Metamath Proof Explorer


Theorem lgsdir2lem3

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

Ref Expression
Assertion lgsdir2lem3 A ¬ 2 A A mod 8 1 7 3 5

Proof

Step Hyp Ref Expression
1 simpl A ¬ 2 A A
2 8nn 8
3 zmodfz A 8 A mod 8 0 8 1
4 1 2 3 sylancl A ¬ 2 A A mod 8 0 8 1
5 8m1e7 8 1 = 7
6 5 oveq2i 0 8 1 = 0 7
7 4 6 eleqtrdi A ¬ 2 A A mod 8 0 7
8 neg1z 1
9 z0even 2 0
10 1pneg1e0 1 + -1 = 0
11 ax-1cn 1
12 neg1cn 1
13 11 12 addcomi 1 + -1 = - 1 + 1
14 10 13 eqtr3i 0 = - 1 + 1
15 9 14 breqtri 2 - 1 + 1
16 noel ¬ A mod 8
17 16 pm2.21i A mod 8 A mod 8 1 7 3 5
18 neg1lt0 1 < 0
19 0z 0
20 fzn 0 1 1 < 0 0 1 =
21 19 8 20 mp2an 1 < 0 0 1 =
22 18 21 mpbi 0 1 =
23 17 22 eleq2s A mod 8 0 1 A mod 8 1 7 3 5
24 23 a1i A ¬ 2 A A mod 8 0 1 A mod 8 1 7 3 5
25 8 15 24 3pm3.2i 1 2 - 1 + 1 A ¬ 2 A A mod 8 0 1 A mod 8 1 7 3 5
26 1e0p1 1 = 0 + 1
27 ssun1 1 7 1 7 3 5
28 1ex 1 V
29 28 prid1 1 1 7
30 27 29 sselii 1 1 7 3 5
31 25 14 26 30 lgsdir2lem2 1 2 1 + 1 A ¬ 2 A A mod 8 0 1 A mod 8 1 7 3 5
32 df-2 2 = 1 + 1
33 df-3 3 = 2 + 1
34 ssun2 3 5 1 7 3 5
35 3ex 3 V
36 35 prid1 3 3 5
37 34 36 sselii 3 1 7 3 5
38 31 32 33 37 lgsdir2lem2 3 2 3 + 1 A ¬ 2 A A mod 8 0 3 A mod 8 1 7 3 5
39 df-4 4 = 3 + 1
40 df-5 5 = 4 + 1
41 5nn 5
42 41 elexi 5 V
43 42 prid2 5 3 5
44 34 43 sselii 5 1 7 3 5
45 38 39 40 44 lgsdir2lem2 5 2 5 + 1 A ¬ 2 A A mod 8 0 5 A mod 8 1 7 3 5
46 df-6 6 = 5 + 1
47 df-7 7 = 6 + 1
48 7nn 7
49 48 elexi 7 V
50 49 prid2 7 1 7
51 27 50 sselii 7 1 7 3 5
52 45 46 47 51 lgsdir2lem2 7 2 7 + 1 A ¬ 2 A A mod 8 0 7 A mod 8 1 7 3 5
53 52 simp3i A ¬ 2 A A mod 8 0 7 A mod 8 1 7 3 5
54 7 53 mpd A ¬ 2 A A mod 8 1 7 3 5