Metamath Proof Explorer


Theorem lcmcllem

Description: Lemma for lcmn0cl and dvdslcm . (Contributed by Steve Rodriguez, 20-Jan-2020) (Proof shortened by AV, 16-Sep-2020)

Ref Expression
Assertion lcmcllem M N ¬ M = 0 N = 0 M lcm N n | M n N n

Proof

Step Hyp Ref Expression
1 lcmn0val M N ¬ M = 0 N = 0 M lcm N = sup n | M n N n <
2 ssrab2 n | M n N n
3 nnuz = 1
4 2 3 sseqtri n | M n N n 1
5 zmulcl M N M N
6 5 adantr M N ¬ M = 0 N = 0 M N
7 zcn M M
8 zcn N N
9 7 8 anim12i M N M N
10 ioran ¬ M = 0 N = 0 ¬ M = 0 ¬ N = 0
11 df-ne M 0 ¬ M = 0
12 df-ne N 0 ¬ N = 0
13 11 12 anbi12i M 0 N 0 ¬ M = 0 ¬ N = 0
14 10 13 sylbb2 ¬ M = 0 N = 0 M 0 N 0
15 mulne0 M M 0 N N 0 M N 0
16 15 an4s M N M 0 N 0 M N 0
17 9 14 16 syl2an M N ¬ M = 0 N = 0 M N 0
18 nnabscl M N M N 0 M N
19 6 17 18 syl2anc M N ¬ M = 0 N = 0 M N
20 dvdsmul1 M N M M N
21 dvdsabsb M M N M M N M M N
22 5 21 syldan M N M M N M M N
23 20 22 mpbid M N M M N
24 dvdsmul2 M N N M N
25 dvdsabsb N M N N M N N M N
26 5 25 sylan2 N M N N M N N M N
27 26 anabss7 M N N M N N M N
28 24 27 mpbid M N N M N
29 23 28 jca M N M M N N M N
30 29 adantr M N ¬ M = 0 N = 0 M M N N M N
31 breq2 n = M N M n M M N
32 breq2 n = M N N n N M N
33 31 32 anbi12d n = M N M n N n M M N N M N
34 33 rspcev M N M M N N M N n M n N n
35 19 30 34 syl2anc M N ¬ M = 0 N = 0 n M n N n
36 rabn0 n | M n N n n M n N n
37 35 36 sylibr M N ¬ M = 0 N = 0 n | M n N n
38 infssuzcl n | M n N n 1 n | M n N n sup n | M n N n < n | M n N n
39 4 37 38 sylancr M N ¬ M = 0 N = 0 sup n | M n N n < n | M n N n
40 1 39 eqeltrd M N ¬ M = 0 N = 0 M lcm N n | M n N n