Metamath Proof Explorer


Theorem ftalem6

Description: Lemma for fta : Discharge the auxiliary variables in ftalem5 . (Contributed by Mario Carneiro, 20-Sep-2014) (Proof shortened by AV, 28-Sep-2020)

Ref Expression
Hypotheses ftalem.1 ⊢ A = coeff ⁡ F
ftalem.2 ⊢ N = deg ⁡ F
ftalem.3 ⊢ φ → F ∈ Poly ⁡ S
ftalem.4 ⊢ φ → N ∈ ℕ
ftalem6.5 ⊢ φ → F ⁡ 0 ≠ 0
Assertion ftalem6 ⊢ φ → ∃ x ∈ ℂ F ⁡ x < F ⁡ 0

Proof

Step Hyp Ref Expression
1 ftalem.1 ⊢ A = coeff ⁡ F
2 ftalem.2 ⊢ N = deg ⁡ F
3 ftalem.3 ⊢ φ → F ∈ Poly ⁡ S
4 ftalem.4 ⊢ φ → N ∈ ℕ
5 ftalem6.5 ⊢ φ → F ⁡ 0 ≠ 0
6 fveq2 ⊢ k = n → A ⁡ k = A ⁡ n
7 6 neeq1d ⊢ k = n → A ⁡ k ≠ 0 ↔ A ⁡ n ≠ 0
8 7 cbvrabv ⊢ k ∈ ℕ | A ⁡ k ≠ 0 = n ∈ ℕ | A ⁡ n ≠ 0
9 8 infeq1i ⊢ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < = inf n ∈ ℕ | A ⁡ n ≠ 0 ℝ <
10 eqid ⊢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < = − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ <
11 fveq2 ⊢ r = s → A ⁡ r = A ⁡ s
12 oveq2 ⊢ r = s → − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r = − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < s
13 11 12 oveq12d ⊢ r = s → A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r = A ⁡ s ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < s
14 13 fveq2d ⊢ r = s → A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r = A ⁡ s ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < s
15 14 cbvsumv ⊢ ∑ r = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r = ∑ s = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ s ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < s
16 15 oveq1i ⊢ ∑ r = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r + 1 = ∑ s = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ s ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < s + 1
17 16 oveq2i ⊢ F ⁡ 0 ∑ r = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r + 1 = F ⁡ 0 ∑ s = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ s ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < s + 1
18 eqid ⊢ if 1 ≤ F ⁡ 0 ∑ r = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r + 1 1 F ⁡ 0 ∑ r = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r + 1 = if 1 ≤ F ⁡ 0 ∑ r = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r + 1 1 F ⁡ 0 ∑ r = inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < + 1 N A ⁡ r ⁢ − F ⁡ 0 A ⁡ inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < 1 inf k ∈ ℕ | A ⁡ k ≠ 0 ℝ < r + 1
19 1 2 3 4 5 9 10 17 18 ftalem5 ⊢ φ → ∃ x ∈ ℂ F ⁡ x < F ⁡ 0