Metamath Proof Explorer


Theorem 2503prm

Description: 2503 is a prime number. (Contributed by Mario Carneiro, 3-Mar-2014) (Proof shortened by Mario Carneiro, 20-Apr-2015)

Ref Expression
Hypothesis 2503prm.1 ⊢ N = 2503
Assertion 2503prm ⊢ N ∈ ℙ

Proof

Step Hyp Ref Expression
1 2503prm.1 ⊢ N = 2503
2 139prm ⊢ 139 ∈ ℙ
3 1nn0 ⊢ 1 ∈ ℕ 0
4 8nn ⊢ 8 ∈ ℕ
5 3 4 decnncl ⊢ 18 ∈ ℕ
6 2nn0 ⊢ 2 ∈ ℕ 0
7 5nn0 ⊢ 5 ∈ ℕ 0
8 6 7 deccl ⊢ 25 ∈ ℕ 0
9 0nn0 ⊢ 0 ∈ ℕ 0
10 8 9 deccl ⊢ 250 ∈ ℕ 0
11 2p1e3 ⊢ 2 + 1 = 3
12 eqid ⊢ 2502 = 2502
13 10 6 11 12 decsuc ⊢ 2502 + 1 = 2503
14 1 13 eqtr4i ⊢ N = 2502 + 1
15 14 oveq1i ⊢ N − 1 = 2502 + 1 - 1
16 8nn0 ⊢ 8 ∈ ℕ 0
17 3 16 deccl ⊢ 18 ∈ ℕ 0
18 3nn0 ⊢ 3 ∈ ℕ 0
19 3 18 deccl ⊢ 13 ∈ ℕ 0
20 9nn0 ⊢ 9 ∈ ℕ 0
21 eqid ⊢ 139 = 139
22 6nn0 ⊢ 6 ∈ ℕ 0
23 3 22 deccl ⊢ 16 ∈ ℕ 0
24 eqid ⊢ 13 = 13
25 eqid ⊢ 16 = 16
26 7nn0 ⊢ 7 ∈ ℕ 0
27 eqid ⊢ 18 = 18
28 6cn ⊢ 6 ∈ ℂ
29 ax-1cn ⊢ 1 ∈ ℂ
30 6p1e7 ⊢ 6 + 1 = 7
31 28 29 30 addcomli ⊢ 1 + 6 = 7
32 26 dec0h ⊢ 7 = 07
33 31 32 eqtri ⊢ 1 + 6 = 07
34 29 mulridi ⊢ 1 ⋅ 1 = 1
35 29 addlidi ⊢ 0 + 1 = 1
36 34 35 oveq12i ⊢ 1 ⋅ 1 + 0 + 1 = 1 + 1
37 1p1e2 ⊢ 1 + 1 = 2
38 36 37 eqtri ⊢ 1 ⋅ 1 + 0 + 1 = 2
39 8cn ⊢ 8 ∈ ℂ
40 39 mulridi ⊢ 8 ⋅ 1 = 8
41 40 oveq1i ⊢ 8 ⋅ 1 + 7 = 8 + 7
42 8p7e15 ⊢ 8 + 7 = 15
43 41 42 eqtri ⊢ 8 ⋅ 1 + 7 = 15
44 3 16 9 26 27 33 3 7 3 38 43 decmac ⊢ 18 ⋅ 1 + 1 + 6 = 25
45 22 dec0h ⊢ 6 = 06
46 3cn ⊢ 3 ∈ ℂ
47 46 mullidi ⊢ 1 ⋅ 3 = 3
48 46 addlidi ⊢ 0 + 3 = 3
49 47 48 oveq12i ⊢ 1 ⋅ 3 + 0 + 3 = 3 + 3
50 3p3e6 ⊢ 3 + 3 = 6
51 49 50 eqtri ⊢ 1 ⋅ 3 + 0 + 3 = 6
52 4nn0 ⊢ 4 ∈ ℕ 0
53 8t3e24 ⊢ 8 ⋅ 3 = 24
54 4cn ⊢ 4 ∈ ℂ
55 6p4e10 ⊢ 6 + 4 = 10
56 28 54 55 addcomli ⊢ 4 + 6 = 10
57 6 52 22 53 11 56 decaddci2 ⊢ 8 ⋅ 3 + 6 = 30
58 3 16 9 22 27 45 18 9 18 51 57 decmac ⊢ 18 ⋅ 3 + 6 = 60
59 3 18 3 22 24 25 17 9 22 44 58 decma2c ⊢ 18 ⋅ 13 + 16 = 250
60 9cn ⊢ 9 ∈ ℂ
61 60 mullidi ⊢ 1 ⋅ 9 = 9
62 61 oveq1i ⊢ 1 ⋅ 9 + 7 = 9 + 7
63 9p7e16 ⊢ 9 + 7 = 16
64 62 63 eqtri ⊢ 1 ⋅ 9 + 7 = 16
65 9t8e72 ⊢ 9 ⋅ 8 = 72
66 60 39 65 mulcomli ⊢ 8 ⋅ 9 = 72
67 20 3 16 27 6 26 64 66 decmul1c ⊢ 18 ⋅ 9 = 162
68 17 19 20 21 6 23 59 67 decmul2c ⊢ 18 ⋅ 139 = 2502
69 10 6 deccl ⊢ 2502 ∈ ℕ 0
70 69 nn0cni ⊢ 2502 ∈ ℂ
71 70 29 pncan3oi ⊢ 2502 + 1 - 1 = 2502
72 68 71 eqtr4i ⊢ 18 ⋅ 139 = 2502 + 1 - 1
73 15 72 eqtr4i ⊢ N − 1 = 18 ⋅ 139
74 10 18 deccl ⊢ 2503 ∈ ℕ 0
75 1 74 eqeltri ⊢ N ∈ ℕ 0
76 75 nn0cni ⊢ N ∈ ℂ
77 npcan ⊢ N ∈ ℂ ∧ 1 ∈ ℂ → N - 1 + 1 = N
78 76 29 77 mp2an ⊢ N - 1 + 1 = N
79 78 eqcomi ⊢ N = N - 1 + 1
80 1nn ⊢ 1 ∈ ℕ
81 2nn ⊢ 2 ∈ ℕ
82 19 20 deccl ⊢ 139 ∈ ℕ 0
83 82 numexp1 ⊢ 139 1 = 139
84 83 oveq2i ⊢ 18 ⁢ 139 1 = 18 ⋅ 139
85 73 84 eqtr4i ⊢ N − 1 = 18 ⁢ 139 1
86 8lt10 ⊢ 8 < 10
87 1lt10 ⊢ 1 < 10
88 80 18 3 87 declti ⊢ 1 < 13
89 3 19 16 20 86 88 decltc ⊢ 18 < 139
90 89 83 breqtrri ⊢ 18 < 139 1
91 1 2503lem2 ⊢ 2 N − 1 mod N = 1 mod N
92 1 2503lem3 ⊢ 2 18 − 1 gcd N = 1
93 2 5 73 79 5 80 81 85 90 91 92 pockthi ⊢ N ∈ ℙ