Metamath Proof Explorer


Theorem rpnnen2lem9

Description: Lemma for rpnnen2 . (Contributed by Mario Carneiro, 13-May-2013) (Revised by Mario Carneiro, 30-Apr-2014)

Ref Expression
Hypothesis rpnnen2.1 F = x 𝒫 n if n x 1 3 n 0
Assertion rpnnen2lem9 M k M F M k = 0 + 1 3 M + 1 1 1 3

Proof

Step Hyp Ref Expression
1 rpnnen2.1 F = x 𝒫 n if n x 1 3 n 0
2 eqid M = M
3 nnz M M
4 eqidd M k M F M k = F M k
5 eluznn M k M k
6 difss M
7 1 rpnnen2lem2 M F M :
8 6 7 ax-mp F M :
9 8 ffvelrni k F M k
10 9 recnd k F M k
11 5 10 syl M k M F M k
12 1 rpnnen2lem5 M M seq M + F M dom
13 6 12 mpan M seq M + F M dom
14 2 3 4 11 13 isum1p M k M F M k = F M M + k M + 1 F M k
15 1 rpnnen2lem1 M M F M M = if M M 1 3 M 0
16 6 15 mpan M F M M = if M M 1 3 M 0
17 neldifsnd M ¬ M M
18 17 iffalsed M if M M 1 3 M 0 = 0
19 16 18 eqtrd M F M M = 0
20 eqid M + 1 = M + 1
21 peano2nn M M + 1
22 21 nnzd M M + 1
23 eqidd M k M + 1 F M k = F M k
24 eluznn M + 1 k M + 1 k
25 21 24 sylan M k M + 1 k
26 25 10 syl M k M + 1 F M k
27 1re 1
28 3nn 3
29 nndivre 1 3 1 3
30 27 28 29 mp2an 1 3
31 30 recni 1 3
32 31 a1i M 1 3
33 0re 0
34 3re 3
35 3pos 0 < 3
36 34 35 recgt0ii 0 < 1 3
37 33 30 36 ltleii 0 1 3
38 absid 1 3 0 1 3 1 3 = 1 3
39 30 37 38 mp2an 1 3 = 1 3
40 1lt3 1 < 3
41 recgt1 3 0 < 3 1 < 3 1 3 < 1
42 34 35 41 mp2an 1 < 3 1 3 < 1
43 40 42 mpbi 1 3 < 1
44 39 43 eqbrtri 1 3 < 1
45 44 a1i M 1 3 < 1
46 21 nnnn0d M M + 1 0
47 1 rpnnen2lem1 M k F M k = if k M 1 3 k 0
48 6 47 mpan k F M k = if k M 1 3 k 0
49 25 48 syl M k M + 1 F M k = if k M 1 3 k 0
50 nnre M M
51 50 adantr M k M + 1 M
52 eluzle k M + 1 M + 1 k
53 52 adantl M k M + 1 M + 1 k
54 nnltp1le M k M < k M + 1 k
55 25 54 syldan M k M + 1 M < k M + 1 k
56 53 55 mpbird M k M + 1 M < k
57 51 56 gtned M k M + 1 k M
58 eldifsn k M k k M
59 25 57 58 sylanbrc M k M + 1 k M
60 59 iftrued M k M + 1 if k M 1 3 k 0 = 1 3 k
61 49 60 eqtrd M k M + 1 F M k = 1 3 k
62 32 45 46 61 geolim2 M seq M + 1 + F M 1 3 M + 1 1 1 3
63 20 22 23 26 62 isumclim M k M + 1 F M k = 1 3 M + 1 1 1 3
64 19 63 oveq12d M F M M + k M + 1 F M k = 0 + 1 3 M + 1 1 1 3
65 14 64 eqtrd M k M F M k = 0 + 1 3 M + 1 1 1 3