Metamath Proof Explorer


Theorem clim2div

Description: The limit of an infinite product with an initial segment removed. (Contributed by Scott Fenton, 20-Dec-2017)

Ref Expression
Hypotheses clim2div.1 ⊢ Z = ℤ ≥ M
clim2div.2 ⊢ φ → N ∈ Z
clim2div.3 ⊢ φ ∧ k ∈ Z → F ⁡ k ∈ ℂ
clim2div.4 ⊢ φ → seq M × F ⇝ A
clim2div.5 ⊢ φ → seq M × F ⁡ N ≠ 0
Assertion clim2div ⊢ φ → seq N + 1 × F ⇝ A seq M × F ⁡ N

Proof

Step Hyp Ref Expression
1 clim2div.1 ⊢ Z = ℤ ≥ M
2 clim2div.2 ⊢ φ → N ∈ Z
3 clim2div.3 ⊢ φ ∧ k ∈ Z → F ⁡ k ∈ ℂ
4 clim2div.4 ⊢ φ → seq M × F ⇝ A
5 clim2div.5 ⊢ φ → seq M × F ⁡ N ≠ 0
6 eqid ⊢ ℤ ≥ N + 1 = ℤ ≥ N + 1
7 eluzelz ⊢ N ∈ ℤ ≥ M → N ∈ ℤ
8 7 1 eleq2s ⊢ N ∈ Z → N ∈ ℤ
9 2 8 syl ⊢ φ → N ∈ ℤ
10 9 peano2zd ⊢ φ → N + 1 ∈ ℤ
11 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
12 11 1 eleq2s ⊢ N ∈ Z → M ∈ ℤ
13 2 12 syl ⊢ φ → M ∈ ℤ
14 1 13 3 prodf ⊢ φ → seq M × F : Z ⟶ ℂ
15 14 2 ffvelcdmd ⊢ φ → seq M × F ⁡ N ∈ ℂ
16 15 5 reccld ⊢ φ → 1 seq M × F ⁡ N ∈ ℂ
17 seqex ⊢ seq N + 1 × F ∈ V
18 17 a1i ⊢ φ → seq N + 1 × F ∈ V
19 2 1 eleqtrdi ⊢ φ → N ∈ ℤ ≥ M
20 peano2uz ⊢ N ∈ ℤ ≥ M → N + 1 ∈ ℤ ≥ M
21 19 20 syl ⊢ φ → N + 1 ∈ ℤ ≥ M
22 21 1 eleqtrrdi ⊢ φ → N + 1 ∈ Z
23 1 uztrn2 ⊢ N + 1 ∈ Z ∧ j ∈ ℤ ≥ N + 1 → j ∈ Z
24 22 23 sylan ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → j ∈ Z
25 14 ffvelcdmda ⊢ φ ∧ j ∈ Z → seq M × F ⁡ j ∈ ℂ
26 24 25 syldan ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M × F ⁡ j ∈ ℂ
27 mulcl ⊢ k ∈ ℂ ∧ x ∈ ℂ → k ⁢ x ∈ ℂ
28 27 adantl ⊢ φ ∧ j ∈ ℤ ≥ N + 1 ∧ k ∈ ℂ ∧ x ∈ ℂ → k ⁢ x ∈ ℂ
29 mulass ⊢ k ∈ ℂ ∧ x ∈ ℂ ∧ y ∈ ℂ → k ⁢ x ⁢ y = k ⁢ x ⁢ y
30 29 adantl ⊢ φ ∧ j ∈ ℤ ≥ N + 1 ∧ k ∈ ℂ ∧ x ∈ ℂ ∧ y ∈ ℂ → k ⁢ x ⁢ y = k ⁢ x ⁢ y
31 simpr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → j ∈ ℤ ≥ N + 1
32 19 adantr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → N ∈ ℤ ≥ M
33 elfzuz ⊢ k ∈ M … j → k ∈ ℤ ≥ M
34 33 1 eleqtrrdi ⊢ k ∈ M … j → k ∈ Z
35 34 3 sylan2 ⊢ φ ∧ k ∈ M … j → F ⁡ k ∈ ℂ
36 35 adantlr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 ∧ k ∈ M … j → F ⁡ k ∈ ℂ
37 28 30 31 32 36 seqsplit ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M × F ⁡ j = seq M × F ⁡ N ⁢ seq N + 1 × F ⁡ j
38 37 eqcomd ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M × F ⁡ N ⁢ seq N + 1 × F ⁡ j = seq M × F ⁡ j
39 15 adantr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M × F ⁡ N ∈ ℂ
40 1 uztrn2 ⊢ N + 1 ∈ Z ∧ k ∈ ℤ ≥ N + 1 → k ∈ Z
41 22 40 sylan ⊢ φ ∧ k ∈ ℤ ≥ N + 1 → k ∈ Z
42 41 3 syldan ⊢ φ ∧ k ∈ ℤ ≥ N + 1 → F ⁡ k ∈ ℂ
43 6 10 42 prodf ⊢ φ → seq N + 1 × F : ℤ ≥ N + 1 ⟶ ℂ
44 43 ffvelcdmda ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq N + 1 × F ⁡ j ∈ ℂ
45 5 adantr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M × F ⁡ N ≠ 0
46 26 39 44 45 divmuld ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M × F ⁡ j seq M × F ⁡ N = seq N + 1 × F ⁡ j ↔ seq M × F ⁡ N ⁢ seq N + 1 × F ⁡ j = seq M × F ⁡ j
47 38 46 mpbird ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M × F ⁡ j seq M × F ⁡ N = seq N + 1 × F ⁡ j
48 26 39 45 divrec2d ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M × F ⁡ j seq M × F ⁡ N = 1 seq M × F ⁡ N ⁢ seq M × F ⁡ j
49 47 48 eqtr3d ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq N + 1 × F ⁡ j = 1 seq M × F ⁡ N ⁢ seq M × F ⁡ j
50 6 10 4 16 18 26 49 climmulc2 ⊢ φ → seq N + 1 × F ⇝ 1 seq M × F ⁡ N ⁢ A
51 climcl ⊢ seq M × F ⇝ A → A ∈ ℂ
52 4 51 syl ⊢ φ → A ∈ ℂ
53 52 15 5 divrec2d ⊢ φ → A seq M × F ⁡ N = 1 seq M × F ⁡ N ⁢ A
54 50 53 breqtrrd ⊢ φ → seq N + 1 × F ⇝ A seq M × F ⁡ N