Metamath Proof Explorer


Theorem dvres

Description: Restriction of a derivative. Note that our definition of derivative df-dv would still make sense if we demanded that x be an element of the domain instead of an interior point of the domain, but then it is possible for a non-differentiable function to have two different derivatives at a single point x when restricted to different subsets containing x ; a classic example is the absolute value function restricted to [ 0 , +oo ) and ( -oo , 0 ] . (Contributed by Mario Carneiro, 8-Aug-2014) (Revised by Mario Carneiro, 9-Feb-2015)

Ref Expression
Hypotheses dvres.k K = TopOpen fld
dvres.t T = K 𝑡 S
Assertion dvres S F : A A S B S S D F B = F S int T B

Proof

Step Hyp Ref Expression
1 dvres.k K = TopOpen fld
2 dvres.t T = K 𝑡 S
3 reldv Rel F B S
4 relres Rel F S int T B
5 simpll S F : A A S B S S
6 simplr S F : A A S B S F : A
7 inss1 A B A
8 fssres F : A A B A F A B : A B
9 6 7 8 sylancl S F : A A S B S F A B : A B
10 resres F A B = F A B
11 ffn F : A F Fn A
12 fnresdm F Fn A F A = F
13 6 11 12 3syl S F : A A S B S F A = F
14 13 reseq1d S F : A A S B S F A B = F B
15 10 14 eqtr3id S F : A A S B S F A B = F B
16 15 feq1d S F : A A S B S F A B : A B F B : A B
17 9 16 mpbid S F : A A S B S F B : A B
18 simprl S F : A A S B S A S
19 7 18 sstrid S F : A A S B S A B S
20 5 17 19 dvcl S F : A A S B S x F B S y y
21 20 ex S F : A A S B S x F B S y y
22 5 6 18 dvcl S F : A A S B S x F S y y
23 22 ex S F : A A S B S x F S y y
24 23 adantld S F : A A S B S x int T B x F S y y
25 eqid z A x F z F x z x = z A x F z F x z x
26 5 adantr S F : A A S B S y S
27 6 adantr S F : A A S B S y F : A
28 18 adantr S F : A A S B S y A S
29 simplrr S F : A A S B S y B S
30 simpr S F : A A S B S y y
31 1 2 25 26 27 28 29 30 dvreslem S F : A A S B S y x F B S y x int T B x F S y
32 31 ex S F : A A S B S y x F B S y x int T B x F S y
33 21 24 32 pm5.21ndd S F : A A S B S x F B S y x int T B x F S y
34 vex y V
35 34 brresi x F S int T B y x int T B x F S y
36 33 35 bitr4di S F : A A S B S x F B S y x F S int T B y
37 3 4 36 eqbrrdiv S F : A A S B S S D F B = F S int T B