Metamath Proof Explorer


Theorem 2lgslem1c

Description: Lemma 3 for 2lgslem1 . (Contributed by AV, 19-Jun-2021)

Ref Expression
Assertion 2lgslem1c ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P → P 4 ≤ P − 1 2

Proof

Step Hyp Ref Expression
1 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
2 nnnn0 ⊢ P ∈ ℕ → P ∈ ℕ 0
3 oddnn02np1 ⊢ P ∈ ℕ 0 → ¬ 2 ∥ P ↔ ∃ n ∈ ℕ 0 2 ⁢ n + 1 = P
4 1 2 3 3syl ⊢ P ∈ ℙ → ¬ 2 ∥ P ↔ ∃ n ∈ ℕ 0 2 ⁢ n + 1 = P
5 iftrue ⊢ 2 ∥ n → if 2 ∥ n n 2 n − 1 2 = n 2
6 5 adantr ⊢ 2 ∥ n ∧ n ∈ ℕ 0 → if 2 ∥ n n 2 n − 1 2 = n 2
7 2nn ⊢ 2 ∈ ℕ
8 nn0ledivnn ⊢ n ∈ ℕ 0 ∧ 2 ∈ ℕ → n 2 ≤ n
9 7 8 mpan2 ⊢ n ∈ ℕ 0 → n 2 ≤ n
10 9 adantl ⊢ 2 ∥ n ∧ n ∈ ℕ 0 → n 2 ≤ n
11 6 10 eqbrtrd ⊢ 2 ∥ n ∧ n ∈ ℕ 0 → if 2 ∥ n n 2 n − 1 2 ≤ n
12 iffalse ⊢ ¬ 2 ∥ n → if 2 ∥ n n 2 n − 1 2 = n − 1 2
13 12 adantr ⊢ ¬ 2 ∥ n ∧ n ∈ ℕ 0 → if 2 ∥ n n 2 n − 1 2 = n − 1 2
14 nn0re ⊢ n ∈ ℕ 0 → n ∈ ℝ
15 peano2rem ⊢ n ∈ ℝ → n − 1 ∈ ℝ
16 15 rehalfcld ⊢ n ∈ ℝ → n − 1 2 ∈ ℝ
17 14 16 syl ⊢ n ∈ ℕ 0 → n − 1 2 ∈ ℝ
18 14 rehalfcld ⊢ n ∈ ℕ 0 → n 2 ∈ ℝ
19 14 lem1d ⊢ n ∈ ℕ 0 → n − 1 ≤ n
20 14 15 syl ⊢ n ∈ ℕ 0 → n − 1 ∈ ℝ
21 2re ⊢ 2 ∈ ℝ
22 2pos ⊢ 0 < 2
23 21 22 pm3.2i ⊢ 2 ∈ ℝ ∧ 0 < 2
24 23 a1i ⊢ n ∈ ℕ 0 → 2 ∈ ℝ ∧ 0 < 2
25 lediv1 ⊢ n − 1 ∈ ℝ ∧ n ∈ ℝ ∧ 2 ∈ ℝ ∧ 0 < 2 → n − 1 ≤ n ↔ n − 1 2 ≤ n 2
26 20 14 24 25 syl3anc ⊢ n ∈ ℕ 0 → n − 1 ≤ n ↔ n − 1 2 ≤ n 2
27 19 26 mpbid ⊢ n ∈ ℕ 0 → n − 1 2 ≤ n 2
28 17 18 14 27 9 letrd ⊢ n ∈ ℕ 0 → n − 1 2 ≤ n
29 28 adantl ⊢ ¬ 2 ∥ n ∧ n ∈ ℕ 0 → n − 1 2 ≤ n
30 13 29 eqbrtrd ⊢ ¬ 2 ∥ n ∧ n ∈ ℕ 0 → if 2 ∥ n n 2 n − 1 2 ≤ n
31 11 30 pm2.61ian ⊢ n ∈ ℕ 0 → if 2 ∥ n n 2 n − 1 2 ≤ n
32 31 ad2antlr ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → if 2 ∥ n n 2 n − 1 2 ≤ n
33 nn0z ⊢ n ∈ ℕ 0 → n ∈ ℤ
34 33 adantl ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 → n ∈ ℤ
35 eqcom ⊢ 2 ⁢ n + 1 = P ↔ P = 2 ⁢ n + 1
36 35 biimpi ⊢ 2 ⁢ n + 1 = P → P = 2 ⁢ n + 1
37 flodddiv4 ⊢ n ∈ ℤ ∧ P = 2 ⁢ n + 1 → P 4 = if 2 ∥ n n 2 n − 1 2
38 34 36 37 syl2an ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → P 4 = if 2 ∥ n n 2 n − 1 2
39 oveq1 ⊢ P = 2 ⁢ n + 1 → P − 1 = 2 ⁢ n + 1 - 1
40 39 eqcoms ⊢ 2 ⁢ n + 1 = P → P − 1 = 2 ⁢ n + 1 - 1
41 40 adantl ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → P − 1 = 2 ⁢ n + 1 - 1
42 2nn0 ⊢ 2 ∈ ℕ 0
43 42 a1i ⊢ n ∈ ℕ 0 → 2 ∈ ℕ 0
44 id ⊢ n ∈ ℕ 0 → n ∈ ℕ 0
45 43 44 nn0mulcld ⊢ n ∈ ℕ 0 → 2 ⁢ n ∈ ℕ 0
46 45 nn0cnd ⊢ n ∈ ℕ 0 → 2 ⁢ n ∈ ℂ
47 pncan1 ⊢ 2 ⁢ n ∈ ℂ → 2 ⁢ n + 1 - 1 = 2 ⁢ n
48 46 47 syl ⊢ n ∈ ℕ 0 → 2 ⁢ n + 1 - 1 = 2 ⁢ n
49 48 ad2antlr ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → 2 ⁢ n + 1 - 1 = 2 ⁢ n
50 41 49 eqtrd ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → P − 1 = 2 ⁢ n
51 50 oveq1d ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → P − 1 2 = 2 ⁢ n 2
52 nn0cn ⊢ n ∈ ℕ 0 → n ∈ ℂ
53 2cnd ⊢ n ∈ ℕ 0 → 2 ∈ ℂ
54 2ne0 ⊢ 2 ≠ 0
55 54 a1i ⊢ n ∈ ℕ 0 → 2 ≠ 0
56 52 53 55 divcan3d ⊢ n ∈ ℕ 0 → 2 ⁢ n 2 = n
57 56 ad2antlr ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → 2 ⁢ n 2 = n
58 51 57 eqtrd ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → P − 1 2 = n
59 32 38 58 3brtr4d ⊢ P ∈ ℙ ∧ n ∈ ℕ 0 ∧ 2 ⁢ n + 1 = P → P 4 ≤ P − 1 2
60 59 rexlimdva2 ⊢ P ∈ ℙ → ∃ n ∈ ℕ 0 2 ⁢ n + 1 = P → P 4 ≤ P − 1 2
61 4 60 sylbid ⊢ P ∈ ℙ → ¬ 2 ∥ P → P 4 ≤ P − 1 2
62 61 imp ⊢ P ∈ ℙ ∧ ¬ 2 ∥ P → P 4 ≤ P − 1 2