Metamath Proof Explorer


Theorem dchrmusumlema

Description: Lemma for dchrmusum and dchrisumn0 . Apply dchrisum for the function 1 / y . (Contributed by Mario Carneiro, 4-May-2016)

Ref Expression
Hypotheses rpvmasum.z Z = /N
rpvmasum.l L = ℤRHom Z
rpvmasum.a φ N
rpvmasum.g G = DChr N
rpvmasum.d D = Base G
rpvmasum.1 1 ˙ = 0 G
dchrisum.b φ X D
dchrisum.n1 φ X 1 ˙
dchrisumn0.f F = a X L a a
Assertion dchrmusumlema φ t c 0 +∞ seq 1 + F t y 1 +∞ seq 1 + F y t c y

Proof

Step Hyp Ref Expression
1 rpvmasum.z Z = /N
2 rpvmasum.l L = ℤRHom Z
3 rpvmasum.a φ N
4 rpvmasum.g G = DChr N
5 rpvmasum.d D = Base G
6 rpvmasum.1 1 ˙ = 0 G
7 dchrisum.b φ X D
8 dchrisum.n1 φ X 1 ˙
9 dchrisumn0.f F = a X L a a
10 oveq2 n = x 1 n = 1 x
11 1nn 1
12 11 a1i φ 1
13 rpreccl n + 1 n +
14 13 adantl φ n + 1 n +
15 14 rpred φ n + 1 n
16 simp3r φ n + x + 1 n n x n x
17 rpregt0 n + n 0 < n
18 rpregt0 x + x 0 < x
19 lerec n 0 < n x 0 < x n x 1 x 1 n
20 17 18 19 syl2an n + x + n x 1 x 1 n
21 20 3ad2ant2 φ n + x + 1 n n x n x 1 x 1 n
22 16 21 mpbid φ n + x + 1 n n x 1 x 1 n
23 ax-1cn 1
24 divrcnv 1 n + 1 n 0
25 23 24 mp1i φ n + 1 n 0
26 2fveq3 a = n X L a = X L n
27 oveq2 a = n 1 a = 1 n
28 26 27 oveq12d a = n X L a 1 a = X L n 1 n
29 28 cbvmptv a X L a 1 a = n X L n 1 n
30 1 2 3 4 5 6 7 8 10 12 15 22 25 29 dchrisum φ t c 0 +∞ seq 1 + a X L a 1 a t x 1 +∞ seq 1 + a X L a 1 a x t c 1 x
31 7 adantr φ n X D
32 nnz n n
33 32 adantl φ n n
34 4 1 5 2 31 33 dchrzrhcl φ n X L n
35 nncn n n
36 35 adantl φ n n
37 nnne0 n n 0
38 37 adantl φ n n 0
39 34 36 38 divrecd φ n X L n n = X L n 1 n
40 39 mpteq2dva φ n X L n n = n X L n 1 n
41 id a = n a = n
42 26 41 oveq12d a = n X L a a = X L n n
43 42 cbvmptv a X L a a = n X L n n
44 9 43 eqtri F = n X L n n
45 40 44 29 3eqtr4g φ F = a X L a 1 a
46 45 adantr φ c 0 +∞ F = a X L a 1 a
47 46 seqeq3d φ c 0 +∞ seq 1 + F = seq 1 + a X L a 1 a
48 47 breq1d φ c 0 +∞ seq 1 + F t seq 1 + a X L a 1 a t
49 2fveq3 y = x seq 1 + F y = seq 1 + F x
50 49 fvoveq1d y = x seq 1 + F y t = seq 1 + F x t
51 oveq2 y = x c y = c x
52 50 51 breq12d y = x seq 1 + F y t c y seq 1 + F x t c x
53 52 cbvralvw y 1 +∞ seq 1 + F y t c y x 1 +∞ seq 1 + F x t c x
54 45 seqeq3d φ seq 1 + F = seq 1 + a X L a 1 a
55 54 fveq1d φ seq 1 + F x = seq 1 + a X L a 1 a x
56 55 fvoveq1d φ seq 1 + F x t = seq 1 + a X L a 1 a x t
57 56 ad2antrr φ c 0 +∞ x 1 +∞ seq 1 + F x t = seq 1 + a X L a 1 a x t
58 elrege0 c 0 +∞ c 0 c
59 58 simplbi c 0 +∞ c
60 59 recnd c 0 +∞ c
61 60 ad2antlr φ c 0 +∞ x 1 +∞ c
62 1re 1
63 elicopnf 1 x 1 +∞ x 1 x
64 62 63 ax-mp x 1 +∞ x 1 x
65 64 simplbi x 1 +∞ x
66 65 adantl φ c 0 +∞ x 1 +∞ x
67 66 recnd φ c 0 +∞ x 1 +∞ x
68 0red φ c 0 +∞ x 1 +∞ 0
69 1red φ c 0 +∞ x 1 +∞ 1
70 0lt1 0 < 1
71 70 a1i φ c 0 +∞ x 1 +∞ 0 < 1
72 64 simprbi x 1 +∞ 1 x
73 72 adantl φ c 0 +∞ x 1 +∞ 1 x
74 68 69 66 71 73 ltletrd φ c 0 +∞ x 1 +∞ 0 < x
75 74 gt0ne0d φ c 0 +∞ x 1 +∞ x 0
76 61 67 75 divrecd φ c 0 +∞ x 1 +∞ c x = c 1 x
77 57 76 breq12d φ c 0 +∞ x 1 +∞ seq 1 + F x t c x seq 1 + a X L a 1 a x t c 1 x
78 77 ralbidva φ c 0 +∞ x 1 +∞ seq 1 + F x t c x x 1 +∞ seq 1 + a X L a 1 a x t c 1 x
79 53 78 syl5bb φ c 0 +∞ y 1 +∞ seq 1 + F y t c y x 1 +∞ seq 1 + a X L a 1 a x t c 1 x
80 48 79 anbi12d φ c 0 +∞ seq 1 + F t y 1 +∞ seq 1 + F y t c y seq 1 + a X L a 1 a t x 1 +∞ seq 1 + a X L a 1 a x t c 1 x
81 80 rexbidva φ c 0 +∞ seq 1 + F t y 1 +∞ seq 1 + F y t c y c 0 +∞ seq 1 + a X L a 1 a t x 1 +∞ seq 1 + a X L a 1 a x t c 1 x
82 81 exbidv φ t c 0 +∞ seq 1 + F t y 1 +∞ seq 1 + F y t c y t c 0 +∞ seq 1 + a X L a 1 a t x 1 +∞ seq 1 + a X L a 1 a x t c 1 x
83 30 82 mpbird φ t c 0 +∞ seq 1 + F t y 1 +∞ seq 1 + F y t c y