Metamath Proof Explorer


Theorem 2lgslem1a

Description: Lemma 1 for 2lgslem1 . (Contributed by AV, 18-Jun-2021)

Ref Expression
Assertion 2lgslem1a ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P → x ∈ ℤ | ∃ i ∈ 1 … P − 1 2 x = i ⋅ 2 ∧ P 2 < x mod P = x ∈ ℤ | ∃ i ∈ P 4 + 1 … P − 1 2 x = i ⋅ 2

Proof

Step Hyp Ref Expression
1 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
2 1 nnnn0d ⊢ P ∈ ℙ → P ∈ ℕ 0
3 2 ad2antrr ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → P ∈ ℕ 0
4 4nn ⊢ 4 ∈ ℕ
5 3 4 jctir ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → P ∈ ℕ 0 ∧ 4 ∈ ℕ
6 fldivnn0 ⊢ P ∈ ℕ 0 ∧ 4 ∈ ℕ → P 4 ∈ ℕ 0
7 nn0p1nn ⊢ P 4 ∈ ℕ 0 → P 4 + 1 ∈ ℕ
8 5 6 7 3syl ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → P 4 + 1 ∈ ℕ
9 elnnuz ⊢ P 4 + 1 ∈ ℕ ↔ P 4 + 1 ∈ ℤ ≥ 1
10 8 9 sylib ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → P 4 + 1 ∈ ℤ ≥ 1
11 fzss1 ⊢ P 4 + 1 ∈ ℤ ≥ 1 → P 4 + 1 … P − 1 2 ⊆ 1 … P − 1 2
12 rexss ⊢ P 4 + 1 … P − 1 2 ⊆ 1 … P − 1 2 → ∃ i ∈ P 4 + 1 … P − 1 2 x = i ⋅ 2 ↔ ∃ i ∈ 1 … P − 1 2 i ∈ P 4 + 1 … P − 1 2 ∧ x = i ⋅ 2
13 10 11 12 3syl ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → ∃ i ∈ P 4 + 1 … P − 1 2 x = i ⋅ 2 ↔ ∃ i ∈ 1 … P − 1 2 i ∈ P 4 + 1 … P − 1 2 ∧ x = i ⋅ 2
14 ancom ⊢ i ∈ P 4 + 1 … P − 1 2 ∧ x = i ⋅ 2 ↔ x = i ⋅ 2 ∧ i ∈ P 4 + 1 … P − 1 2
15 2 4 jctir ⊢ P ∈ ℙ → P ∈ ℕ 0 ∧ 4 ∈ ℕ
16 15 6 syl ⊢ P ∈ ℙ → P 4 ∈ ℕ 0
17 16 nn0zd ⊢ P ∈ ℙ → P 4 ∈ ℤ
18 17 ad2antrr ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → P 4 ∈ ℤ
19 elfzelz ⊢ i ∈ 1 … P − 1 2 → i ∈ ℤ
20 zltp1le ⊢ P 4 ∈ ℤ ∧ i ∈ ℤ → P 4 < i ↔ P 4 + 1 ≤ i
21 18 19 20 syl2an ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → P 4 < i ↔ P 4 + 1 ≤ i
22 21 bicomd ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → P 4 + 1 ≤ i ↔ P 4 < i
23 22 anbi1d ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → P 4 + 1 ≤ i ∧ i ≤ P − 1 2 ↔ P 4 < i ∧ i ≤ P − 1 2
24 19 adantl ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → i ∈ ℤ
25 17 peano2zd ⊢ P ∈ ℙ → P 4 + 1 ∈ ℤ
26 25 adantr ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P → P 4 + 1 ∈ ℤ
27 26 ad2antrr ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → P 4 + 1 ∈ ℤ
28 prmz ⊢ P ∈ ℙ → P ∈ ℤ
29 oddm1d2 ⊢ P ∈ ℤ → ¬ 2 ∥ P ↔ P − 1 2 ∈ ℤ
30 28 29 syl ⊢ P ∈ ℙ → ¬ 2 ∥ P ↔ P − 1 2 ∈ ℤ
31 30 biimpa ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P → P − 1 2 ∈ ℤ
32 31 ad2antrr ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → P − 1 2 ∈ ℤ
33 elfz ⊢ i ∈ ℤ ∧ P 4 + 1 ∈ ℤ ∧ P − 1 2 ∈ ℤ → i ∈ P 4 + 1 … P − 1 2 ↔ P 4 + 1 ≤ i ∧ i ≤ P − 1 2
34 24 27 32 33 syl3anc ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → i ∈ P 4 + 1 … P − 1 2 ↔ P 4 + 1 ≤ i ∧ i ≤ P − 1 2
35 elfzle2 ⊢ i ∈ 1 … P − 1 2 → i ≤ P − 1 2
36 35 adantl ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → i ≤ P − 1 2
37 36 biantrud ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → P 4 < i ↔ P 4 < i ∧ i ≤ P − 1 2
38 23 34 37 3bitr4d ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → i ∈ P 4 + 1 … P − 1 2 ↔ P 4 < i
39 28 ad2antrr ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → P ∈ ℤ
40 2lgslem1a2 ⊢ P ∈ ℤ ∧ i ∈ ℤ → P 4 < i ↔ P 2 < i ⋅ 2
41 39 19 40 syl2an ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → P 4 < i ↔ P 2 < i ⋅ 2
42 38 41 bitrd ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → i ∈ P 4 + 1 … P − 1 2 ↔ P 2 < i ⋅ 2
43 2lgslem1a1 ⊢ P ∈ ℕ ∧ ¬ 2 ∥ P → ∀ k ∈ 1 … P − 1 2 k ⋅ 2 = k ⋅ 2 mod P
44 1 43 sylan ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P → ∀ k ∈ 1 … P − 1 2 k ⋅ 2 = k ⋅ 2 mod P
45 44 adantr ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → ∀ k ∈ 1 … P − 1 2 k ⋅ 2 = k ⋅ 2 mod P
46 oveq1 ⊢ k = i → k ⋅ 2 = i ⋅ 2
47 46 oveq1d ⊢ k = i → k ⋅ 2 mod P = i ⋅ 2 mod P
48 46 47 eqeq12d ⊢ k = i → k ⋅ 2 = k ⋅ 2 mod P ↔ i ⋅ 2 = i ⋅ 2 mod P
49 48 rspccva ⊢ ∀ k ∈ 1 … P − 1 2 k ⋅ 2 = k ⋅ 2 mod P ∧ i ∈ 1 … P − 1 2 → i ⋅ 2 = i ⋅ 2 mod P
50 45 49 sylan ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → i ⋅ 2 = i ⋅ 2 mod P
51 50 breq2d ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → P 2 < i ⋅ 2 ↔ P 2 < i ⋅ 2 mod P
52 42 51 bitrd ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → i ∈ P 4 + 1 … P − 1 2 ↔ P 2 < i ⋅ 2 mod P
53 oveq1 ⊢ x = i ⋅ 2 → x mod P = i ⋅ 2 mod P
54 53 eqcomd ⊢ x = i ⋅ 2 → i ⋅ 2 mod P = x mod P
55 54 breq2d ⊢ x = i ⋅ 2 → P 2 < i ⋅ 2 mod P ↔ P 2 < x mod P
56 52 55 sylan9bb ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 ∧ x = i ⋅ 2 → i ∈ P 4 + 1 … P − 1 2 ↔ P 2 < x mod P
57 56 pm5.32da ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → x = i ⋅ 2 ∧ i ∈ P 4 + 1 … P − 1 2 ↔ x = i ⋅ 2 ∧ P 2 < x mod P
58 14 57 bitrid ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ ∧ i ∈ 1 … P − 1 2 → i ∈ P 4 + 1 … P − 1 2 ∧ x = i ⋅ 2 ↔ x = i ⋅ 2 ∧ P 2 < x mod P
59 58 rexbidva ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → ∃ i ∈ 1 … P − 1 2 i ∈ P 4 + 1 … P − 1 2 ∧ x = i ⋅ 2 ↔ ∃ i ∈ 1 … P − 1 2 x = i ⋅ 2 ∧ P 2 < x mod P
60 13 59 bitrd ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → ∃ i ∈ P 4 + 1 … P − 1 2 x = i ⋅ 2 ↔ ∃ i ∈ 1 … P − 1 2 x = i ⋅ 2 ∧ P 2 < x mod P
61 60 bicomd ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P ∧ x ∈ ℤ → ∃ i ∈ 1 … P − 1 2 x = i ⋅ 2 ∧ P 2 < x mod P ↔ ∃ i ∈ P 4 + 1 … P − 1 2 x = i ⋅ 2
62 61 rabbidva ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P → x ∈ ℤ | ∃ i ∈ 1 … P − 1 2 x = i ⋅ 2 ∧ P 2 < x mod P = x ∈ ℤ | ∃ i ∈ P 4 + 1 … P − 1 2 x = i ⋅ 2