Metamath Proof Explorer


Theorem plysub

Description: The difference of two polynomials is a polynomial. (Contributed by Mario Carneiro, 21-Jul-2014)

Ref Expression
Hypotheses plyadd.1 ⊢ φ → F ∈ Poly ⁡ S
plyadd.2 ⊢ φ → G ∈ Poly ⁡ S
plyadd.3 ⊢ φ ∧ x ∈ S ∧ y ∈ S → x + y ∈ S
plymul.4 ⊢ φ ∧ x ∈ S ∧ y ∈ S → x ⁢ y ∈ S
plysub.5 ⊢ φ → − 1 ∈ S
Assertion plysub ⊢ φ → F − f G ∈ Poly ⁡ S

Proof

Step Hyp Ref Expression
1 plyadd.1 ⊢ φ → F ∈ Poly ⁡ S
2 plyadd.2 ⊢ φ → G ∈ Poly ⁡ S
3 plyadd.3 ⊢ φ ∧ x ∈ S ∧ y ∈ S → x + y ∈ S
4 plymul.4 ⊢ φ ∧ x ∈ S ∧ y ∈ S → x ⁢ y ∈ S
5 plysub.5 ⊢ φ → − 1 ∈ S
6 cnex ⊢ ℂ ∈ V
7 plyf ⊢ F ∈ Poly ⁡ S → F : ℂ ⟶ ℂ
8 1 7 syl ⊢ φ → F : ℂ ⟶ ℂ
9 plyf ⊢ G ∈ Poly ⁡ S → G : ℂ ⟶ ℂ
10 2 9 syl ⊢ φ → G : ℂ ⟶ ℂ
11 ofnegsub ⊢ ℂ ∈ V ∧ F : ℂ ⟶ ℂ ∧ G : ℂ ⟶ ℂ → F + f ℂ × − 1 × f G = F − f G
12 6 8 10 11 mp3an2i ⊢ φ → F + f ℂ × − 1 × f G = F − f G
13 plybss ⊢ F ∈ Poly ⁡ S → S ⊆ ℂ
14 1 13 syl ⊢ φ → S ⊆ ℂ
15 plyconst ⊢ S ⊆ ℂ ∧ − 1 ∈ S → ℂ × − 1 ∈ Poly ⁡ S
16 14 5 15 syl2anc ⊢ φ → ℂ × − 1 ∈ Poly ⁡ S
17 16 2 3 4 plymul ⊢ φ → ℂ × − 1 × f G ∈ Poly ⁡ S
18 1 17 3 plyadd ⊢ φ → F + f ℂ × − 1 × f G ∈ Poly ⁡ S
19 12 18 eqeltrrd ⊢ φ → F − f G ∈ Poly ⁡ S