Metamath Proof Explorer


Theorem faclbnd4lem3

Description: Lemma for faclbnd4 . The N = 0 case. (Contributed by NM, 23-Dec-2005)

Ref Expression
Assertion faclbnd4lem3 M 0 K 0 N = 0 N K M N 2 K 2 M M + K N !

Proof

Step Hyp Ref Expression
1 elnn0 K 0 K K = 0
2 0exp K 0 K = 0
3 2 adantl M 0 K 0 K = 0
4 nnnn0 K K 0
5 2nn0 2 0
6 nn0sqcl K 0 K 2 0
7 nn0expcl 2 0 K 2 0 2 K 2 0
8 5 6 7 sylancr K 0 2 K 2 0
9 8 adantl M 0 K 0 2 K 2 0
10 nn0addcl M 0 K 0 M + K 0
11 nn0expcl M 0 M + K 0 M M + K 0
12 10 11 syldan M 0 K 0 M M + K 0
13 9 12 nn0mulcld M 0 K 0 2 K 2 M M + K 0
14 4 13 sylan2 M 0 K 2 K 2 M M + K 0
15 14 nn0ge0d M 0 K 0 2 K 2 M M + K
16 3 15 eqbrtrd M 0 K 0 K 2 K 2 M M + K
17 1nn 1
18 elnn0 M 0 M M = 0
19 nnnn0 M M 0
20 0nn0 0 0
21 nn0addcl M 0 0 0 M + 0 0
22 19 20 21 sylancl M M + 0 0
23 nnexpcl M M + 0 0 M M + 0
24 22 23 mpdan M M M + 0
25 id M = 0 M = 0
26 oveq1 M = 0 M + 0 = 0 + 0
27 00id 0 + 0 = 0
28 26 27 eqtrdi M = 0 M + 0 = 0
29 25 28 oveq12d M = 0 M M + 0 = 0 0
30 0exp0e1 0 0 = 1
31 29 30 eqtrdi M = 0 M M + 0 = 1
32 31 17 eqeltrdi M = 0 M M + 0
33 24 32 jaoi M M = 0 M M + 0
34 18 33 sylbi M 0 M M + 0
35 nnmulcl 1 M M + 0 1 M M + 0
36 17 34 35 sylancr M 0 1 M M + 0
37 36 nnge1d M 0 1 1 M M + 0
38 37 adantr M 0 K = 0 1 1 M M + 0
39 oveq2 K = 0 0 K = 0 0
40 39 30 eqtrdi K = 0 0 K = 1
41 sq0i K = 0 K 2 = 0
42 41 oveq2d K = 0 2 K 2 = 2 0
43 2cn 2
44 exp0 2 2 0 = 1
45 43 44 ax-mp 2 0 = 1
46 42 45 eqtrdi K = 0 2 K 2 = 1
47 oveq2 K = 0 M + K = M + 0
48 47 oveq2d K = 0 M M + K = M M + 0
49 46 48 oveq12d K = 0 2 K 2 M M + K = 1 M M + 0
50 40 49 breq12d K = 0 0 K 2 K 2 M M + K 1 1 M M + 0
51 50 adantl M 0 K = 0 0 K 2 K 2 M M + K 1 1 M M + 0
52 38 51 mpbird M 0 K = 0 0 K 2 K 2 M M + K
53 16 52 jaodan M 0 K K = 0 0 K 2 K 2 M M + K
54 1 53 sylan2b M 0 K 0 0 K 2 K 2 M M + K
55 nn0cn M 0 M
56 55 exp0d M 0 M 0 = 1
57 56 oveq2d M 0 0 K M 0 = 0 K 1
58 nn0expcl 0 0 K 0 0 K 0
59 20 58 mpan K 0 0 K 0
60 59 nn0cnd K 0 0 K
61 60 mulid1d K 0 0 K 1 = 0 K
62 57 61 sylan9eq M 0 K 0 0 K M 0 = 0 K
63 13 nn0cnd M 0 K 0 2 K 2 M M + K
64 63 mulid1d M 0 K 0 2 K 2 M M + K 1 = 2 K 2 M M + K
65 54 62 64 3brtr4d M 0 K 0 0 K M 0 2 K 2 M M + K 1
66 65 adantr M 0 K 0 N = 0 0 K M 0 2 K 2 M M + K 1
67 oveq1 N = 0 N K = 0 K
68 oveq2 N = 0 M N = M 0
69 67 68 oveq12d N = 0 N K M N = 0 K M 0
70 fveq2 N = 0 N ! = 0 !
71 fac0 0 ! = 1
72 70 71 eqtrdi N = 0 N ! = 1
73 72 oveq2d N = 0 2 K 2 M M + K N ! = 2 K 2 M M + K 1
74 69 73 breq12d N = 0 N K M N 2 K 2 M M + K N ! 0 K M 0 2 K 2 M M + K 1
75 74 adantl M 0 K 0 N = 0 N K M N 2 K 2 M M + K N ! 0 K M 0 2 K 2 M M + K 1
76 66 75 mpbird M 0 K 0 N = 0 N K M N 2 K 2 M M + K N !