Metamath Proof Explorer


Theorem dvres2

Description: Restriction of the base set of a derivative. The primary application of this theorem says that if a function is complex-differentiable then it is also real-differentiable. Unlike dvres , there is no simple reverse relation relating real-differentiable functions to complex differentiability, and indeed there are functions like Re ( x ) which are everywhere real-differentiable but nowhere complex-differentiable.) (Contributed by Mario Carneiro, 9-Feb-2015)

Ref Expression
Assertion dvres2 ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S → F S ′ ↾ B ⊆ B D F ↾ B

Proof

Step Hyp Ref Expression
1 relres ⊢ Rel ⁡ F S ′ ↾ B
2 1 a1i ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S → Rel ⁡ F S ′ ↾ B
3 eqid ⊢ TopOpen ⁡ ℂ fld = TopOpen ⁡ ℂ fld
4 eqid ⊢ TopOpen ⁡ ℂ fld ↾ 𝑡 S = TopOpen ⁡ ℂ fld ↾ 𝑡 S
5 eqid ⊢ z ∈ A ∖ x ⟼ F ⁡ z − F ⁡ x z − x = z ∈ A ∖ x ⟼ F ⁡ z − F ⁡ x z − x
6 simp1l ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y → S ⊆ ℂ
7 simp1r ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y → F : A ⟶ ℂ
8 simp2l ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y → A ⊆ S
9 simp2r ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y → B ⊆ S
10 simp3r ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y → x F S ′ y
11 6 7 8 dvcl ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y ∧ x F S ′ y → y ∈ ℂ
12 10 11 mpdan ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y → y ∈ ℂ
13 simp3l ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y → x ∈ B
14 3 4 5 6 7 8 9 12 10 13 dvres2lem ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S ∧ x ∈ B ∧ x F S ′ y → x F ↾ B B ′ y
15 14 3expia ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S → x ∈ B ∧ x F S ′ y → x F ↾ B B ′ y
16 vex ⊢ y ∈ V
17 16 brresi ⊢ x F S ′ ↾ B y ↔ x ∈ B ∧ x F S ′ y
18 df-br ⊢ x F S ′ ↾ B y ↔ x y ∈ F S ′ ↾ B
19 17 18 bitr3i ⊢ x ∈ B ∧ x F S ′ y ↔ x y ∈ F S ′ ↾ B
20 df-br ⊢ x F ↾ B B ′ y ↔ x y ∈ F ↾ B B ′
21 15 19 20 3imtr3g ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S → x y ∈ F S ′ ↾ B → x y ∈ F ↾ B B ′
22 2 21 relssdv ⊢ S ⊆ ℂ ∧ F : A ⟶ ℂ ∧ A ⊆ S ∧ B ⊆ S → F S ′ ↾ B ⊆ B D F ↾ B