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 ∈ ℝ
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 ∈ ℕ
14 nnrp ⊢ 8 ∈ ℕ → 8 ∈ ℝ +
15 13 14 ax-mp ⊢ 8 ∈ ℝ +
16 0le2 ⊢ 0 ≤ 2
17 2lt8 ⊢ 2 < 8
18 modid ⊢ 2 ∈ ℝ ∧ 8 ∈ ℝ + ∧ 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