Metamath Proof Explorer


Theorem plyf

Description: A polynomial is a function on the complex numbers. (Contributed by Mario Carneiro, 22-Jul-2014)

Ref Expression
Assertion plyf ⊢ F ∈ Poly ⁡ S → F : ℂ ⟶ ℂ

Proof

Step Hyp Ref Expression
1 elply ⊢ F ∈ Poly ⁡ S ↔ S ⊆ ℂ ∧ ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
2 1 simprbi ⊢ F ∈ Poly ⁡ S → ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
3 fzfid ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ → 0 … n ∈ Fin
4 plybss ⊢ F ∈ Poly ⁡ S → S ⊆ ℂ
5 0cnd ⊢ F ∈ Poly ⁡ S → 0 ∈ ℂ
6 5 snssd ⊢ F ∈ Poly ⁡ S → 0 ⊆ ℂ
7 4 6 unssd ⊢ F ∈ Poly ⁡ S → S ∪ 0 ⊆ ℂ
8 7 ad2antrr ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ → S ∪ 0 ⊆ ℂ
9 8 adantr ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … n → S ∪ 0 ⊆ ℂ
10 simplrr ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ → a ∈ S ∪ 0 ℕ 0
11 cnex ⊢ ℂ ∈ V
12 ssexg ⊢ S ∪ 0 ⊆ ℂ ∧ ℂ ∈ V → S ∪ 0 ∈ V
13 8 11 12 sylancl ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ → S ∪ 0 ∈ V
14 nn0ex ⊢ ℕ 0 ∈ V
15 elmapg ⊢ S ∪ 0 ∈ V ∧ ℕ 0 ∈ V → a ∈ S ∪ 0 ℕ 0 ↔ a : ℕ 0 ⟶ S ∪ 0
16 13 14 15 sylancl ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ → a ∈ S ∪ 0 ℕ 0 ↔ a : ℕ 0 ⟶ S ∪ 0
17 10 16 mpbid ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ → a : ℕ 0 ⟶ S ∪ 0
18 elfznn0 ⊢ k ∈ 0 … n → k ∈ ℕ 0
19 ffvelcdm ⊢ a : ℕ 0 ⟶ S ∪ 0 ∧ k ∈ ℕ 0 → a ⁡ k ∈ S ∪ 0
20 17 18 19 syl2an ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … n → a ⁡ k ∈ S ∪ 0
21 9 20 sseldd ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … n → a ⁡ k ∈ ℂ
22 simpr ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ → z ∈ ℂ
23 expcl ⊢ z ∈ ℂ ∧ k ∈ ℕ 0 → z k ∈ ℂ
24 22 18 23 syl2an ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … n → z k ∈ ℂ
25 21 24 mulcld ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ ∧ k ∈ 0 … n → a ⁡ k ⁢ z k ∈ ℂ
26 3 25 fsumcl ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ → ∑ k = 0 n a ⁡ k ⁢ z k ∈ ℂ
27 26 fmpttd ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 → z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k : ℂ ⟶ ℂ
28 feq1 ⊢ F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k → F : ℂ ⟶ ℂ ↔ z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k : ℂ ⟶ ℂ
29 27 28 syl5ibrcom ⊢ F ∈ Poly ⁡ S ∧ n ∈ ℕ 0 ∧ a ∈ S ∪ 0 ℕ 0 → F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k → F : ℂ ⟶ ℂ
30 29 rexlimdvva ⊢ F ∈ Poly ⁡ S → ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 F = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k → F : ℂ ⟶ ℂ
31 2 30 mpd ⊢ F ∈ Poly ⁡ S → F : ℂ ⟶ ℂ