Metamath Proof Explorer


Theorem isprm6

Description: A number is prime iff it satisfies Euclid's lemma euclemma . (Contributed by Mario Carneiro, 6-Sep-2015)

Ref Expression
Assertion isprm6 ⊢ P ∈ ℙ ↔ P ∈ ℤ ≥ 2 ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y

Proof

Step Hyp Ref Expression
1 prmuz2 ⊢ P ∈ ℙ → P ∈ ℤ ≥ 2
2 euclemma ⊢ P ∈ ℙ ∧ x ∈ ℤ ∧ y ∈ ℤ → P ∥ x ⁢ y ↔ P ∥ x ∨ P ∥ y
3 2 3expb ⊢ P ∈ ℙ ∧ x ∈ ℤ ∧ y ∈ ℤ → P ∥ x ⁢ y ↔ P ∥ x ∨ P ∥ y
4 3 biimpd ⊢ P ∈ ℙ ∧ x ∈ ℤ ∧ y ∈ ℤ → P ∥ x ⁢ y → P ∥ x ∨ P ∥ y
5 4 ralrimivva ⊢ P ∈ ℙ → ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y
6 1 5 jca ⊢ P ∈ ℙ → P ∈ ℤ ≥ 2 ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y
7 simpl ⊢ P ∈ ℤ ≥ 2 ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y → P ∈ ℤ ≥ 2
8 eluz2nn ⊢ P ∈ ℤ ≥ 2 → P ∈ ℕ
9 8 adantr ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∈ ℕ
10 9 nnzd ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∈ ℤ
11 iddvds ⊢ P ∈ ℤ → P ∥ P
12 10 11 syl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∥ P
13 nncn ⊢ P ∈ ℕ → P ∈ ℂ
14 9 13 syl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∈ ℂ
15 nncn ⊢ z ∈ ℕ → z ∈ ℂ
16 15 ad2antrl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ∈ ℂ
17 nnne0 ⊢ z ∈ ℕ → z ≠ 0
18 17 ad2antrl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ≠ 0
19 14 16 18 divcan1d ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P z ⁢ z = P
20 12 19 breqtrrd ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∥ P z ⁢ z
21 20 adantr ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y → P ∥ P z ⁢ z
22 simprr ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ∥ P
23 simprl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ∈ ℕ
24 nndivdvds ⊢ P ∈ ℕ ∧ z ∈ ℕ → z ∥ P ↔ P z ∈ ℕ
25 9 23 24 syl2anc ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ∥ P ↔ P z ∈ ℕ
26 22 25 mpbid ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P z ∈ ℕ
27 26 nnzd ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P z ∈ ℤ
28 nnz ⊢ z ∈ ℕ → z ∈ ℤ
29 28 ad2antrl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ∈ ℤ
30 27 29 jca ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P z ∈ ℤ ∧ z ∈ ℤ
31 oveq1 ⊢ x = P z → x ⁢ y = P z ⁢ y
32 31 breq2d ⊢ x = P z → P ∥ x ⁢ y ↔ P ∥ P z ⁢ y
33 breq2 ⊢ x = P z → P ∥ x ↔ P ∥ P z
34 33 orbi1d ⊢ x = P z → P ∥ x ∨ P ∥ y ↔ P ∥ P z ∨ P ∥ y
35 32 34 imbi12d ⊢ x = P z → P ∥ x ⁢ y → P ∥ x ∨ P ∥ y ↔ P ∥ P z ⁢ y → P ∥ P z ∨ P ∥ y
36 oveq2 ⊢ y = z → P z ⁢ y = P z ⁢ z
37 36 breq2d ⊢ y = z → P ∥ P z ⁢ y ↔ P ∥ P z ⁢ z
38 breq2 ⊢ y = z → P ∥ y ↔ P ∥ z
39 38 orbi2d ⊢ y = z → P ∥ P z ∨ P ∥ y ↔ P ∥ P z ∨ P ∥ z
40 37 39 imbi12d ⊢ y = z → P ∥ P z ⁢ y → P ∥ P z ∨ P ∥ y ↔ P ∥ P z ⁢ z → P ∥ P z ∨ P ∥ z
41 35 40 rspc2va ⊢ P z ∈ ℤ ∧ z ∈ ℤ ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y → P ∥ P z ⁢ z → P ∥ P z ∨ P ∥ z
42 30 41 sylan ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y → P ∥ P z ⁢ z → P ∥ P z ∨ P ∥ z
43 21 42 mpd ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y → P ∥ P z ∨ P ∥ z
44 dvdsle ⊢ P ∈ ℤ ∧ P z ∈ ℕ → P ∥ P z → P ≤ P z
45 10 26 44 syl2anc ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∥ P z → P ≤ P z
46 14 div1d ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P 1 = P
47 46 breq1d ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P 1 ≤ P z ↔ P ≤ P z
48 45 47 sylibrd ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∥ P z → P 1 ≤ P z
49 nnrp ⊢ z ∈ ℕ → z ∈ ℝ +
50 49 rpregt0d ⊢ z ∈ ℕ → z ∈ ℝ ∧ 0 < z
51 50 ad2antrl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ∈ ℝ ∧ 0 < z
52 1rp ⊢ 1 ∈ ℝ +
53 rpregt0 ⊢ 1 ∈ ℝ + → 1 ∈ ℝ ∧ 0 < 1
54 52 53 mp1i ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → 1 ∈ ℝ ∧ 0 < 1
55 nnrp ⊢ P ∈ ℕ → P ∈ ℝ +
56 9 55 syl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∈ ℝ +
57 56 rpregt0d ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∈ ℝ ∧ 0 < P
58 lediv2 ⊢ z ∈ ℝ ∧ 0 < z ∧ 1 ∈ ℝ ∧ 0 < 1 ∧ P ∈ ℝ ∧ 0 < P → z ≤ 1 ↔ P 1 ≤ P z
59 51 54 57 58 syl3anc ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ≤ 1 ↔ P 1 ≤ P z
60 48 59 sylibrd ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∥ P z → z ≤ 1
61 nnle1eq1 ⊢ z ∈ ℕ → z ≤ 1 ↔ z = 1
62 61 ad2antrl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ≤ 1 ↔ z = 1
63 60 62 sylibd ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∥ P z → z = 1
64 nnnn0 ⊢ z ∈ ℕ → z ∈ ℕ 0
65 64 ad2antrl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → z ∈ ℕ 0
66 65 adantr ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ P ∥ z → z ∈ ℕ 0
67 nnnn0 ⊢ P ∈ ℕ → P ∈ ℕ 0
68 9 67 syl ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∈ ℕ 0
69 68 adantr ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ P ∥ z → P ∈ ℕ 0
70 simplrr ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ P ∥ z → z ∥ P
71 simpr ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ P ∥ z → P ∥ z
72 dvdseq ⊢ z ∈ ℕ 0 ∧ P ∈ ℕ 0 ∧ z ∥ P ∧ P ∥ z → z = P
73 66 69 70 71 72 syl22anc ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ P ∥ z → z = P
74 73 ex ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∥ z → z = P
75 63 74 orim12d ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P → P ∥ P z ∨ P ∥ z → z = 1 ∨ z = P
76 75 imp ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ P ∥ P z ∨ P ∥ z → z = 1 ∨ z = P
77 43 76 syldan ⊢ P ∈ ℤ ≥ 2 ∧ z ∈ ℕ ∧ z ∥ P ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y → z = 1 ∨ z = P
78 77 an32s ⊢ P ∈ ℤ ≥ 2 ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y ∧ z ∈ ℕ ∧ z ∥ P → z = 1 ∨ z = P
79 78 expr ⊢ P ∈ ℤ ≥ 2 ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y ∧ z ∈ ℕ → z ∥ P → z = 1 ∨ z = P
80 79 ralrimiva ⊢ P ∈ ℤ ≥ 2 ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y → ∀ z ∈ ℕ z ∥ P → z = 1 ∨ z = P
81 isprm2 ⊢ P ∈ ℙ ↔ P ∈ ℤ ≥ 2 ∧ ∀ z ∈ ℕ z ∥ P → z = 1 ∨ z = P
82 7 80 81 sylanbrc ⊢ P ∈ ℤ ≥ 2 ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y → P ∈ ℙ
83 6 82 impbii ⊢ P ∈ ℙ ↔ P ∈ ℤ ≥ 2 ∧ ∀ x ∈ ℤ ∀ y ∈ ℤ P ∥ x ⁢ y → P ∥ x ∨ P ∥ y