Metamath Proof Explorer


Theorem lgsdir2lem1

Description: Lemma for lgsdir2 . (Contributed by Mario Carneiro, 4-Feb-2015)

Ref Expression
Assertion lgsdir2lem1 1 mod 8 = 1 -1 mod 8 = 7 3 mod 8 = 3 -3 mod 8 = 5

Proof

Step Hyp Ref Expression
1 1re 1
2 8re 8
3 8pos 0 < 8
4 2 3 elrpii 8 +
5 0le1 0 1
6 1lt8 1 < 8
7 modid 1 8 + 0 1 1 < 8 1 mod 8 = 1
8 1 4 5 6 7 mp4an 1 mod 8 = 1
9 8cn 8
10 9 mulid2i 1 8 = 8
11 10 oveq2i - 1 + 1 8 = - 1 + 8
12 ax-1cn 1
13 12 negcli 1
14 9 12 negsubi 8 + -1 = 8 1
15 8m1e7 8 1 = 7
16 14 15 eqtri 8 + -1 = 7
17 9 13 16 addcomli - 1 + 8 = 7
18 11 17 eqtri - 1 + 1 8 = 7
19 18 oveq1i - 1 + 1 8 mod 8 = 7 mod 8
20 1 renegcli 1
21 1z 1
22 modcyc 1 8 + 1 - 1 + 1 8 mod 8 = -1 mod 8
23 20 4 21 22 mp3an - 1 + 1 8 mod 8 = -1 mod 8
24 7re 7
25 0re 0
26 7pos 0 < 7
27 25 24 26 ltleii 0 7
28 7lt8 7 < 8
29 modid 7 8 + 0 7 7 < 8 7 mod 8 = 7
30 24 4 27 28 29 mp4an 7 mod 8 = 7
31 19 23 30 3eqtr3i -1 mod 8 = 7
32 8 31 pm3.2i 1 mod 8 = 1 -1 mod 8 = 7
33 3re 3
34 3pos 0 < 3
35 25 33 34 ltleii 0 3
36 3lt8 3 < 8
37 modid 3 8 + 0 3 3 < 8 3 mod 8 = 3
38 33 4 35 36 37 mp4an 3 mod 8 = 3
39 10 oveq2i - 3 + 1 8 = - 3 + 8
40 3cn 3
41 40 negcli 3
42 9 40 negsubi 8 + -3 = 8 3
43 5cn 5
44 5p3e8 5 + 3 = 8
45 43 40 44 addcomli 3 + 5 = 8
46 9 40 43 45 subaddrii 8 3 = 5
47 42 46 eqtri 8 + -3 = 5
48 9 41 47 addcomli - 3 + 8 = 5
49 39 48 eqtri - 3 + 1 8 = 5
50 49 oveq1i - 3 + 1 8 mod 8 = 5 mod 8
51 33 renegcli 3
52 modcyc 3 8 + 1 - 3 + 1 8 mod 8 = -3 mod 8
53 51 4 21 52 mp3an - 3 + 1 8 mod 8 = -3 mod 8
54 5re 5
55 5pos 0 < 5
56 25 54 55 ltleii 0 5
57 5lt8 5 < 8
58 modid 5 8 + 0 5 5 < 8 5 mod 8 = 5
59 54 4 56 57 58 mp4an 5 mod 8 = 5
60 50 53 59 3eqtr3i -3 mod 8 = 5
61 38 60 pm3.2i 3 mod 8 = 3 -3 mod 8 = 5
62 32 61 pm3.2i 1 mod 8 = 1 -1 mod 8 = 7 3 mod 8 = 3 -3 mod 8 = 5