Metamath Proof Explorer


Theorem 2lgslem3a

Description: Lemma for 2lgslem3a1 . (Contributed by AV, 14-Jul-2021)

Ref Expression
Hypothesis 2lgslem2.n ⊢ N = P − 1 2 − P 4
Assertion 2lgslem3a ⊢ K ∈ ℕ 0 ∧ P = 8 ⁢ K + 1 → N = 2 ⁢ K

Proof

Step Hyp Ref Expression
1 2lgslem2.n ⊢ N = P − 1 2 − P 4
2 oveq1 ⊢ P = 8 ⁢ K + 1 → P − 1 = 8 ⁢ K + 1 - 1
3 2 oveq1d ⊢ P = 8 ⁢ K + 1 → P − 1 2 = 8 ⁢ K + 1 - 1 2
4 fvoveq1 ⊢ P = 8 ⁢ K + 1 → P 4 = 8 ⁢ K + 1 4
5 3 4 oveq12d ⊢ P = 8 ⁢ K + 1 → P − 1 2 − P 4 = 8 ⁢ K + 1 - 1 2 − 8 ⁢ K + 1 4
6 1 5 eqtrid ⊢ P = 8 ⁢ K + 1 → N = 8 ⁢ K + 1 - 1 2 − 8 ⁢ K + 1 4
7 8nn0 ⊢ 8 ∈ ℕ 0
8 7 a1i ⊢ K ∈ ℕ 0 → 8 ∈ ℕ 0
9 id ⊢ K ∈ ℕ 0 → K ∈ ℕ 0
10 8 9 nn0mulcld ⊢ K ∈ ℕ 0 → 8 ⁢ K ∈ ℕ 0
11 10 nn0cnd ⊢ K ∈ ℕ 0 → 8 ⁢ K ∈ ℂ
12 pncan1 ⊢ 8 ⁢ K ∈ ℂ → 8 ⁢ K + 1 - 1 = 8 ⁢ K
13 11 12 syl ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 - 1 = 8 ⁢ K
14 13 oveq1d ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 - 1 2 = 8 ⁢ K 2
15 2t4e8 ⊢ 2 ⋅ 4 = 8
16 15 eqcomi ⊢ 8 = 2 ⋅ 4
17 16 a1i ⊢ K ∈ ℕ 0 → 8 = 2 ⋅ 4
18 17 oveq1d ⊢ K ∈ ℕ 0 → 8 ⁢ K = 2 ⋅ 4 ⁢ K
19 2cn ⊢ 2 ∈ ℂ
20 19 a1i ⊢ K ∈ ℕ 0 → 2 ∈ ℂ
21 4cn ⊢ 4 ∈ ℂ
22 21 a1i ⊢ K ∈ ℕ 0 → 4 ∈ ℂ
23 nn0cn ⊢ K ∈ ℕ 0 → K ∈ ℂ
24 20 22 23 mulassd ⊢ K ∈ ℕ 0 → 2 ⋅ 4 ⁢ K = 2 ⁢ 4 ⁢ K
25 18 24 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K = 2 ⁢ 4 ⁢ K
26 25 oveq1d ⊢ K ∈ ℕ 0 → 8 ⁢ K 2 = 2 ⁢ 4 ⁢ K 2
27 4nn0 ⊢ 4 ∈ ℕ 0
28 27 a1i ⊢ K ∈ ℕ 0 → 4 ∈ ℕ 0
29 28 9 nn0mulcld ⊢ K ∈ ℕ 0 → 4 ⁢ K ∈ ℕ 0
30 29 nn0cnd ⊢ K ∈ ℕ 0 → 4 ⁢ K ∈ ℂ
31 2ne0 ⊢ 2 ≠ 0
32 31 a1i ⊢ K ∈ ℕ 0 → 2 ≠ 0
33 30 20 32 divcan3d ⊢ K ∈ ℕ 0 → 2 ⁢ 4 ⁢ K 2 = 4 ⁢ K
34 14 26 33 3eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 - 1 2 = 4 ⁢ K
35 1cnd ⊢ K ∈ ℕ 0 → 1 ∈ ℂ
36 4ne0 ⊢ 4 ≠ 0
37 21 36 pm3.2i ⊢ 4 ∈ ℂ ∧ 4 ≠ 0
38 37 a1i ⊢ K ∈ ℕ 0 → 4 ∈ ℂ ∧ 4 ≠ 0
39 divdir ⊢ 8 ⁢ K ∈ ℂ ∧ 1 ∈ ℂ ∧ 4 ∈ ℂ ∧ 4 ≠ 0 → 8 ⁢ K + 1 4 = 8 ⁢ K 4 + 1 4
40 11 35 38 39 syl3anc ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 4 = 8 ⁢ K 4 + 1 4
41 8cn ⊢ 8 ∈ ℂ
42 41 a1i ⊢ K ∈ ℕ 0 → 8 ∈ ℂ
43 div23 ⊢ 8 ∈ ℂ ∧ K ∈ ℂ ∧ 4 ∈ ℂ ∧ 4 ≠ 0 → 8 ⁢ K 4 = 8 4 ⁢ K
44 42 23 38 43 syl3anc ⊢ K ∈ ℕ 0 → 8 ⁢ K 4 = 8 4 ⁢ K
45 4t2e8 ⊢ 4 ⋅ 2 = 8
46 45 eqcomi ⊢ 8 = 4 ⋅ 2
47 46 oveq1i ⊢ 8 4 = 4 ⋅ 2 4
48 19 21 36 divcan3i ⊢ 4 ⋅ 2 4 = 2
49 47 48 eqtri ⊢ 8 4 = 2
50 49 a1i ⊢ K ∈ ℕ 0 → 8 4 = 2
51 50 oveq1d ⊢ K ∈ ℕ 0 → 8 4 ⁢ K = 2 ⁢ K
52 44 51 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K 4 = 2 ⁢ K
53 52 oveq1d ⊢ K ∈ ℕ 0 → 8 ⁢ K 4 + 1 4 = 2 ⁢ K + 1 4
54 40 53 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 4 = 2 ⁢ K + 1 4
55 54 fveq2d ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 4 = 2 ⁢ K + 1 4
56 1lt4 ⊢ 1 < 4
57 2nn0 ⊢ 2 ∈ ℕ 0
58 57 a1i ⊢ K ∈ ℕ 0 → 2 ∈ ℕ 0
59 58 9 nn0mulcld ⊢ K ∈ ℕ 0 → 2 ⁢ K ∈ ℕ 0
60 59 nn0zd ⊢ K ∈ ℕ 0 → 2 ⁢ K ∈ ℤ
61 1nn0 ⊢ 1 ∈ ℕ 0
62 61 a1i ⊢ K ∈ ℕ 0 → 1 ∈ ℕ 0
63 4nn ⊢ 4 ∈ ℕ
64 63 a1i ⊢ K ∈ ℕ 0 → 4 ∈ ℕ
65 adddivflid ⊢ 2 ⁢ K ∈ ℤ ∧ 1 ∈ ℕ 0 ∧ 4 ∈ ℕ → 1 < 4 ↔ 2 ⁢ K + 1 4 = 2 ⁢ K
66 60 62 64 65 syl3anc ⊢ K ∈ ℕ 0 → 1 < 4 ↔ 2 ⁢ K + 1 4 = 2 ⁢ K
67 56 66 mpbii ⊢ K ∈ ℕ 0 → 2 ⁢ K + 1 4 = 2 ⁢ K
68 55 67 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 4 = 2 ⁢ K
69 34 68 oveq12d ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 - 1 2 − 8 ⁢ K + 1 4 = 4 ⁢ K − 2 ⁢ K
70 2t2e4 ⊢ 2 ⋅ 2 = 4
71 70 eqcomi ⊢ 4 = 2 ⋅ 2
72 71 a1i ⊢ K ∈ ℕ 0 → 4 = 2 ⋅ 2
73 72 oveq1d ⊢ K ∈ ℕ 0 → 4 ⁢ K = 2 ⋅ 2 ⁢ K
74 20 20 23 mulassd ⊢ K ∈ ℕ 0 → 2 ⋅ 2 ⁢ K = 2 ⁢ 2 ⁢ K
75 73 74 eqtrd ⊢ K ∈ ℕ 0 → 4 ⁢ K = 2 ⁢ 2 ⁢ K
76 75 oveq1d ⊢ K ∈ ℕ 0 → 4 ⁢ K − 2 ⁢ K = 2 ⁢ 2 ⁢ K − 2 ⁢ K
77 59 nn0cnd ⊢ K ∈ ℕ 0 → 2 ⁢ K ∈ ℂ
78 2txmxeqx ⊢ 2 ⁢ K ∈ ℂ → 2 ⁢ 2 ⁢ K − 2 ⁢ K = 2 ⁢ K
79 77 78 syl ⊢ K ∈ ℕ 0 → 2 ⁢ 2 ⁢ K − 2 ⁢ K = 2 ⁢ K
80 69 76 79 3eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 1 - 1 2 − 8 ⁢ K + 1 4 = 2 ⁢ K
81 6 80 sylan9eqr ⊢ K ∈ ℕ 0 ∧ P = 8 ⁢ K + 1 → N = 2 ⁢ K