Metamath Proof Explorer


Theorem ply1termlem

Description: Lemma for ply1term . (Contributed by Mario Carneiro, 26-Jul-2014)

Ref Expression
Hypothesis ply1term.1 ⊢ F = z ∈ ℂ ⟼ A ⁢ z N
Assertion ply1termlem ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → F = z ∈ ℂ ⟼ ∑ k = 0 N if k = N A 0 ⁢ z k

Proof

Step Hyp Ref Expression
1 ply1term.1 ⊢ F = z ∈ ℂ ⟼ A ⁢ z N
2 simplr ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → N ∈ ℕ 0
3 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
4 2 3 eleqtrdi ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → N ∈ ℤ ≥ 0
5 fzss1 ⊢ N ∈ ℤ ≥ 0 → N … N ⊆ 0 … N
6 4 5 syl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → N … N ⊆ 0 … N
7 elfz1eq ⊢ k ∈ N … N → k = N
8 7 adantl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → k = N
9 iftrue ⊢ k = N → if k = N A 0 = A
10 8 9 syl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → if k = N A 0 = A
11 simpll ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → A ∈ ℂ
12 11 adantr ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → A ∈ ℂ
13 10 12 eqeltrd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → if k = N A 0 ∈ ℂ
14 simplr ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → z ∈ ℂ
15 2 adantr ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → N ∈ ℕ 0
16 8 15 eqeltrd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → k ∈ ℕ 0
17 14 16 expcld ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → z k ∈ ℂ
18 13 17 mulcld ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ N … N → if k = N A 0 ⁢ z k ∈ ℂ
19 eldifn ⊢ k ∈ 0 … N ∖ N … N → ¬ k ∈ N … N
20 19 adantl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → ¬ k ∈ N … N
21 2 adantr ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → N ∈ ℕ 0
22 21 nn0zd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → N ∈ ℤ
23 fzsn ⊢ N ∈ ℤ → N … N = N
24 23 eleq2d ⊢ N ∈ ℤ → k ∈ N … N ↔ k ∈ N
25 elsn2g ⊢ N ∈ ℤ → k ∈ N ↔ k = N
26 24 25 bitrd ⊢ N ∈ ℤ → k ∈ N … N ↔ k = N
27 22 26 syl ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → k ∈ N … N ↔ k = N
28 20 27 mtbid ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → ¬ k = N
29 28 iffalsed ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → if k = N A 0 = 0
30 29 oveq1d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → if k = N A 0 ⁢ z k = 0 ⋅ z k
31 simpr ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → z ∈ ℂ
32 eldifi ⊢ k ∈ 0 … N ∖ N … N → k ∈ 0 … N
33 elfznn0 ⊢ k ∈ 0 … N → k ∈ ℕ 0
34 32 33 syl ⊢ k ∈ 0 … N ∖ N … N → k ∈ ℕ 0
35 expcl ⊢ z ∈ ℂ ∧ k ∈ ℕ 0 → z k ∈ ℂ
36 31 34 35 syl2an ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → z k ∈ ℂ
37 36 mul02d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → 0 ⋅ z k = 0
38 30 37 eqtrd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … N ∖ N … N → if k = N A 0 ⁢ z k = 0
39 fzfid ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → 0 … N ∈ Fin
40 6 18 38 39 fsumss ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → ∑ k = N N if k = N A 0 ⁢ z k = ∑ k = 0 N if k = N A 0 ⁢ z k
41 2 nn0zd ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → N ∈ ℤ
42 31 2 expcld ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → z N ∈ ℂ
43 11 42 mulcld ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → A ⁢ z N ∈ ℂ
44 oveq2 ⊢ k = N → z k = z N
45 9 44 oveq12d ⊢ k = N → if k = N A 0 ⁢ z k = A ⁢ z N
46 45 fsum1 ⊢ N ∈ ℤ ∧ A ⁢ z N ∈ ℂ → ∑ k = N N if k = N A 0 ⁢ z k = A ⁢ z N
47 41 43 46 syl2anc ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → ∑ k = N N if k = N A 0 ⁢ z k = A ⁢ z N
48 40 47 eqtr3d ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 ∧ z ∈ ℂ → ∑ k = 0 N if k = N A 0 ⁢ z k = A ⁢ z N
49 48 mpteq2dva ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → z ∈ ℂ ⟼ ∑ k = 0 N if k = N A 0 ⁢ z k = z ∈ ℂ ⟼ A ⁢ z N
50 1 49 eqtr4id ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → F = z ∈ ℂ ⟼ ∑ k = 0 N if k = N A 0 ⁢ z k