Metamath Proof Explorer


Theorem elplyr

Description: Sufficient condition for elementhood in the set of polynomials. (Contributed by Mario Carneiro, 17-Jul-2014) (Revised by Mario Carneiro, 23-Aug-2014)

Ref Expression
Assertion elplyr ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k ∈ Poly ⁡ S

Proof

Step Hyp Ref Expression
1 simp1 ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → S ⊆ ℂ
2 simp2 ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → N ∈ ℕ 0
3 simp3 ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → A : ℕ 0 ⟶ S
4 ssun1 ⊢ S ⊆ S ∪ 0
5 fss ⊢ A : ℕ 0 ⟶ S ∧ S ⊆ S ∪ 0 → A : ℕ 0 ⟶ S ∪ 0
6 3 4 5 sylancl ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → A : ℕ 0 ⟶ S ∪ 0
7 0cnd ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → 0 ∈ ℂ
8 7 snssd ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → 0 ⊆ ℂ
9 1 8 unssd ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → S ∪ 0 ⊆ ℂ
10 cnex ⊢ ℂ ∈ V
11 ssexg ⊢ S ∪ 0 ⊆ ℂ ∧ ℂ ∈ V → S ∪ 0 ∈ V
12 9 10 11 sylancl ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → S ∪ 0 ∈ V
13 nn0ex ⊢ ℕ 0 ∈ V
14 elmapg ⊢ S ∪ 0 ∈ V ∧ ℕ 0 ∈ V → A ∈ S ∪ 0 ℕ 0 ↔ A : ℕ 0 ⟶ S ∪ 0
15 12 13 14 sylancl ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → A ∈ S ∪ 0 ℕ 0 ↔ A : ℕ 0 ⟶ S ∪ 0
16 6 15 mpbird ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → A ∈ S ∪ 0 ℕ 0
17 eqidd ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k
18 oveq2 ⊢ n = N → 0 … n = 0 … N
19 18 sumeq1d ⊢ n = N → ∑ k = 0 n a ⁡ k ⁢ z k = ∑ k = 0 N a ⁡ k ⁢ z k
20 19 mpteq2dv ⊢ n = N → z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 N a ⁡ k ⁢ z k
21 20 eqeq2d ⊢ n = N → z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k ↔ z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 N a ⁡ k ⁢ z k
22 fveq1 ⊢ a = A → a ⁡ k = A ⁡ k
23 22 oveq1d ⊢ a = A → a ⁡ k ⁢ z k = A ⁡ k ⁢ z k
24 23 sumeq2sdv ⊢ a = A → ∑ k = 0 N a ⁡ k ⁢ z k = ∑ k = 0 N A ⁡ k ⁢ z k
25 24 mpteq2dv ⊢ a = A → z ∈ ℂ ⟼ ∑ k = 0 N a ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k
26 25 eqeq2d ⊢ a = A → z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 N a ⁡ k ⁢ z k ↔ z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k
27 21 26 rspc2ev ⊢ N ∈ ℕ 0 ∧ A ∈ S ∪ 0 ℕ 0 ∧ z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k → ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
28 2 16 17 27 syl3anc ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
29 elply ⊢ z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k ∈ Poly ⁡ S ↔ S ⊆ ℂ ∧ ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
30 1 28 29 sylanbrc ⊢ S ⊆ ℂ ∧ N ∈ ℕ 0 ∧ A : ℕ 0 ⟶ S → z ∈ ℂ ⟼ ∑ k = 0 N A ⁡ k ⁢ z k ∈ Poly ⁡ S