Metamath Proof Explorer


Theorem 2lgslem3b

Description: Lemma for 2lgslem3b1 . (Contributed by AV, 16-Jul-2021)

Ref Expression
Hypothesis 2lgslem2.n ⊢ N = P − 1 2 − P 4
Assertion 2lgslem3b ⊢ K ∈ ℕ 0 ∧ P = 8 ⁢ K + 3 → N = 2 ⁢ K + 1

Proof

Step Hyp Ref Expression
1 2lgslem2.n ⊢ N = P − 1 2 − P 4
2 oveq1 ⊢ P = 8 ⁢ K + 3 → P − 1 = 8 ⁢ K + 3 - 1
3 2 oveq1d ⊢ P = 8 ⁢ K + 3 → P − 1 2 = 8 ⁢ K + 3 - 1 2
4 fvoveq1 ⊢ P = 8 ⁢ K + 3 → P 4 = 8 ⁢ K + 3 4
5 3 4 oveq12d ⊢ P = 8 ⁢ K + 3 → P − 1 2 − P 4 = 8 ⁢ K + 3 - 1 2 − 8 ⁢ K + 3 4
6 1 5 eqtrid ⊢ P = 8 ⁢ K + 3 → N = 8 ⁢ K + 3 - 1 2 − 8 ⁢ K + 3 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 3cn ⊢ 3 ∈ ℂ
13 12 a1i ⊢ K ∈ ℕ 0 → 3 ∈ ℂ
14 1cnd ⊢ K ∈ ℕ 0 → 1 ∈ ℂ
15 11 13 14 addsubassd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 - 1 = 8 ⁢ K + 3 - 1
16 4t2e8 ⊢ 4 ⋅ 2 = 8
17 16 eqcomi ⊢ 8 = 4 ⋅ 2
18 17 a1i ⊢ K ∈ ℕ 0 → 8 = 4 ⋅ 2
19 18 oveq1d ⊢ K ∈ ℕ 0 → 8 ⁢ K = 4 ⋅ 2 ⁢ K
20 4cn ⊢ 4 ∈ ℂ
21 20 a1i ⊢ K ∈ ℕ 0 → 4 ∈ ℂ
22 2cn ⊢ 2 ∈ ℂ
23 22 a1i ⊢ K ∈ ℕ 0 → 2 ∈ ℂ
24 nn0cn ⊢ K ∈ ℕ 0 → K ∈ ℂ
25 21 23 24 mul32d ⊢ K ∈ ℕ 0 → 4 ⋅ 2 ⁢ K = 4 ⁢ K ⋅ 2
26 19 25 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K = 4 ⁢ K ⋅ 2
27 3m1e2 ⊢ 3 − 1 = 2
28 27 a1i ⊢ K ∈ ℕ 0 → 3 − 1 = 2
29 26 28 oveq12d ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 - 1 = 4 ⁢ K ⋅ 2 + 2
30 15 29 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 - 1 = 4 ⁢ K ⋅ 2 + 2
31 30 oveq1d ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 - 1 2 = 4 ⁢ K ⋅ 2 + 2 2
32 4nn0 ⊢ 4 ∈ ℕ 0
33 32 a1i ⊢ K ∈ ℕ 0 → 4 ∈ ℕ 0
34 33 9 nn0mulcld ⊢ K ∈ ℕ 0 → 4 ⁢ K ∈ ℕ 0
35 34 nn0cnd ⊢ K ∈ ℕ 0 → 4 ⁢ K ∈ ℂ
36 35 23 mulcld ⊢ K ∈ ℕ 0 → 4 ⁢ K ⋅ 2 ∈ ℂ
37 2rp ⊢ 2 ∈ ℝ +
38 37 a1i ⊢ K ∈ ℕ 0 → 2 ∈ ℝ +
39 38 rpcnne0d ⊢ K ∈ ℕ 0 → 2 ∈ ℂ ∧ 2 ≠ 0
40 divdir ⊢ 4 ⁢ K ⋅ 2 ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → 4 ⁢ K ⋅ 2 + 2 2 = 4 ⁢ K ⋅ 2 2 + 2 2
41 36 23 39 40 syl3anc ⊢ K ∈ ℕ 0 → 4 ⁢ K ⋅ 2 + 2 2 = 4 ⁢ K ⋅ 2 2 + 2 2
42 2ne0 ⊢ 2 ≠ 0
43 42 a1i ⊢ K ∈ ℕ 0 → 2 ≠ 0
44 35 23 43 divcan4d ⊢ K ∈ ℕ 0 → 4 ⁢ K ⋅ 2 2 = 4 ⁢ K
45 2div2e1 ⊢ 2 2 = 1
46 45 a1i ⊢ K ∈ ℕ 0 → 2 2 = 1
47 44 46 oveq12d ⊢ K ∈ ℕ 0 → 4 ⁢ K ⋅ 2 2 + 2 2 = 4 ⁢ K + 1
48 31 41 47 3eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 - 1 2 = 4 ⁢ K + 1
49 4ne0 ⊢ 4 ≠ 0
50 20 49 pm3.2i ⊢ 4 ∈ ℂ ∧ 4 ≠ 0
51 50 a1i ⊢ K ∈ ℕ 0 → 4 ∈ ℂ ∧ 4 ≠ 0
52 divdir ⊢ 8 ⁢ K ∈ ℂ ∧ 3 ∈ ℂ ∧ 4 ∈ ℂ ∧ 4 ≠ 0 → 8 ⁢ K + 3 4 = 8 ⁢ K 4 + 3 4
53 11 13 51 52 syl3anc ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 4 = 8 ⁢ K 4 + 3 4
54 8cn ⊢ 8 ∈ ℂ
55 54 a1i ⊢ K ∈ ℕ 0 → 8 ∈ ℂ
56 div23 ⊢ 8 ∈ ℂ ∧ K ∈ ℂ ∧ 4 ∈ ℂ ∧ 4 ≠ 0 → 8 ⁢ K 4 = 8 4 ⁢ K
57 55 24 51 56 syl3anc ⊢ K ∈ ℕ 0 → 8 ⁢ K 4 = 8 4 ⁢ K
58 17 oveq1i ⊢ 8 4 = 4 ⋅ 2 4
59 22 20 49 divcan3i ⊢ 4 ⋅ 2 4 = 2
60 58 59 eqtri ⊢ 8 4 = 2
61 60 a1i ⊢ K ∈ ℕ 0 → 8 4 = 2
62 61 oveq1d ⊢ K ∈ ℕ 0 → 8 4 ⁢ K = 2 ⁢ K
63 57 62 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K 4 = 2 ⁢ K
64 63 oveq1d ⊢ K ∈ ℕ 0 → 8 ⁢ K 4 + 3 4 = 2 ⁢ K + 3 4
65 53 64 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 4 = 2 ⁢ K + 3 4
66 65 fveq2d ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 4 = 2 ⁢ K + 3 4
67 3lt4 ⊢ 3 < 4
68 2nn0 ⊢ 2 ∈ ℕ 0
69 68 a1i ⊢ K ∈ ℕ 0 → 2 ∈ ℕ 0
70 69 9 nn0mulcld ⊢ K ∈ ℕ 0 → 2 ⁢ K ∈ ℕ 0
71 70 nn0zd ⊢ K ∈ ℕ 0 → 2 ⁢ K ∈ ℤ
72 3nn0 ⊢ 3 ∈ ℕ 0
73 72 a1i ⊢ K ∈ ℕ 0 → 3 ∈ ℕ 0
74 4nn ⊢ 4 ∈ ℕ
75 74 a1i ⊢ K ∈ ℕ 0 → 4 ∈ ℕ
76 adddivflid ⊢ 2 ⁢ K ∈ ℤ ∧ 3 ∈ ℕ 0 ∧ 4 ∈ ℕ → 3 < 4 ↔ 2 ⁢ K + 3 4 = 2 ⁢ K
77 71 73 75 76 syl3anc ⊢ K ∈ ℕ 0 → 3 < 4 ↔ 2 ⁢ K + 3 4 = 2 ⁢ K
78 67 77 mpbii ⊢ K ∈ ℕ 0 → 2 ⁢ K + 3 4 = 2 ⁢ K
79 66 78 eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 4 = 2 ⁢ K
80 48 79 oveq12d ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 - 1 2 − 8 ⁢ K + 3 4 = 4 ⁢ K + 1 - 2 ⁢ K
81 70 nn0cnd ⊢ K ∈ ℕ 0 → 2 ⁢ K ∈ ℂ
82 35 14 81 addsubd ⊢ K ∈ ℕ 0 → 4 ⁢ K + 1 - 2 ⁢ K = 4 ⁢ K - 2 ⁢ K + 1
83 2t2e4 ⊢ 2 ⋅ 2 = 4
84 83 eqcomi ⊢ 4 = 2 ⋅ 2
85 84 a1i ⊢ K ∈ ℕ 0 → 4 = 2 ⋅ 2
86 85 oveq1d ⊢ K ∈ ℕ 0 → 4 ⁢ K = 2 ⋅ 2 ⁢ K
87 23 23 24 mulassd ⊢ K ∈ ℕ 0 → 2 ⋅ 2 ⁢ K = 2 ⁢ 2 ⁢ K
88 86 87 eqtrd ⊢ K ∈ ℕ 0 → 4 ⁢ K = 2 ⁢ 2 ⁢ K
89 88 oveq1d ⊢ K ∈ ℕ 0 → 4 ⁢ K − 2 ⁢ K = 2 ⁢ 2 ⁢ K − 2 ⁢ K
90 2txmxeqx ⊢ 2 ⁢ K ∈ ℂ → 2 ⁢ 2 ⁢ K − 2 ⁢ K = 2 ⁢ K
91 81 90 syl ⊢ K ∈ ℕ 0 → 2 ⁢ 2 ⁢ K − 2 ⁢ K = 2 ⁢ K
92 89 91 eqtrd ⊢ K ∈ ℕ 0 → 4 ⁢ K − 2 ⁢ K = 2 ⁢ K
93 92 oveq1d ⊢ K ∈ ℕ 0 → 4 ⁢ K - 2 ⁢ K + 1 = 2 ⁢ K + 1
94 80 82 93 3eqtrd ⊢ K ∈ ℕ 0 → 8 ⁢ K + 3 - 1 2 − 8 ⁢ K + 3 4 = 2 ⁢ K + 1
95 6 94 sylan9eqr ⊢ K ∈ ℕ 0 ∧ P = 8 ⁢ K + 3 → N = 2 ⁢ K + 1