Metamath Proof Explorer


Theorem isprm3

Description: The predicate "is a prime number". A prime number is an integer greater than or equal to 2 with no divisors strictly between 1 and itself. (Contributed by Paul Chapman, 26-Oct-2012)

Ref Expression
Assertion isprm3 ⊢ P ∈ ℙ ↔ P ∈ ℤ ≥ 2 ∧ ∀ z ∈ 2 … P − 1 ¬ z ∥ P

Proof

Step Hyp Ref Expression
1 isprm2 ⊢ P ∈ ℙ ↔ P ∈ ℤ ≥ 2 ∧ ∀ z ∈ ℕ z ∥ P → z = 1 ∨ z = P
2 iman ⊢ z ∈ ℕ → z = 1 ∨ z = P ↔ ¬ z ∈ ℕ ∧ ¬ z = 1 ∨ z = P
3 eluz2nn ⊢ P ∈ ℤ ≥ 2 → P ∈ ℕ
4 nnz ⊢ z ∈ ℕ → z ∈ ℤ
5 dvdsle ⊢ z ∈ ℤ ∧ P ∈ ℕ → z ∥ P → z ≤ P
6 4 5 sylan ⊢ z ∈ ℕ ∧ P ∈ ℕ → z ∥ P → z ≤ P
7 nnge1 ⊢ z ∈ ℕ → 1 ≤ z
8 7 adantr ⊢ z ∈ ℕ ∧ P ∈ ℕ → 1 ≤ z
9 6 8 jctild ⊢ z ∈ ℕ ∧ P ∈ ℕ → z ∥ P → 1 ≤ z ∧ z ≤ P
10 3 9 sylan2 ⊢ z ∈ ℕ ∧ P ∈ ℤ ≥ 2 → z ∥ P → 1 ≤ z ∧ z ≤ P
11 zre ⊢ z ∈ ℤ → z ∈ ℝ
12 nnre ⊢ P ∈ ℕ → P ∈ ℝ
13 1re ⊢ 1 ∈ ℝ
14 leltne ⊢ 1 ∈ ℝ ∧ z ∈ ℝ ∧ 1 ≤ z → 1 < z ↔ z ≠ 1
15 13 14 mp3an1 ⊢ z ∈ ℝ ∧ 1 ≤ z → 1 < z ↔ z ≠ 1
16 15 3adant2 ⊢ z ∈ ℝ ∧ P ∈ ℝ ∧ 1 ≤ z → 1 < z ↔ z ≠ 1
17 16 3expia ⊢ z ∈ ℝ ∧ P ∈ ℝ → 1 ≤ z → 1 < z ↔ z ≠ 1
18 leltne ⊢ z ∈ ℝ ∧ P ∈ ℝ ∧ z ≤ P → z < P ↔ P ≠ z
19 18 3expia ⊢ z ∈ ℝ ∧ P ∈ ℝ → z ≤ P → z < P ↔ P ≠ z
20 17 19 anim12d ⊢ z ∈ ℝ ∧ P ∈ ℝ → 1 ≤ z ∧ z ≤ P → 1 < z ↔ z ≠ 1 ∧ z < P ↔ P ≠ z
21 11 12 20 syl2an ⊢ z ∈ ℤ ∧ P ∈ ℕ → 1 ≤ z ∧ z ≤ P → 1 < z ↔ z ≠ 1 ∧ z < P ↔ P ≠ z
22 pm4.38 ⊢ 1 < z ↔ z ≠ 1 ∧ z < P ↔ P ≠ z → 1 < z ∧ z < P ↔ z ≠ 1 ∧ P ≠ z
23 df-ne ⊢ z ≠ 1 ↔ ¬ z = 1
24 nesym ⊢ P ≠ z ↔ ¬ z = P
25 23 24 anbi12i ⊢ z ≠ 1 ∧ P ≠ z ↔ ¬ z = 1 ∧ ¬ z = P
26 ioran ⊢ ¬ z = 1 ∨ z = P ↔ ¬ z = 1 ∧ ¬ z = P
27 25 26 bitr4i ⊢ z ≠ 1 ∧ P ≠ z ↔ ¬ z = 1 ∨ z = P
28 22 27 bitrdi ⊢ 1 < z ↔ z ≠ 1 ∧ z < P ↔ P ≠ z → 1 < z ∧ z < P ↔ ¬ z = 1 ∨ z = P
29 21 28 syl6 ⊢ z ∈ ℤ ∧ P ∈ ℕ → 1 ≤ z ∧ z ≤ P → 1 < z ∧ z < P ↔ ¬ z = 1 ∨ z = P
30 4 3 29 syl2an ⊢ z ∈ ℕ ∧ P ∈ ℤ ≥ 2 → 1 ≤ z ∧ z ≤ P → 1 < z ∧ z < P ↔ ¬ z = 1 ∨ z = P
31 10 30 syld ⊢ z ∈ ℕ ∧ P ∈ ℤ ≥ 2 → z ∥ P → 1 < z ∧ z < P ↔ ¬ z = 1 ∨ z = P
32 31 imp ⊢ z ∈ ℕ ∧ P ∈ ℤ ≥ 2 ∧ z ∥ P → 1 < z ∧ z < P ↔ ¬ z = 1 ∨ z = P
33 eluzelz ⊢ P ∈ ℤ ≥ 2 → P ∈ ℤ
34 1z ⊢ 1 ∈ ℤ
35 zltp1le ⊢ 1 ∈ ℤ ∧ z ∈ ℤ → 1 < z ↔ 1 + 1 ≤ z
36 34 35 mpan ⊢ z ∈ ℤ → 1 < z ↔ 1 + 1 ≤ z
37 df-2 ⊢ 2 = 1 + 1
38 37 breq1i ⊢ 2 ≤ z ↔ 1 + 1 ≤ z
39 36 38 bitr4di ⊢ z ∈ ℤ → 1 < z ↔ 2 ≤ z
40 39 adantr ⊢ z ∈ ℤ ∧ P ∈ ℤ → 1 < z ↔ 2 ≤ z
41 zltlem1 ⊢ z ∈ ℤ ∧ P ∈ ℤ → z < P ↔ z ≤ P − 1
42 40 41 anbi12d ⊢ z ∈ ℤ ∧ P ∈ ℤ → 1 < z ∧ z < P ↔ 2 ≤ z ∧ z ≤ P − 1
43 peano2zm ⊢ P ∈ ℤ → P − 1 ∈ ℤ
44 2z ⊢ 2 ∈ ℤ
45 elfz ⊢ z ∈ ℤ ∧ 2 ∈ ℤ ∧ P − 1 ∈ ℤ → z ∈ 2 … P − 1 ↔ 2 ≤ z ∧ z ≤ P − 1
46 44 45 mp3an2 ⊢ z ∈ ℤ ∧ P − 1 ∈ ℤ → z ∈ 2 … P − 1 ↔ 2 ≤ z ∧ z ≤ P − 1
47 43 46 sylan2 ⊢ z ∈ ℤ ∧ P ∈ ℤ → z ∈ 2 … P − 1 ↔ 2 ≤ z ∧ z ≤ P − 1
48 42 47 bitr4d ⊢ z ∈ ℤ ∧ P ∈ ℤ → 1 < z ∧ z < P ↔ z ∈ 2 … P − 1
49 4 33 48 syl2an ⊢ z ∈ ℕ ∧ P ∈ ℤ ≥ 2 → 1 < z ∧ z < P ↔ z ∈ 2 … P − 1
50 49 adantr ⊢ z ∈ ℕ ∧ P ∈ ℤ ≥ 2 ∧ z ∥ P → 1 < z ∧ z < P ↔ z ∈ 2 … P − 1
51 32 50 bitr3d ⊢ z ∈ ℕ ∧ P ∈ ℤ ≥ 2 ∧ z ∥ P → ¬ z = 1 ∨ z = P ↔ z ∈ 2 … P − 1
52 51 anasss ⊢ z ∈ ℕ ∧ P ∈ ℤ ≥ 2 ∧ z ∥ P → ¬ z = 1 ∨ z = P ↔ z ∈ 2 … P − 1
53 52 expcom ⊢ P ∈ ℤ ≥ 2 ∧ z ∥ P → z ∈ ℕ → ¬ z = 1 ∨ z = P ↔ z ∈ 2 … P − 1
54 53 pm5.32d ⊢ P ∈ ℤ ≥ 2 ∧ z ∥ P → z ∈ ℕ ∧ ¬ z = 1 ∨ z = P ↔ z ∈ ℕ ∧ z ∈ 2 … P − 1
55 fzssuz ⊢ 2 … P − 1 ⊆ ℤ ≥ 2
56 2eluzge1 ⊢ 2 ∈ ℤ ≥ 1
57 uzss ⊢ 2 ∈ ℤ ≥ 1 → ℤ ≥ 2 ⊆ ℤ ≥ 1
58 56 57 ax-mp ⊢ ℤ ≥ 2 ⊆ ℤ ≥ 1
59 55 58 sstri ⊢ 2 … P − 1 ⊆ ℤ ≥ 1
60 nnuz ⊢ ℕ = ℤ ≥ 1
61 59 60 sseqtrri ⊢ 2 … P − 1 ⊆ ℕ
62 61 sseli ⊢ z ∈ 2 … P − 1 → z ∈ ℕ
63 62 pm4.71ri ⊢ z ∈ 2 … P − 1 ↔ z ∈ ℕ ∧ z ∈ 2 … P − 1
64 54 63 bitr4di ⊢ P ∈ ℤ ≥ 2 ∧ z ∥ P → z ∈ ℕ ∧ ¬ z = 1 ∨ z = P ↔ z ∈ 2 … P − 1
65 64 notbid ⊢ P ∈ ℤ ≥ 2 ∧ z ∥ P → ¬ z ∈ ℕ ∧ ¬ z = 1 ∨ z = P ↔ ¬ z ∈ 2 … P − 1
66 2 65 bitrid ⊢ P ∈ ℤ ≥ 2 ∧ z ∥ P → z ∈ ℕ → z = 1 ∨ z = P ↔ ¬ z ∈ 2 … P − 1
67 66 pm5.74da ⊢ P ∈ ℤ ≥ 2 → z ∥ P → z ∈ ℕ → z = 1 ∨ z = P ↔ z ∥ P → ¬ z ∈ 2 … P − 1
68 bi2.04 ⊢ z ∥ P → z ∈ ℕ → z = 1 ∨ z = P ↔ z ∈ ℕ → z ∥ P → z = 1 ∨ z = P
69 con2b ⊢ z ∥ P → ¬ z ∈ 2 … P − 1 ↔ z ∈ 2 … P − 1 → ¬ z ∥ P
70 67 68 69 3bitr3g ⊢ P ∈ ℤ ≥ 2 → z ∈ ℕ → z ∥ P → z = 1 ∨ z = P ↔ z ∈ 2 … P − 1 → ¬ z ∥ P
71 70 ralbidv2 ⊢ P ∈ ℤ ≥ 2 → ∀ z ∈ ℕ z ∥ P → z = 1 ∨ z = P ↔ ∀ z ∈ 2 … P − 1 ¬ z ∥ P
72 71 pm5.32i ⊢ P ∈ ℤ ≥ 2 ∧ ∀ z ∈ ℕ z ∥ P → z = 1 ∨ z = P ↔ P ∈ ℤ ≥ 2 ∧ ∀ z ∈ 2 … P − 1 ¬ z ∥ P
73 1 72 bitri ⊢ P ∈ ℙ ↔ P ∈ ℤ ≥ 2 ∧ ∀ z ∈ 2 … P − 1 ¬ z ∥ P