Metamath Proof Explorer


Theorem divcncf

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

Ref Expression
Hypotheses divcncf.1 ⊢ φ → x ∈ X ⟼ A : X ⟶cn ℂ
divcncf.2 ⊢ φ → x ∈ X ⟼ B : X ⟶cn ℂ ∖ 0
Assertion divcncf ⊢ φ → x ∈ X ⟼ A B : X ⟶cn ℂ

Proof

Step Hyp Ref Expression
1 divcncf.1 ⊢ φ → x ∈ X ⟼ A : X ⟶cn ℂ
2 divcncf.2 ⊢ φ → x ∈ X ⟼ B : X ⟶cn ℂ ∖ 0
3 cncff ⊢ x ∈ X ⟼ A : X ⟶cn ℂ → x ∈ X ⟼ A : X ⟶ ℂ
4 1 3 syl ⊢ φ → x ∈ X ⟼ A : X ⟶ ℂ
5 4 fvmptelcdm ⊢ φ ∧ x ∈ X → A ∈ ℂ
6 cncff ⊢ x ∈ X ⟼ B : X ⟶cn ℂ ∖ 0 → x ∈ X ⟼ B : X ⟶ ℂ ∖ 0
7 2 6 syl ⊢ φ → x ∈ X ⟼ B : X ⟶ ℂ ∖ 0
8 7 fvmptelcdm ⊢ φ ∧ x ∈ X → B ∈ ℂ ∖ 0
9 8 eldifad ⊢ φ ∧ x ∈ X → B ∈ ℂ
10 eldifsni ⊢ B ∈ ℂ ∖ 0 → B ≠ 0
11 8 10 syl ⊢ φ ∧ x ∈ X → B ≠ 0
12 5 9 11 divrecd ⊢ φ ∧ x ∈ X → A B = A ⁢ 1 B
13 12 mpteq2dva ⊢ φ → x ∈ X ⟼ A B = x ∈ X ⟼ A ⁢ 1 B
14 8 ralrimiva ⊢ φ → ∀ x ∈ X B ∈ ℂ ∖ 0
15 eqidd ⊢ φ → x ∈ X ⟼ B = x ∈ X ⟼ B
16 eqidd ⊢ φ → y ∈ ℂ ∖ 0 ⟼ 1 y = y ∈ ℂ ∖ 0 ⟼ 1 y
17 14 15 16 fmptcos ⊢ φ → y ∈ ℂ ∖ 0 ⟼ 1 y ∘ x ∈ X ⟼ B = x ∈ X ⟼ ⦋ B / y⦌ 1 y
18 csbov2g ⊢ B ∈ ℂ → ⦋ B / y⦌ 1 y = 1 ⦋ B / y⦌ y
19 9 18 syl ⊢ φ ∧ x ∈ X → ⦋ B / y⦌ 1 y = 1 ⦋ B / y⦌ y
20 csbvarg ⊢ B ∈ ℂ → ⦋ B / y⦌ y = B
21 9 20 syl ⊢ φ ∧ x ∈ X → ⦋ B / y⦌ y = B
22 21 oveq2d ⊢ φ ∧ x ∈ X → 1 ⦋ B / y⦌ y = 1 B
23 19 22 eqtrd ⊢ φ ∧ x ∈ X → ⦋ B / y⦌ 1 y = 1 B
24 23 mpteq2dva ⊢ φ → x ∈ X ⟼ ⦋ B / y⦌ 1 y = x ∈ X ⟼ 1 B
25 17 24 eqtr2d ⊢ φ → x ∈ X ⟼ 1 B = y ∈ ℂ ∖ 0 ⟼ 1 y ∘ x ∈ X ⟼ B
26 ax-1cn ⊢ 1 ∈ ℂ
27 eqid ⊢ y ∈ ℂ ∖ 0 ⟼ 1 y = y ∈ ℂ ∖ 0 ⟼ 1 y
28 27 cdivcncf ⊢ 1 ∈ ℂ → y ∈ ℂ ∖ 0 ⟼ 1 y : ℂ ∖ 0 ⟶cn ℂ
29 26 28 mp1i ⊢ φ → y ∈ ℂ ∖ 0 ⟼ 1 y : ℂ ∖ 0 ⟶cn ℂ
30 2 29 cncfco ⊢ φ → y ∈ ℂ ∖ 0 ⟼ 1 y ∘ x ∈ X ⟼ B : X ⟶cn ℂ
31 25 30 eqeltrd ⊢ φ → x ∈ X ⟼ 1 B : X ⟶cn ℂ
32 1 31 mulcncf ⊢ φ → x ∈ X ⟼ A ⁢ 1 B : X ⟶cn ℂ
33 13 32 eqeltrd ⊢ φ → x ∈ X ⟼ A B : X ⟶cn ℂ