Metamath Proof Explorer


Theorem breprexpnat

Description: Express the S th power of the finite series in terms of the number of representations of integers m as sums of S terms of elements of A , bounded by N . Proposition of Nathanson p. 123. (Contributed by Thierry Arnoux, 11-Dec-2021)

Ref Expression
Hypotheses breprexp.n φ N 0
breprexp.s φ S 0
breprexp.z φ Z
breprexpnat.a φ A
breprexpnat.p P = b A 1 N Z b
breprexpnat.r R = A 1 N repr S m
Assertion breprexpnat φ P S = m = 0 S N R Z m

Proof

Step Hyp Ref Expression
1 breprexp.n φ N 0
2 breprexp.s φ S 0
3 breprexp.z φ Z
4 breprexpnat.a φ A
5 breprexpnat.p P = b A 1 N Z b
6 breprexpnat.r R = A 1 N repr S m
7 fvex 𝟙 A V
8 7 fconst 0 ..^ S × 𝟙 A : 0 ..^ S 𝟙 A
9 nnex V
10 indf V A 𝟙 A : 0 1
11 9 4 10 sylancr φ 𝟙 A : 0 1
12 0cn 0
13 ax-1cn 1
14 prssi 0 1 0 1
15 12 13 14 mp2an 0 1
16 fss 𝟙 A : 0 1 0 1 𝟙 A :
17 11 15 16 sylancl φ 𝟙 A :
18 cnex V
19 18 9 elmap 𝟙 A 𝟙 A :
20 17 19 sylibr φ 𝟙 A
21 7 snss 𝟙 A 𝟙 A
22 20 21 sylib φ 𝟙 A
23 fss 0 ..^ S × 𝟙 A : 0 ..^ S 𝟙 A 𝟙 A 0 ..^ S × 𝟙 A : 0 ..^ S
24 8 22 23 sylancr φ 0 ..^ S × 𝟙 A : 0 ..^ S
25 1 2 3 24 breprexp φ a 0 ..^ S b = 1 N 0 ..^ S × 𝟙 A a b Z b = m = 0 S N c 1 N repr S m a 0 ..^ S 0 ..^ S × 𝟙 A a c a Z m
26 7 fvconst2 a 0 ..^ S 0 ..^ S × 𝟙 A a = 𝟙 A
27 26 ad2antlr φ a 0 ..^ S b 1 N 0 ..^ S × 𝟙 A a = 𝟙 A
28 27 fveq1d φ a 0 ..^ S b 1 N 0 ..^ S × 𝟙 A a b = 𝟙 A b
29 28 oveq1d φ a 0 ..^ S b 1 N 0 ..^ S × 𝟙 A a b Z b = 𝟙 A b Z b
30 29 sumeq2dv φ a 0 ..^ S b = 1 N 0 ..^ S × 𝟙 A a b Z b = b = 1 N 𝟙 A b Z b
31 9 a1i φ a 0 ..^ S V
32 fzfi 1 N Fin
33 32 a1i φ a 0 ..^ S 1 N Fin
34 fz1ssnn 1 N
35 34 a1i φ a 0 ..^ S 1 N
36 4 adantr φ a 0 ..^ S A
37 3 ad2antrr φ a 0 ..^ S b 1 N Z
38 nnssnn0 0
39 34 38 sstri 1 N 0
40 simpr φ a 0 ..^ S b 1 N b 1 N
41 39 40 sselid φ a 0 ..^ S b 1 N b 0
42 37 41 expcld φ a 0 ..^ S b 1 N Z b
43 31 33 35 36 42 indsumin φ a 0 ..^ S b = 1 N 𝟙 A b Z b = b 1 N A Z b
44 incom 1 N A = A 1 N
45 44 a1i φ a 0 ..^ S 1 N A = A 1 N
46 45 sumeq1d φ a 0 ..^ S b 1 N A Z b = b A 1 N Z b
47 30 43 46 3eqtrd φ a 0 ..^ S b = 1 N 0 ..^ S × 𝟙 A a b Z b = b A 1 N Z b
48 47 prodeq2dv φ a 0 ..^ S b = 1 N 0 ..^ S × 𝟙 A a b Z b = a 0 ..^ S b A 1 N Z b
49 fzofi 0 ..^ S Fin
50 49 a1i φ 0 ..^ S Fin
51 inss2 A 1 N 1 N
52 ssfi 1 N Fin A 1 N 1 N A 1 N Fin
53 32 51 52 mp2an A 1 N Fin
54 53 a1i φ A 1 N Fin
55 3 adantr φ b A 1 N Z
56 51 39 sstri A 1 N 0
57 simpr φ b A 1 N b A 1 N
58 56 57 sselid φ b A 1 N b 0
59 55 58 expcld φ b A 1 N Z b
60 54 59 fsumcl φ b A 1 N Z b
61 fprodconst 0 ..^ S Fin b A 1 N Z b a 0 ..^ S b A 1 N Z b = b A 1 N Z b 0 ..^ S
62 50 60 61 syl2anc φ a 0 ..^ S b A 1 N Z b = b A 1 N Z b 0 ..^ S
63 hashfzo0 S 0 0 ..^ S = S
64 2 63 syl φ 0 ..^ S = S
65 64 oveq2d φ b A 1 N Z b 0 ..^ S = b A 1 N Z b S
66 48 62 65 3eqtrd φ a 0 ..^ S b = 1 N 0 ..^ S × 𝟙 A a b Z b = b A 1 N Z b S
67 34 a1i φ m 0 S N 1 N
68 fzssz 0 S N
69 simpr φ m 0 S N m 0 S N
70 68 69 sselid φ m 0 S N m
71 2 adantr φ m 0 S N S 0
72 32 a1i φ m 0 S N 1 N Fin
73 67 70 71 72 reprfi φ m 0 S N 1 N repr S m Fin
74 3 adantr φ m 0 S N Z
75 fz0ssnn0 0 S N 0
76 75 69 sselid φ m 0 S N m 0
77 74 76 expcld φ m 0 S N Z m
78 49 a1i φ m 0 S N c 1 N repr S m 0 ..^ S Fin
79 11 ad3antrrr φ m 0 S N c 1 N repr S m a 0 ..^ S 𝟙 A : 0 1
80 34 a1i φ m 0 S N c 1 N repr S m 1 N
81 70 adantr φ m 0 S N c 1 N repr S m m
82 71 adantr φ m 0 S N c 1 N repr S m S 0
83 simpr φ m 0 S N c 1 N repr S m c 1 N repr S m
84 80 81 82 83 reprf φ m 0 S N c 1 N repr S m c : 0 ..^ S 1 N
85 84 ffvelrnda φ m 0 S N c 1 N repr S m a 0 ..^ S c a 1 N
86 34 85 sselid φ m 0 S N c 1 N repr S m a 0 ..^ S c a
87 79 86 ffvelrnd φ m 0 S N c 1 N repr S m a 0 ..^ S 𝟙 A c a 0 1
88 15 87 sselid φ m 0 S N c 1 N repr S m a 0 ..^ S 𝟙 A c a
89 78 88 fprodcl φ m 0 S N c 1 N repr S m a 0 ..^ S 𝟙 A c a
90 73 77 89 fsummulc1 φ m 0 S N c 1 N repr S m a 0 ..^ S 𝟙 A c a Z m = c 1 N repr S m a 0 ..^ S 𝟙 A c a Z m
91 4 adantr φ m 0 S N A
92 91 70 71 72 67 hashreprin φ m 0 S N A 1 N repr S m = c 1 N repr S m a 0 ..^ S 𝟙 A c a
93 92 oveq1d φ m 0 S N A 1 N repr S m Z m = c 1 N repr S m a 0 ..^ S 𝟙 A c a Z m
94 26 fveq1d a 0 ..^ S 0 ..^ S × 𝟙 A a c a = 𝟙 A c a
95 94 adantl φ m 0 S N a 0 ..^ S 0 ..^ S × 𝟙 A a c a = 𝟙 A c a
96 95 prodeq2dv φ m 0 S N a 0 ..^ S 0 ..^ S × 𝟙 A a c a = a 0 ..^ S 𝟙 A c a
97 96 adantr φ m 0 S N c 1 N repr S m a 0 ..^ S 0 ..^ S × 𝟙 A a c a = a 0 ..^ S 𝟙 A c a
98 97 oveq1d φ m 0 S N c 1 N repr S m a 0 ..^ S 0 ..^ S × 𝟙 A a c a Z m = a 0 ..^ S 𝟙 A c a Z m
99 98 sumeq2dv φ m 0 S N c 1 N repr S m a 0 ..^ S 0 ..^ S × 𝟙 A a c a Z m = c 1 N repr S m a 0 ..^ S 𝟙 A c a Z m
100 90 93 99 3eqtr4rd φ m 0 S N c 1 N repr S m a 0 ..^ S 0 ..^ S × 𝟙 A a c a Z m = A 1 N repr S m Z m
101 100 sumeq2dv φ m = 0 S N c 1 N repr S m a 0 ..^ S 0 ..^ S × 𝟙 A a c a Z m = m = 0 S N A 1 N repr S m Z m
102 25 66 101 3eqtr3d φ b A 1 N Z b S = m = 0 S N A 1 N repr S m Z m
103 5 oveq1i P S = b A 1 N Z b S
104 6 oveq1i R Z m = A 1 N repr S m Z m
105 104 a1i m 0 S N R Z m = A 1 N repr S m Z m
106 105 sumeq2i m = 0 S N R Z m = m = 0 S N A 1 N repr S m Z m
107 102 103 106 3eqtr4g φ P S = m = 0 S N R Z m