Metamath Proof Explorer


Theorem lcmdvds

Description: The lcm of two integers divides any integer the two divide. (Contributed by Steve Rodriguez, 20-Jan-2020)

Ref Expression
Assertion lcmdvds K M N M K N K M lcm N K

Proof

Step Hyp Ref Expression
1 id 0 K 0 K
2 breq1 M = 0 M K 0 K
3 2 adantl N M = 0 M K 0 K
4 oveq1 M = 0 M lcm N = 0 lcm N
5 0z 0
6 lcmcom 0 N 0 lcm N = N lcm 0
7 5 6 mpan N 0 lcm N = N lcm 0
8 lcm0val N N lcm 0 = 0
9 7 8 eqtrd N 0 lcm N = 0
10 4 9 sylan9eqr N M = 0 M lcm N = 0
11 10 breq1d N M = 0 M lcm N K 0 K
12 3 11 imbi12d N M = 0 M K M lcm N K 0 K 0 K
13 1 12 mpbiri N M = 0 M K M lcm N K
14 13 3ad2antl3 K M N M = 0 M K M lcm N K
15 14 adantrd K M N M = 0 M K N K M lcm N K
16 15 ex K M N M = 0 M K N K M lcm N K
17 breq1 N = 0 N K 0 K
18 17 adantl M N = 0 N K 0 K
19 oveq2 N = 0 M lcm N = M lcm 0
20 lcm0val M M lcm 0 = 0
21 19 20 sylan9eqr M N = 0 M lcm N = 0
22 21 breq1d M N = 0 M lcm N K 0 K
23 18 22 imbi12d M N = 0 N K M lcm N K 0 K 0 K
24 1 23 mpbiri M N = 0 N K M lcm N K
25 24 3ad2antl2 K M N N = 0 N K M lcm N K
26 25 adantld K M N N = 0 M K N K M lcm N K
27 26 ex K M N N = 0 M K N K M lcm N K
28 neanior M 0 N 0 ¬ M = 0 N = 0
29 lcmcl M N M lcm N 0
30 29 nn0zd M N M lcm N
31 dvds0 M lcm N M lcm N 0
32 30 31 syl M N M lcm N 0
33 32 a1d M N M 0 N 0 M lcm N 0
34 33 adantr M N K = 0 M 0 N 0 M lcm N 0
35 breq2 K = 0 M K M 0
36 breq2 K = 0 N K N 0
37 35 36 anbi12d K = 0 M K N K M 0 N 0
38 breq2 K = 0 M lcm N K M lcm N 0
39 37 38 imbi12d K = 0 M K N K M lcm N K M 0 N 0 M lcm N 0
40 39 adantl M N K = 0 M K N K M lcm N K M 0 N 0 M lcm N 0
41 34 40 mpbird M N K = 0 M K N K M lcm N K
42 41 adantrl M N K K = 0 M K N K M lcm N K
43 42 adantllr M M 0 N K K = 0 M K N K M lcm N K
44 43 adantlrr M M 0 N N 0 K K = 0 M K N K M lcm N K
45 44 anassrs M M 0 N N 0 K K = 0 M K N K M lcm N K
46 nnabscl M M 0 M
47 nnabscl N N 0 N
48 nnabscl K K 0 K
49 lcmgcdlem M N M lcm N M gcd N = M N K M K N K M lcm N K
50 49 simprd M N K M K N K M lcm N K
51 48 50 sylani M N K K 0 M K N K M lcm N K
52 46 47 51 syl2an M M 0 N N 0 K K 0 M K N K M lcm N K
53 52 expdimp M M 0 N N 0 K K 0 M K N K M lcm N K
54 dvdsabsb M K M K M K
55 zabscl K K
56 absdvdsb M K M K M K
57 55 56 sylan2 M K M K M K
58 54 57 bitrd M K M K M K
59 58 adantlr M N K M K M K
60 dvdsabsb N K N K N K
61 absdvdsb N K N K N K
62 55 61 sylan2 N K N K N K
63 60 62 bitrd N K N K N K
64 63 adantll M N K N K N K
65 59 64 anbi12d M N K M K N K M K N K
66 65 bicomd M N K M K N K M K N K
67 lcmabs M N M lcm N = M lcm N
68 67 breq1d M N M lcm N K M lcm N K
69 68 adantr M N K M lcm N K M lcm N K
70 dvdsabsb M lcm N K M lcm N K M lcm N K
71 30 70 sylan M N K M lcm N K M lcm N K
72 69 71 bitr4d M N K M lcm N K M lcm N K
73 66 72 imbi12d M N K M K N K M lcm N K M K N K M lcm N K
74 73 adantrr M N K K 0 M K N K M lcm N K M K N K M lcm N K
75 74 adantllr M M 0 N K K 0 M K N K M lcm N K M K N K M lcm N K
76 75 adantlrr M M 0 N N 0 K K 0 M K N K M lcm N K M K N K M lcm N K
77 53 76 mpbid M M 0 N N 0 K K 0 M K N K M lcm N K
78 77 anassrs M M 0 N N 0 K K 0 M K N K M lcm N K
79 45 78 pm2.61dane M M 0 N N 0 K M K N K M lcm N K
80 79 ex M M 0 N N 0 K M K N K M lcm N K
81 80 an4s M N M 0 N 0 K M K N K M lcm N K
82 28 81 sylan2br M N ¬ M = 0 N = 0 K M K N K M lcm N K
83 82 impancom M N K ¬ M = 0 N = 0 M K N K M lcm N K
84 83 3impa M N K ¬ M = 0 N = 0 M K N K M lcm N K
85 84 3comr K M N ¬ M = 0 N = 0 M K N K M lcm N K
86 16 27 85 ecase3d K M N M K N K M lcm N K