Metamath Proof Explorer


Theorem summolem3

Description: Lemma for summo . (Contributed by Mario Carneiro, 29-Mar-2014)

Ref Expression
Hypotheses summo.1 F = k if k A B 0
summo.2 φ k A B
summo.3 G = n f n / k B
summolem3.4 H = n K n / k B
summolem3.5 φ M N
summolem3.6 φ f : 1 M 1-1 onto A
summolem3.7 φ K : 1 N 1-1 onto A
Assertion summolem3 φ seq 1 + G M = seq 1 + H N

Proof

Step Hyp Ref Expression
1 summo.1 F = k if k A B 0
2 summo.2 φ k A B
3 summo.3 G = n f n / k B
4 summolem3.4 H = n K n / k B
5 summolem3.5 φ M N
6 summolem3.6 φ f : 1 M 1-1 onto A
7 summolem3.7 φ K : 1 N 1-1 onto A
8 addcl m j m + j
9 8 adantl φ m j m + j
10 addcom m j m + j = j + m
11 10 adantl φ m j m + j = j + m
12 addass m j y m + j + y = m + j + y
13 12 adantl φ m j y m + j + y = m + j + y
14 5 simpld φ M
15 nnuz = 1
16 14 15 eleqtrdi φ M 1
17 ssidd φ
18 f1ocnv f : 1 M 1-1 onto A f -1 : A 1-1 onto 1 M
19 6 18 syl φ f -1 : A 1-1 onto 1 M
20 f1oco f -1 : A 1-1 onto 1 M K : 1 N 1-1 onto A f -1 K : 1 N 1-1 onto 1 M
21 19 7 20 syl2anc φ f -1 K : 1 N 1-1 onto 1 M
22 ovex 1 N V
23 22 f1oen f -1 K : 1 N 1-1 onto 1 M 1 N 1 M
24 21 23 syl φ 1 N 1 M
25 fzfi 1 N Fin
26 fzfi 1 M Fin
27 hashen 1 N Fin 1 M Fin 1 N = 1 M 1 N 1 M
28 25 26 27 mp2an 1 N = 1 M 1 N 1 M
29 24 28 sylibr φ 1 N = 1 M
30 5 simprd φ N
31 nnnn0 N N 0
32 hashfz1 N 0 1 N = N
33 30 31 32 3syl φ 1 N = N
34 nnnn0 M M 0
35 hashfz1 M 0 1 M = M
36 14 34 35 3syl φ 1 M = M
37 29 33 36 3eqtr3rd φ M = N
38 37 oveq2d φ 1 M = 1 N
39 38 f1oeq2d φ f -1 K : 1 M 1-1 onto 1 M f -1 K : 1 N 1-1 onto 1 M
40 21 39 mpbird φ f -1 K : 1 M 1-1 onto 1 M
41 fveq2 n = m f n = f m
42 41 csbeq1d n = m f n / k B = f m / k B
43 elfznn m 1 M m
44 43 adantl φ m 1 M m
45 f1of f : 1 M 1-1 onto A f : 1 M A
46 6 45 syl φ f : 1 M A
47 46 ffvelrnda φ m 1 M f m A
48 2 ralrimiva φ k A B
49 48 adantr φ m 1 M k A B
50 nfcsb1v _ k f m / k B
51 50 nfel1 k f m / k B
52 csbeq1a k = f m B = f m / k B
53 52 eleq1d k = f m B f m / k B
54 51 53 rspc f m A k A B f m / k B
55 47 49 54 sylc φ m 1 M f m / k B
56 3 42 44 55 fvmptd3 φ m 1 M G m = f m / k B
57 56 55 eqeltrd φ m 1 M G m
58 38 f1oeq2d φ K : 1 M 1-1 onto A K : 1 N 1-1 onto A
59 7 58 mpbird φ K : 1 M 1-1 onto A
60 f1of K : 1 M 1-1 onto A K : 1 M A
61 59 60 syl φ K : 1 M A
62 fvco3 K : 1 M A i 1 M f -1 K i = f -1 K i
63 61 62 sylan φ i 1 M f -1 K i = f -1 K i
64 63 fveq2d φ i 1 M f f -1 K i = f f -1 K i
65 6 adantr φ i 1 M f : 1 M 1-1 onto A
66 61 ffvelrnda φ i 1 M K i A
67 f1ocnvfv2 f : 1 M 1-1 onto A K i A f f -1 K i = K i
68 65 66 67 syl2anc φ i 1 M f f -1 K i = K i
69 64 68 eqtr2d φ i 1 M K i = f f -1 K i
70 69 csbeq1d φ i 1 M K i / k B = f f -1 K i / k B
71 70 fveq2d φ i 1 M I K i / k B = I f f -1 K i / k B
72 elfznn i 1 M i
73 72 adantl φ i 1 M i
74 fveq2 n = i K n = K i
75 74 csbeq1d n = i K n / k B = K i / k B
76 75 4 fvmpti i H i = I K i / k B
77 73 76 syl φ i 1 M H i = I K i / k B
78 f1of f -1 K : 1 M 1-1 onto 1 M f -1 K : 1 M 1 M
79 40 78 syl φ f -1 K : 1 M 1 M
80 79 ffvelrnda φ i 1 M f -1 K i 1 M
81 elfznn f -1 K i 1 M f -1 K i
82 fveq2 n = f -1 K i f n = f f -1 K i
83 82 csbeq1d n = f -1 K i f n / k B = f f -1 K i / k B
84 83 3 fvmpti f -1 K i G f -1 K i = I f f -1 K i / k B
85 80 81 84 3syl φ i 1 M G f -1 K i = I f f -1 K i / k B
86 71 77 85 3eqtr4d φ i 1 M H i = G f -1 K i
87 9 11 13 16 17 40 57 86 seqf1o φ seq 1 + H M = seq 1 + G M
88 37 fveq2d φ seq 1 + H M = seq 1 + H N
89 87 88 eqtr3d φ seq 1 + G M = seq 1 + H N