Metamath Proof Explorer


Theorem perfectlem1

Description: Lemma for perfect . (Contributed by Mario Carneiro, 7-Jun-2016)

Ref Expression
Hypotheses perfectlem.1 ⊢ φ → A ∈ ℕ
perfectlem.2 ⊢ φ → B ∈ ℕ
perfectlem.3 ⊢ φ → ¬ 2 ∥ B
perfectlem.4 ⊢ φ → 1 σ 2 A ⁢ B = 2 ⁢ 2 A ⁢ B
Assertion perfectlem1 ⊢ φ → 2 A + 1 ∈ ℕ ∧ 2 A + 1 − 1 ∈ ℕ ∧ B 2 A + 1 − 1 ∈ ℕ

Proof

Step Hyp Ref Expression
1 perfectlem.1 ⊢ φ → A ∈ ℕ
2 perfectlem.2 ⊢ φ → B ∈ ℕ
3 perfectlem.3 ⊢ φ → ¬ 2 ∥ B
4 perfectlem.4 ⊢ φ → 1 σ 2 A ⁢ B = 2 ⁢ 2 A ⁢ B
5 2nn ⊢ 2 ∈ ℕ
6 1 nnnn0d ⊢ φ → A ∈ ℕ 0
7 peano2nn0 ⊢ A ∈ ℕ 0 → A + 1 ∈ ℕ 0
8 6 7 syl ⊢ φ → A + 1 ∈ ℕ 0
9 nnexpcl ⊢ 2 ∈ ℕ ∧ A + 1 ∈ ℕ 0 → 2 A + 1 ∈ ℕ
10 5 8 9 sylancr ⊢ φ → 2 A + 1 ∈ ℕ
11 2re ⊢ 2 ∈ ℝ
12 1 peano2nnd ⊢ φ → A + 1 ∈ ℕ
13 1lt2 ⊢ 1 < 2
14 13 a1i ⊢ φ → 1 < 2
15 expgt1 ⊢ 2 ∈ ℝ ∧ A + 1 ∈ ℕ ∧ 1 < 2 → 1 < 2 A + 1
16 11 12 14 15 mp3an2i ⊢ φ → 1 < 2 A + 1
17 1nn ⊢ 1 ∈ ℕ
18 nnsub ⊢ 1 ∈ ℕ ∧ 2 A + 1 ∈ ℕ → 1 < 2 A + 1 ↔ 2 A + 1 − 1 ∈ ℕ
19 17 10 18 sylancr ⊢ φ → 1 < 2 A + 1 ↔ 2 A + 1 − 1 ∈ ℕ
20 16 19 mpbid ⊢ φ → 2 A + 1 − 1 ∈ ℕ
21 10 nnzd ⊢ φ → 2 A + 1 ∈ ℤ
22 peano2zm ⊢ 2 A + 1 ∈ ℤ → 2 A + 1 − 1 ∈ ℤ
23 21 22 syl ⊢ φ → 2 A + 1 − 1 ∈ ℤ
24 1nn0 ⊢ 1 ∈ ℕ 0
25 sgmnncl ⊢ 1 ∈ ℕ 0 ∧ B ∈ ℕ → 1 σ B ∈ ℕ
26 24 2 25 sylancr ⊢ φ → 1 σ B ∈ ℕ
27 26 nnzd ⊢ φ → 1 σ B ∈ ℤ
28 dvdsmul1 ⊢ 2 A + 1 − 1 ∈ ℤ ∧ 1 σ B ∈ ℤ → 2 A + 1 − 1 ∥ 2 A + 1 − 1 ⁢ 1 σ B
29 23 27 28 syl2anc ⊢ φ → 2 A + 1 − 1 ∥ 2 A + 1 − 1 ⁢ 1 σ B
30 2cn ⊢ 2 ∈ ℂ
31 expp1 ⊢ 2 ∈ ℂ ∧ A ∈ ℕ 0 → 2 A + 1 = 2 A ⋅ 2
32 30 6 31 sylancr ⊢ φ → 2 A + 1 = 2 A ⋅ 2
33 nnexpcl ⊢ 2 ∈ ℕ ∧ A ∈ ℕ 0 → 2 A ∈ ℕ
34 5 6 33 sylancr ⊢ φ → 2 A ∈ ℕ
35 34 nncnd ⊢ φ → 2 A ∈ ℂ
36 mulcom ⊢ 2 A ∈ ℂ ∧ 2 ∈ ℂ → 2 A ⋅ 2 = 2 ⁢ 2 A
37 35 30 36 sylancl ⊢ φ → 2 A ⋅ 2 = 2 ⁢ 2 A
38 32 37 eqtrd ⊢ φ → 2 A + 1 = 2 ⁢ 2 A
39 38 oveq1d ⊢ φ → 2 A + 1 ⁢ B = 2 ⁢ 2 A ⁢ B
40 30 a1i ⊢ φ → 2 ∈ ℂ
41 2 nncnd ⊢ φ → B ∈ ℂ
42 40 35 41 mulassd ⊢ φ → 2 ⁢ 2 A ⁢ B = 2 ⁢ 2 A ⁢ B
43 ax-1cn ⊢ 1 ∈ ℂ
44 43 a1i ⊢ φ → 1 ∈ ℂ
45 2prm ⊢ 2 ∈ ℙ
46 2 nnzd ⊢ φ → B ∈ ℤ
47 coprm ⊢ 2 ∈ ℙ ∧ B ∈ ℤ → ¬ 2 ∥ B ↔ 2 gcd B = 1
48 45 46 47 sylancr ⊢ φ → ¬ 2 ∥ B ↔ 2 gcd B = 1
49 3 48 mpbid ⊢ φ → 2 gcd B = 1
50 2z ⊢ 2 ∈ ℤ
51 rpexp1i ⊢ 2 ∈ ℤ ∧ B ∈ ℤ ∧ A ∈ ℕ 0 → 2 gcd B = 1 → 2 A gcd B = 1
52 50 46 6 51 mp3an2i ⊢ φ → 2 gcd B = 1 → 2 A gcd B = 1
53 49 52 mpd ⊢ φ → 2 A gcd B = 1
54 sgmmul ⊢ 1 ∈ ℂ ∧ 2 A ∈ ℕ ∧ B ∈ ℕ ∧ 2 A gcd B = 1 → 1 σ 2 A ⁢ B = 1 σ 2 A ⁢ 1 σ B
55 44 34 2 53 54 syl13anc ⊢ φ → 1 σ 2 A ⁢ B = 1 σ 2 A ⁢ 1 σ B
56 1 nncnd ⊢ φ → A ∈ ℂ
57 pncan ⊢ A ∈ ℂ ∧ 1 ∈ ℂ → A + 1 - 1 = A
58 56 43 57 sylancl ⊢ φ → A + 1 - 1 = A
59 58 oveq2d ⊢ φ → 2 A + 1 - 1 = 2 A
60 59 oveq2d ⊢ φ → 1 σ 2 A + 1 - 1 = 1 σ 2 A
61 1sgm2ppw ⊢ A + 1 ∈ ℕ → 1 σ 2 A + 1 - 1 = 2 A + 1 − 1
62 12 61 syl ⊢ φ → 1 σ 2 A + 1 - 1 = 2 A + 1 − 1
63 60 62 eqtr3d ⊢ φ → 1 σ 2 A = 2 A + 1 − 1
64 63 oveq1d ⊢ φ → 1 σ 2 A ⁢ 1 σ B = 2 A + 1 − 1 ⁢ 1 σ B
65 55 4 64 3eqtr3d ⊢ φ → 2 ⁢ 2 A ⁢ B = 2 A + 1 − 1 ⁢ 1 σ B
66 39 42 65 3eqtrd ⊢ φ → 2 A + 1 ⁢ B = 2 A + 1 − 1 ⁢ 1 σ B
67 29 66 breqtrrd ⊢ φ → 2 A + 1 − 1 ∥ 2 A + 1 ⁢ B
68 23 21 gcdcomd ⊢ φ → 2 A + 1 − 1 gcd 2 A + 1 = 2 A + 1 gcd 2 A + 1 − 1
69 iddvdsexp ⊢ 2 ∈ ℤ ∧ A + 1 ∈ ℕ → 2 ∥ 2 A + 1
70 50 12 69 sylancr ⊢ φ → 2 ∥ 2 A + 1
71 n2dvds1 ⊢ ¬ 2 ∥ 1
72 50 a1i ⊢ φ → 2 ∈ ℤ
73 1zzd ⊢ φ → 1 ∈ ℤ
74 72 21 73 3jca ⊢ φ → 2 ∈ ℤ ∧ 2 A + 1 ∈ ℤ ∧ 1 ∈ ℤ
75 dvdssub2 ⊢ 2 ∈ ℤ ∧ 2 A + 1 ∈ ℤ ∧ 1 ∈ ℤ ∧ 2 ∥ 2 A + 1 − 1 → 2 ∥ 2 A + 1 ↔ 2 ∥ 1
76 74 75 sylan ⊢ φ ∧ 2 ∥ 2 A + 1 − 1 → 2 ∥ 2 A + 1 ↔ 2 ∥ 1
77 71 76 mtbiri ⊢ φ ∧ 2 ∥ 2 A + 1 − 1 → ¬ 2 ∥ 2 A + 1
78 77 ex ⊢ φ → 2 ∥ 2 A + 1 − 1 → ¬ 2 ∥ 2 A + 1
79 70 78 mt2d ⊢ φ → ¬ 2 ∥ 2 A + 1 − 1
80 coprm ⊢ 2 ∈ ℙ ∧ 2 A + 1 − 1 ∈ ℤ → ¬ 2 ∥ 2 A + 1 − 1 ↔ 2 gcd 2 A + 1 − 1 = 1
81 45 23 80 sylancr ⊢ φ → ¬ 2 ∥ 2 A + 1 − 1 ↔ 2 gcd 2 A + 1 − 1 = 1
82 79 81 mpbid ⊢ φ → 2 gcd 2 A + 1 − 1 = 1
83 rpexp1i ⊢ 2 ∈ ℤ ∧ 2 A + 1 − 1 ∈ ℤ ∧ A + 1 ∈ ℕ 0 → 2 gcd 2 A + 1 − 1 = 1 → 2 A + 1 gcd 2 A + 1 − 1 = 1
84 50 23 8 83 mp3an2i ⊢ φ → 2 gcd 2 A + 1 − 1 = 1 → 2 A + 1 gcd 2 A + 1 − 1 = 1
85 82 84 mpd ⊢ φ → 2 A + 1 gcd 2 A + 1 − 1 = 1
86 68 85 eqtrd ⊢ φ → 2 A + 1 − 1 gcd 2 A + 1 = 1
87 coprmdvds ⊢ 2 A + 1 − 1 ∈ ℤ ∧ 2 A + 1 ∈ ℤ ∧ B ∈ ℤ → 2 A + 1 − 1 ∥ 2 A + 1 ⁢ B ∧ 2 A + 1 − 1 gcd 2 A + 1 = 1 → 2 A + 1 − 1 ∥ B
88 23 21 46 87 syl3anc ⊢ φ → 2 A + 1 − 1 ∥ 2 A + 1 ⁢ B ∧ 2 A + 1 − 1 gcd 2 A + 1 = 1 → 2 A + 1 − 1 ∥ B
89 67 86 88 mp2and ⊢ φ → 2 A + 1 − 1 ∥ B
90 nndivdvds ⊢ B ∈ ℕ ∧ 2 A + 1 − 1 ∈ ℕ → 2 A + 1 − 1 ∥ B ↔ B 2 A + 1 − 1 ∈ ℕ
91 2 20 90 syl2anc ⊢ φ → 2 A + 1 − 1 ∥ B ↔ B 2 A + 1 − 1 ∈ ℕ
92 89 91 mpbid ⊢ φ → B 2 A + 1 − 1 ∈ ℕ
93 10 20 92 3jca ⊢ φ → 2 A + 1 ∈ ℕ ∧ 2 A + 1 − 1 ∈ ℕ ∧ B 2 A + 1 − 1 ∈ ℕ