Metamath Proof Explorer


Theorem gcdcllem3

Description: Lemma for gcdn0cl , gcddvds and dvdslegcd . (Contributed by Paul Chapman, 21-Mar-2011)

Ref Expression
Hypotheses gcdcllem2.1 S = z | n M N z n
gcdcllem2.2 R = z | z M z N
Assertion gcdcllem3 M N ¬ M = 0 N = 0 sup R < sup R < M sup R < N K K M K N K sup R <

Proof

Step Hyp Ref Expression
1 gcdcllem2.1 S = z | n M N z n
2 gcdcllem2.2 R = z | z M z N
3 2 ssrab3 R
4 prssi M N M N
5 neorian M 0 N 0 ¬ M = 0 N = 0
6 prid1g M M M N
7 neeq1 n = M n 0 M 0
8 7 rspcev M M N M 0 n M N n 0
9 6 8 sylan M M 0 n M N n 0
10 9 adantlr M N M 0 n M N n 0
11 prid2g N N M N
12 neeq1 n = N n 0 N 0
13 12 rspcev N M N N 0 n M N n 0
14 11 13 sylan N N 0 n M N n 0
15 14 adantll M N N 0 n M N n 0
16 10 15 jaodan M N M 0 N 0 n M N n 0
17 5 16 sylan2br M N ¬ M = 0 N = 0 n M N n 0
18 1 gcdcllem1 M N n M N n 0 S x y S y x
19 4 17 18 syl2an2r M N ¬ M = 0 N = 0 S x y S y x
20 1 2 gcdcllem2 M N R = S
21 neeq1 R = S R S
22 raleq R = S y R y x y S y x
23 22 rexbidv R = S x y R y x x y S y x
24 21 23 anbi12d R = S R x y R y x S x y S y x
25 20 24 syl M N R x y R y x S x y S y x
26 25 adantr M N ¬ M = 0 N = 0 R x y R y x S x y S y x
27 19 26 mpbird M N ¬ M = 0 N = 0 R x y R y x
28 suprzcl2 R R x y R y x sup R < R
29 3 28 mp3an1 R x y R y x sup R < R
30 27 29 syl M N ¬ M = 0 N = 0 sup R < R
31 3 30 sselid M N ¬ M = 0 N = 0 sup R <
32 27 simprd M N ¬ M = 0 N = 0 x y R y x
33 1dvds M 1 M
34 1dvds N 1 N
35 33 34 anim12i M N 1 M 1 N
36 1z 1
37 breq1 z = 1 z M 1 M
38 breq1 z = 1 z N 1 N
39 37 38 anbi12d z = 1 z M z N 1 M 1 N
40 39 2 elrab2 1 R 1 1 M 1 N
41 36 40 mpbiran 1 R 1 M 1 N
42 35 41 sylibr M N 1 R
43 42 adantr M N ¬ M = 0 N = 0 1 R
44 suprzub R x y R y x 1 R 1 sup R <
45 3 32 43 44 mp3an2i M N ¬ M = 0 N = 0 1 sup R <
46 elnnz1 sup R < sup R < 1 sup R <
47 31 45 46 sylanbrc M N ¬ M = 0 N = 0 sup R <
48 breq1 x = sup R < x M sup R < M
49 breq1 x = sup R < x N sup R < N
50 48 49 anbi12d x = sup R < x M x N sup R < M sup R < N
51 breq1 z = x z M x M
52 breq1 z = x z N x N
53 51 52 anbi12d z = x z M z N x M x N
54 53 cbvrabv z | z M z N = x | x M x N
55 2 54 eqtri R = x | x M x N
56 50 55 elrab2 sup R < R sup R < sup R < M sup R < N
57 30 56 sylib M N ¬ M = 0 N = 0 sup R < sup R < M sup R < N
58 57 simprd M N ¬ M = 0 N = 0 sup R < M sup R < N
59 breq1 z = K z M K M
60 breq1 z = K z N K N
61 59 60 anbi12d z = K z M z N K M K N
62 61 2 elrab2 K R K K M K N
63 62 biimpri K K M K N K R
64 63 3impb K K M K N K R
65 suprzub R x y R y x K R K sup R <
66 65 3expia R x y R y x K R K sup R <
67 3 66 mpan x y R y x K R K sup R <
68 32 64 67 syl2im M N ¬ M = 0 N = 0 K K M K N K sup R <
69 47 58 68 3jca M N ¬ M = 0 N = 0 sup R < sup R < M sup R < N K K M K N K sup R <