Metamath Proof Explorer


Theorem plyval

Description: Value of the polynomial set function. (Contributed by Mario Carneiro, 17-Jul-2014)

Ref Expression
Assertion plyval ⊢ S ⊆ ℂ → Poly ⁡ S = f | ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k

Proof

Step Hyp Ref Expression
1 cnex ⊢ ℂ ∈ V
2 1 elpw2 ⊢ S ∈ 𝒫 ℂ ↔ S ⊆ ℂ
3 uneq1 ⊢ x = S → x ∪ 0 = S ∪ 0
4 3 oveq1d ⊢ x = S → x ∪ 0 ℕ 0 = S ∪ 0 ℕ 0
5 4 rexeqdv ⊢ x = S → ∃ a ∈ x ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k ↔ ∃ a ∈ S ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
6 5 rexbidv ⊢ x = S → ∃ n ∈ ℕ 0 ∃ a ∈ x ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k ↔ ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
7 6 abbidv ⊢ x = S → f | ∃ n ∈ ℕ 0 ∃ a ∈ x ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k = f | ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
8 df-ply ⊢ Poly = x ∈ 𝒫 ℂ ⟼ f | ∃ n ∈ ℕ 0 ∃ a ∈ x ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
9 nn0ex ⊢ ℕ 0 ∈ V
10 ovex ⊢ S ∪ 0 ℕ 0 ∈ V
11 9 10 ab2rexex ⊢ f | ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k ∈ V
12 7 8 11 fvmpt ⊢ S ∈ 𝒫 ℂ → Poly ⁡ S = f | ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k
13 2 12 sylbir ⊢ S ⊆ ℂ → Poly ⁡ S = f | ∃ n ∈ ℕ 0 ∃ a ∈ S ∪ 0 ℕ 0 f = z ∈ ℂ ⟼ ∑ k = 0 n a ⁡ k ⁢ z k