Metamath Proof Explorer


Theorem 2lgslem4

Description: Lemma 4 for 2lgs : special case of 2lgs for P = 2 . (Contributed by AV, 20-Jun-2021)

Ref Expression
Assertion 2lgslem4 2 / L 2 = 1 2 mod 8 1 7

Proof

Step Hyp Ref Expression
1 2lgs2 2 / L 2 = 0
2 1 eqeq1i 2 / L 2 = 1 0 = 1
3 0ne1 0 1
4 3 neii ¬ 0 = 1
5 1ne2 1 2
6 5 nesymi ¬ 2 = 1
7 2re 2
8 2lt7 2 < 7
9 7 8 ltneii 2 7
10 9 neii ¬ 2 = 7
11 6 10 pm3.2ni ¬ 2 = 1 2 = 7
12 4 11 2false 0 = 1 2 = 1 2 = 7
13 8nn 8
14 nnrp 8 8 +
15 13 14 ax-mp 8 +
16 0le2 0 2
17 2lt8 2 < 8
18 modid 2 8 + 0 2 2 < 8 2 mod 8 = 2
19 7 15 16 17 18 mp4an 2 mod 8 = 2
20 19 eleq1i 2 mod 8 1 7 2 1 7
21 2ex 2 V
22 21 elpr 2 1 7 2 = 1 2 = 7
23 20 22 bitr2i 2 = 1 2 = 7 2 mod 8 1 7
24 2 12 23 3bitri 2 / L 2 = 1 2 mod 8 1 7