Metamath Proof Explorer


Theorem etransclem4

Description: F expressed as a finite product. (Contributed by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses etransclem4.a ⊢ φ → A ⊆ ℂ
etransclem4.p ⊢ φ → P ∈ ℕ
etransclem4.M ⊢ φ → M ∈ ℕ 0
etransclem4.f ⊢ F = x ∈ A ⟼ x P − 1 ⁢ ∏ j = 1 M x − j P
etransclem4.h ⊢ H = j ∈ 0 … M ⟼ x ∈ A ⟼ x − j if j = 0 P − 1 P
etransclem4.e ⊢ E = x ∈ A ⟼ ∏ j = 0 M H ⁡ j ⁡ x
Assertion etransclem4 ⊢ φ → F = E

Proof

Step Hyp Ref Expression
1 etransclem4.a ⊢ φ → A ⊆ ℂ
2 etransclem4.p ⊢ φ → P ∈ ℕ
3 etransclem4.M ⊢ φ → M ∈ ℕ 0
4 etransclem4.f ⊢ F = x ∈ A ⟼ x P − 1 ⁢ ∏ j = 1 M x − j P
5 etransclem4.h ⊢ H = j ∈ 0 … M ⟼ x ∈ A ⟼ x − j if j = 0 P − 1 P
6 etransclem4.e ⊢ E = x ∈ A ⟼ ∏ j = 0 M H ⁡ j ⁡ x
7 simpr ⊢ φ ∧ j ∈ 0 … M → j ∈ 0 … M
8 cnex ⊢ ℂ ∈ V
9 8 ssex ⊢ A ⊆ ℂ → A ∈ V
10 mptexg ⊢ A ∈ V → x ∈ A ⟼ x − j if j = 0 P − 1 P ∈ V
11 1 9 10 3syl ⊢ φ → x ∈ A ⟼ x − j if j = 0 P − 1 P ∈ V
12 11 adantr ⊢ φ ∧ j ∈ 0 … M → x ∈ A ⟼ x − j if j = 0 P − 1 P ∈ V
13 5 fvmpt2 ⊢ j ∈ 0 … M ∧ x ∈ A ⟼ x − j if j = 0 P − 1 P ∈ V → H ⁡ j = x ∈ A ⟼ x − j if j = 0 P − 1 P
14 7 12 13 syl2anc ⊢ φ ∧ j ∈ 0 … M → H ⁡ j = x ∈ A ⟼ x − j if j = 0 P − 1 P
15 ovexd ⊢ φ ∧ j ∈ 0 … M ∧ x ∈ A → x − j if j = 0 P − 1 P ∈ V
16 14 15 fvmpt2d ⊢ φ ∧ j ∈ 0 … M ∧ x ∈ A → H ⁡ j ⁡ x = x − j if j = 0 P − 1 P
17 16 an32s ⊢ φ ∧ x ∈ A ∧ j ∈ 0 … M → H ⁡ j ⁡ x = x − j if j = 0 P − 1 P
18 17 prodeq2dv ⊢ φ ∧ x ∈ A → ∏ j = 0 M H ⁡ j ⁡ x = ∏ j = 0 M x − j if j = 0 P − 1 P
19 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
20 3 19 eleqtrdi ⊢ φ → M ∈ ℤ ≥ 0
21 20 adantr ⊢ φ ∧ x ∈ A → M ∈ ℤ ≥ 0
22 1 sselda ⊢ φ ∧ x ∈ A → x ∈ ℂ
23 22 adantr ⊢ φ ∧ x ∈ A ∧ j ∈ 0 … M → x ∈ ℂ
24 elfzelz ⊢ j ∈ 0 … M → j ∈ ℤ
25 24 zcnd ⊢ j ∈ 0 … M → j ∈ ℂ
26 25 adantl ⊢ φ ∧ x ∈ A ∧ j ∈ 0 … M → j ∈ ℂ
27 23 26 subcld ⊢ φ ∧ x ∈ A ∧ j ∈ 0 … M → x − j ∈ ℂ
28 nnm1nn0 ⊢ P ∈ ℕ → P − 1 ∈ ℕ 0
29 2 28 syl ⊢ φ → P − 1 ∈ ℕ 0
30 2 nnnn0d ⊢ φ → P ∈ ℕ 0
31 29 30 ifcld ⊢ φ → if j = 0 P − 1 P ∈ ℕ 0
32 31 ad2antrr ⊢ φ ∧ x ∈ A ∧ j ∈ 0 … M → if j = 0 P − 1 P ∈ ℕ 0
33 27 32 expcld ⊢ φ ∧ x ∈ A ∧ j ∈ 0 … M → x − j if j = 0 P − 1 P ∈ ℂ
34 oveq2 ⊢ j = 0 → x − j = x − 0
35 iftrue ⊢ j = 0 → if j = 0 P − 1 P = P − 1
36 34 35 oveq12d ⊢ j = 0 → x − j if j = 0 P − 1 P = x − 0 P − 1
37 21 33 36 fprod1p ⊢ φ ∧ x ∈ A → ∏ j = 0 M x − j if j = 0 P − 1 P = x − 0 P − 1 ⁢ ∏ j = 0 + 1 M x − j if j = 0 P − 1 P
38 22 subid1d ⊢ φ ∧ x ∈ A → x − 0 = x
39 38 oveq1d ⊢ φ ∧ x ∈ A → x − 0 P − 1 = x P − 1
40 0p1e1 ⊢ 0 + 1 = 1
41 40 oveq1i ⊢ 0 + 1 … M = 1 … M
42 41 a1i ⊢ φ → 0 + 1 … M = 1 … M
43 0red ⊢ j ∈ 1 … M → 0 ∈ ℝ
44 1red ⊢ j ∈ 1 … M → 1 ∈ ℝ
45 elfzelz ⊢ j ∈ 1 … M → j ∈ ℤ
46 45 zred ⊢ j ∈ 1 … M → j ∈ ℝ
47 0lt1 ⊢ 0 < 1
48 47 a1i ⊢ j ∈ 1 … M → 0 < 1
49 elfzle1 ⊢ j ∈ 1 … M → 1 ≤ j
50 43 44 46 48 49 ltletrd ⊢ j ∈ 1 … M → 0 < j
51 50 gt0ne0d ⊢ j ∈ 1 … M → j ≠ 0
52 51 neneqd ⊢ j ∈ 1 … M → ¬ j = 0
53 52 iffalsed ⊢ j ∈ 1 … M → if j = 0 P − 1 P = P
54 53 oveq2d ⊢ j ∈ 1 … M → x − j if j = 0 P − 1 P = x − j P
55 54 adantl ⊢ φ ∧ j ∈ 1 … M → x − j if j = 0 P − 1 P = x − j P
56 42 55 prodeq12rdv ⊢ φ → ∏ j = 0 + 1 M x − j if j = 0 P − 1 P = ∏ j = 1 M x − j P
57 56 adantr ⊢ φ ∧ x ∈ A → ∏ j = 0 + 1 M x − j if j = 0 P − 1 P = ∏ j = 1 M x − j P
58 39 57 oveq12d ⊢ φ ∧ x ∈ A → x − 0 P − 1 ⁢ ∏ j = 0 + 1 M x − j if j = 0 P − 1 P = x P − 1 ⁢ ∏ j = 1 M x − j P
59 18 37 58 3eqtrrd ⊢ φ ∧ x ∈ A → x P − 1 ⁢ ∏ j = 1 M x − j P = ∏ j = 0 M H ⁡ j ⁡ x
60 59 mpteq2dva ⊢ φ → x ∈ A ⟼ x P − 1 ⁢ ∏ j = 1 M x − j P = x ∈ A ⟼ ∏ j = 0 M H ⁡ j ⁡ x
61 60 4 6 3eqtr4g ⊢ φ → F = E