Metamath Proof Explorer


Theorem dvcl

Description: The derivative function takes values in the complex numbers. (Contributed by Mario Carneiro, 7-Aug-2014) (Revised by Mario Carneiro, 9-Feb-2015)

Ref Expression
Hypotheses dvcl.s ⊢ φ → S ⊆ ℂ
dvcl.f ⊢ φ → F : A ⟶ ℂ
dvcl.a ⊢ φ → A ⊆ S
Assertion dvcl ⊢ φ ∧ B F S ′ C → C ∈ ℂ

Proof

Step Hyp Ref Expression
1 dvcl.s ⊢ φ → S ⊆ ℂ
2 dvcl.f ⊢ φ → F : A ⟶ ℂ
3 dvcl.a ⊢ φ → A ⊆ S
4 limccl ⊢ z ∈ A ∖ B ⟼ F ⁡ z − F ⁡ B z − B lim ℂ B ⊆ ℂ
5 eqid ⊢ TopOpen ⁡ ℂ fld ↾ 𝑡 S = TopOpen ⁡ ℂ fld ↾ 𝑡 S
6 eqid ⊢ TopOpen ⁡ ℂ fld = TopOpen ⁡ ℂ fld
7 eqid ⊢ z ∈ A ∖ B ⟼ F ⁡ z − F ⁡ B z − B = z ∈ A ∖ B ⟼ F ⁡ z − F ⁡ B z − B
8 5 6 7 1 2 3 eldv ⊢ φ → B F S ′ C ↔ B ∈ int ⁡ TopOpen ⁡ ℂ fld ↾ 𝑡 S ⁡ A ∧ C ∈ z ∈ A ∖ B ⟼ F ⁡ z − F ⁡ B z − B lim ℂ B
9 8 simplbda ⊢ φ ∧ B F S ′ C → C ∈ z ∈ A ∖ B ⟼ F ⁡ z − F ⁡ B z − B lim ℂ B
10 4 9 sselid ⊢ φ ∧ B F S ′ C → C ∈ ℂ