Metamath Proof Explorer


Theorem coeeq

Description: If A satisfies the properties of the coefficient function, it must be equal to the coefficient function. (Contributed by Mario Carneiro, 22-Jul-2014) (Revised by Mario Carneiro, 23-Aug-2014)

Ref Expression
Hypotheses coeeq.1 ⊢ φ → F ∈ Poly ⁡ S
coeeq.2 ⊢ φ → N ∈ ℕ 0
coeeq.3 ⊢ φ → A : ℕ 0 ⟶ ℂ
coeeq.4 ⊢ φ → A ℤ ≥ N + 1 = 0
coeeq.5 ⊢ φ → F = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k
Assertion coeeq ⊢ φ → coeff ⁡ F = A

Proof

Step Hyp Ref Expression
1 coeeq.1 ⊢ φ → F ∈ Poly ⁡ S
2 coeeq.2 ⊢ φ → N ∈ ℕ 0
3 coeeq.3 ⊢ φ → A : ℕ 0 ⟶ ℂ
4 coeeq.4 ⊢ φ → A ℤ ≥ N + 1 = 0
5 coeeq.5 ⊢ φ → F = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k
6 coeval ⊢ F ∈ Poly ⁡ S → coeff ⁡ F = ι a ∈ ℂ ℕ 0 | ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
7 1 6 syl ⊢ φ → coeff ⁡ F = ι a ∈ ℂ ℕ 0 | ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
8 fvoveq1 ⊢ n = N → ℤ ≥ n + 1 = ℤ ≥ N + 1
9 8 imaeq2d ⊢ n = N → A ℤ ≥ n + 1 = A ℤ ≥ N + 1
10 9 eqeq1d ⊢ n = N → A ℤ ≥ n + 1 = 0 ↔ A ℤ ≥ N + 1 = 0
11 oveq2 ⊢ n = N → 0 … n = 0 … N
12 11 sumeq1d ⊢ n = N → ∑ k = 0 n A ⁡ k ⁢ z k = ∑ k = 0 N A ⁡ k ⁢ z k
13 12 mpteq2dv ⊢ n = N → z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k
14 13 eqeq2d ⊢ n = N → F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k ↔ F = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k
15 10 14 anbi12d ⊢ n = N → A ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k ↔ A ℤ ≥ N + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k
16 15 rspcev ⊢ N ∈ ℕ 0 ∧ A ℤ ≥ N + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k → ∃ n ∈ ℕ 0 A ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k
17 2 4 5 16 syl12anc ⊢ φ → ∃ n ∈ ℕ 0 A ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k
18 cnex ⊢ ℂ ∈ V
19 nn0ex ⊢ ℕ 0 ∈ V
20 18 19 elmap ⊢ A ∈ ℂ ℕ 0 ↔ A : ℕ 0 ⟶ ℂ
21 3 20 sylibr ⊢ φ → A ∈ ℂ ℕ 0
22 coeeu ⊢ F ∈ Poly ⁡ S → ∃! a ∈ ℂ ℕ 0 ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
23 1 22 syl ⊢ φ → ∃! a ∈ ℂ ℕ 0 ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
24 imaeq1 ⊢ a = A → a ℤ ≥ n + 1 = A ℤ ≥ n + 1
25 24 eqeq1d ⊢ a = A → a ℤ ≥ n + 1 = 0 ↔ A ℤ ≥ n + 1 = 0
26 fveq1 ⊢ a = A → a ⁡ k = A ⁡ k
27 26 oveq1d ⊢ a = A → a ⁡ k ⁢ z k = A ⁡ k ⁢ z k
28 27 sumeq2sdv ⊢ a = A → ∑ k = 0 n a ⁡ k ⁢ z k = ∑ k = 0 n A ⁡ k ⁢ z k
29 28 mpteq2dv ⊢ a = A → z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k
30 29 eqeq2d ⊢ a = A → F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k ↔ F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k
31 25 30 anbi12d ⊢ a = A → a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k ↔ A ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k
32 31 rexbidv ⊢ a = A → ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k ↔ ∃ n ∈ ℕ 0 A ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k
33 32 riota2 ⊢ A ∈ ℂ ℕ 0 ∧ ∃! a ∈ ℂ ℕ 0 ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k → ∃ n ∈ ℕ 0 A ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k ↔ ι a ∈ ℂ ℕ 0 | ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k = A
34 21 23 33 syl2anc ⊢ φ → ∃ n ∈ ℕ 0 A ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n A ⁡ k ⁢ z k ↔ ι a ∈ ℂ ℕ 0 | ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k = A
35 17 34 mpbid ⊢ φ → ι a ∈ ℂ ℕ 0 | ∃ n ∈ ℕ 0 a ℤ ≥ n + 1 = 0 ∧ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k = A
36 7 35 eqtrd ⊢ φ → coeff ⁡ F = A