A ∈ ℂ → ℑ ⁡ A ≤ A
23 13 22 syl ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → ℑ ⁡ A ≤ A
24 11 simprbi ⊢ A ∈ abs -1 0 R → A ∈ 0 R
25 7 24 syl ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → A ∈ 0 R
26 0re ⊢ 0 ∈ ℝ
27 elico2 ⊢ 0 ∈ ℝ ∧ R ∈ ℝ * → A ∈ 0 R ↔ A ∈ ℝ ∧ 0 ≤ A ∧ A < R
28 26 3 27 sylancr ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → A ∈ 0 R ↔ A ∈ ℝ ∧ 0 ≤ A ∧ A < R
29 25 28 mpbid ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → A ∈ ℝ ∧ 0 ≤ A ∧ A < R
30 29 simp3d ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → A < R
31 16 21 18 23 30 lelttrd ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → ℑ ⁡ A < R
32 simplr ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → R < π
33 16 18 20 31 32 lttrd ⊢ R ∈ ℝ + ∧ R < π ∧ A ∈ 0 ball ⁡ abs ∘ − R → ℑ ⁡ A < π