Metamath Proof Explorer


Theorem dveq0

Description: If a continuous function has zero derivative at all points on the interior of a closed interval, then it must be a constant function. (Contributed by Mario Carneiro, 2-Sep-2014) (Proof shortened by Mario Carneiro, 3-Mar-2015)

Ref Expression
Hypotheses dveq0.a φ A
dveq0.b φ B
dveq0.c φ F : A B cn
dveq0.d φ D F = A B × 0
Assertion dveq0 φ F = A B × F A

Proof

Step Hyp Ref Expression
1 dveq0.a φ A
2 dveq0.b φ B
3 dveq0.c φ F : A B cn
4 dveq0.d φ D F = A B × 0
5 cncff F : A B cn F : A B
6 3 5 syl φ F : A B
7 6 ffnd φ F Fn A B
8 fvex F A V
9 fnconstg F A V A B × F A Fn A B
10 8 9 mp1i φ A B × F A Fn A B
11 8 fvconst2 x A B A B × F A x = F A
12 11 adantl φ x A B A B × F A x = F A
13 6 adantr φ x A B F : A B
14 1 adantr φ x A B A
15 14 rexrd φ x A B A *
16 2 adantr φ x A B B
17 16 rexrd φ x A B B *
18 elicc2 A B x A B x A x x B
19 1 2 18 syl2anc φ x A B x A x x B
20 19 biimpa φ x A B x A x x B
21 20 simp1d φ x A B x
22 20 simp2d φ x A B A x
23 20 simp3d φ x A B x B
24 14 21 16 22 23 letrd φ x A B A B
25 lbicc2 A * B * A B A A B
26 15 17 24 25 syl3anc φ x A B A A B
27 13 26 ffvelrnd φ x A B F A
28 6 ffvelrnda φ x A B F x
29 27 28 subcld φ x A B F A F x
30 simpr φ x A B x A B
31 26 30 jca φ x A B A A B x A B
32 4 dmeqd φ dom F = dom A B × 0
33 c0ex 0 V
34 33 snnz 0
35 dmxp 0 dom A B × 0 = A B
36 34 35 ax-mp dom A B × 0 = A B
37 32 36 eqtrdi φ dom F = A B
38 0red φ 0
39 4 fveq1d φ F y = A B × 0 y
40 33 fvconst2 y A B A B × 0 y = 0
41 39 40 sylan9eq φ y A B F y = 0
42 41 abs00bd φ y A B F y = 0
43 0le0 0 0
44 42 43 eqbrtrdi φ y A B F y 0
45 1 2 3 37 38 44 dvlip φ A A B x A B F A F x 0 A x
46 31 45 syldan φ x A B F A F x 0 A x
47 14 recnd φ x A B A
48 21 recnd φ x A B x
49 47 48 subcld φ x A B A x
50 49 abscld φ x A B A x
51 50 recnd φ x A B A x
52 51 mul02d φ x A B 0 A x = 0
53 46 52 breqtrd φ x A B F A F x 0
54 29 absge0d φ x A B 0 F A F x
55 29 abscld φ x A B F A F x
56 0re 0
57 letri3 F A F x 0 F A F x = 0 F A F x 0 0 F A F x
58 55 56 57 sylancl φ x A B F A F x = 0 F A F x 0 0 F A F x
59 53 54 58 mpbir2and φ x A B F A F x = 0
60 29 59 abs00d φ x A B F A F x = 0
61 27 28 60 subeq0d φ x A B F A = F x
62 12 61 eqtr2d φ x A B F x = A B × F A x
63 7 10 62 eqfnfvd φ F = A B × F A