Metamath Proof Explorer


Theorem ply1term

Description: A one-term polynomial. (Contributed by Mario Carneiro, 17-Jul-2014)

Ref Expression
Hypothesis ply1term.1 ⊢ F = z ∈ ℂ ⟼ A ⁢ z N
Assertion ply1term ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → F ∈ Poly ⁡ S

Proof

Step Hyp Ref Expression
1 ply1term.1 ⊢ F = z ∈ ℂ ⟼ A ⁢ z N
2 ssel2 ⊢ S ⊆ ℂ ∧ A ∈ S → A ∈ ℂ
3 1 ply1termlem ⊢ A ∈ ℂ ∧ N ∈ ℕ 0 → F = z ∈ ℂ ⟼ ∑ k = 0 N if k = N A 0 ⁢ z k
4 2 3 stoic3 ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → F = z ∈ ℂ ⟼ ∑ k = 0 N if k = N A 0 ⁢ z k
5 simp1 ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → S ⊆ ℂ
6 0cnd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → 0 ∈ ℂ
7 6 snssd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → 0 ⊆ ℂ
8 5 7 unssd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → S ∪ 0 ⊆ ℂ
9 simp3 ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → N ∈ ℕ 0
10 simpl2 ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 ∧ k ∈ 0 … N → A ∈ S
11 elun1 ⊢ A ∈ S → A ∈ S ∪ 0
12 10 11 syl ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 ∧ k ∈ 0 … N → A ∈ S ∪ 0
13 ssun2 ⊢ 0 ⊆ S ∪ 0
14 c0ex ⊢ 0 ∈ V
15 14 snss ⊢ 0 ∈ S ∪ 0 ↔ 0 ⊆ S ∪ 0
16 13 15 mpbir ⊢ 0 ∈ S ∪ 0
17 ifcl ⊢ A ∈ S ∪ 0 ∧ 0 ∈ S ∪ 0 → if k = N A 0 ∈ S ∪ 0
18 12 16 17 sylancl ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 ∧ k ∈ 0 … N → if k = N A 0 ∈ S ∪ 0
19 8 9 18 elplyd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → z ∈ ℂ ⟼ ∑ k = 0 N if k = N A 0 ⁢ z k ∈ Poly ⁡ S ∪ 0
20 4 19 eqeltrd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → F ∈ Poly ⁡ S ∪ 0
21 plyun0 ⊢ Poly ⁡ S ∪ 0 = Poly ⁡ S
22 20 21 eleqtrdi ⊢ S ⊆ ℂ ∧ A ∈ S ∧ N ∈ ℕ 0 → F ∈ Poly ⁡ S