Metamath Proof Explorer


Theorem subcncff

Description: The subtraction of two continuous complex functions is continuous. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses subcncff.f ⊢ φ → F : X ⟶cn ℂ
subcncff.g ⊢ φ → G : X ⟶cn ℂ
Assertion subcncff ⊢ φ → F − f G : X ⟶cn ℂ

Proof

Step Hyp Ref Expression
1 subcncff.f ⊢ φ → F : X ⟶cn ℂ
2 subcncff.g ⊢ φ → G : X ⟶cn ℂ
3 cncfrss ⊢ F : X ⟶cn ℂ → X ⊆ ℂ
4 cnex ⊢ ℂ ∈ V
5 4 ssex ⊢ X ⊆ ℂ → X ∈ V
6 1 3 5 3syl ⊢ φ → X ∈ V
7 cncff ⊢ F : X ⟶cn ℂ → F : X ⟶ ℂ
8 1 7 syl ⊢ φ → F : X ⟶ ℂ
9 8 ffvelcdmda ⊢ φ ∧ x ∈ X → F ⁡ x ∈ ℂ
10 cncff ⊢ G : X ⟶cn ℂ → G : X ⟶ ℂ
11 2 10 syl ⊢ φ → G : X ⟶ ℂ
12 11 ffvelcdmda ⊢ φ ∧ x ∈ X → G ⁡ x ∈ ℂ
13 8 feqmptd ⊢ φ → F = x ∈ X ⟼ F ⁡ x
14 11 feqmptd ⊢ φ → G = x ∈ X ⟼ G ⁡ x
15 6 9 12 13 14 offval2 ⊢ φ → F − f G = x ∈ X ⟼ F ⁡ x − G ⁡ x
16 13 1 eqeltrrd ⊢ φ → x ∈ X ⟼ F ⁡ x : X ⟶cn ℂ
17 14 2 eqeltrrd ⊢ φ → x ∈ X ⟼ G ⁡ x : X ⟶cn ℂ
18 16 17 subcncf ⊢ φ → x ∈ X ⟼ F ⁡ x − G ⁡ x : X ⟶cn ℂ
19 15 18 eqeltrd ⊢ φ → F − f G : X ⟶cn ℂ