Metamath Proof Explorer


Theorem oddge22np1

Description: An integer greater than one is odd iff it is one plus twice a positive integer. (Contributed by AV, 16-Aug-2021) (Proof shortened by AV, 9-Jul-2022)

Ref Expression
Assertion oddge22np1 ⊢ N ∈ ℤ ≥ 2 → ¬ 2 ∥ N ↔ ∃ n ∈ ℕ 2 ⁢ n + 1 = N

Proof

Step Hyp Ref Expression
1 eleq1 ⊢ 2 ⁢ n + 1 = N → 2 ⁢ n + 1 ∈ ℤ ≥ 2 ↔ N ∈ ℤ ≥ 2
2 nn0z ⊢ n ∈ ℕ 0 → n ∈ ℤ
3 2 adantl ⊢ 2 ⁢ n + 1 ∈ ℤ ≥ 2 ∧ n ∈ ℕ 0 → n ∈ ℤ
4 eluz2 ⊢ 2 ⁢ n + 1 ∈ ℤ ≥ 2 ↔ 2 ∈ ℤ ∧ 2 ⁢ n + 1 ∈ ℤ ∧ 2 ≤ 2 ⁢ n + 1
5 2re ⊢ 2 ∈ ℝ
6 5 a1i ⊢ n ∈ ℕ 0 → 2 ∈ ℝ
7 1red ⊢ n ∈ ℕ 0 → 1 ∈ ℝ
8 2nn0 ⊢ 2 ∈ ℕ 0
9 8 a1i ⊢ n ∈ ℕ 0 → 2 ∈ ℕ 0
10 id ⊢ n ∈ ℕ 0 → n ∈ ℕ 0
11 9 10 nn0mulcld ⊢ n ∈ ℕ 0 → 2 ⁢ n ∈ ℕ 0
12 11 nn0red ⊢ n ∈ ℕ 0 → 2 ⁢ n ∈ ℝ
13 6 7 12 lesubaddd ⊢ n ∈ ℕ 0 → 2 − 1 ≤ 2 ⁢ n ↔ 2 ≤ 2 ⁢ n + 1
14 2m1e1 ⊢ 2 − 1 = 1
15 14 breq1i ⊢ 2 − 1 ≤ 2 ⁢ n ↔ 1 ≤ 2 ⁢ n
16 nn0re ⊢ n ∈ ℕ 0 → n ∈ ℝ
17 2rp ⊢ 2 ∈ ℝ +
18 17 a1i ⊢ n ∈ ℕ 0 → 2 ∈ ℝ +
19 7 16 18 ledivmuld ⊢ n ∈ ℕ 0 → 1 2 ≤ n ↔ 1 ≤ 2 ⁢ n
20 halfgt0 ⊢ 0 < 1 2
21 0red ⊢ n ∈ ℕ 0 → 0 ∈ ℝ
22 halfre ⊢ 1 2 ∈ ℝ
23 22 a1i ⊢ n ∈ ℕ 0 → 1 2 ∈ ℝ
24 ltletr ⊢ 0 ∈ ℝ ∧ 1 2 ∈ ℝ ∧ n ∈ ℝ → 0 < 1 2 ∧ 1 2 ≤ n → 0 < n
25 21 23 16 24 syl3anc ⊢ n ∈ ℕ 0 → 0 < 1 2 ∧ 1 2 ≤ n → 0 < n
26 20 25 mpani ⊢ n ∈ ℕ 0 → 1 2 ≤ n → 0 < n
27 19 26 sylbird ⊢ n ∈ ℕ 0 → 1 ≤ 2 ⁢ n → 0 < n
28 15 27 biimtrid ⊢ n ∈ ℕ 0 → 2 − 1 ≤ 2 ⁢ n → 0 < n
29 13 28 sylbird ⊢ n ∈ ℕ 0 → 2 ≤ 2 ⁢ n + 1 → 0 < n
30 29 com12 ⊢ 2 ≤ 2 ⁢ n + 1 → n ∈ ℕ 0 → 0 < n
31 30 3ad2ant3 ⊢ 2 ∈ ℤ ∧ 2 ⁢ n + 1 ∈ ℤ ∧ 2 ≤ 2 ⁢ n + 1 → n ∈ ℕ 0 → 0 < n
32 4 31 sylbi ⊢ 2 ⁢ n + 1 ∈ ℤ ≥ 2 → n ∈ ℕ 0 → 0 < n
33 32 imp ⊢ 2 ⁢ n + 1 ∈ ℤ ≥ 2 ∧ n ∈ ℕ 0 → 0 < n
34 elnnz ⊢ n ∈ ℕ ↔ n ∈ ℤ ∧ 0 < n
35 3 33 34 sylanbrc ⊢ 2 ⁢ n + 1 ∈ ℤ ≥ 2 ∧ n ∈ ℕ 0 → n ∈ ℕ
36 35 ex ⊢ 2 ⁢ n + 1 ∈ ℤ ≥ 2 → n ∈ ℕ 0 → n ∈ ℕ
37 1 36 biimtrrdi ⊢ 2 ⁢ n + 1 = N → N ∈ ℤ ≥ 2 → n ∈ ℕ 0 → n ∈ ℕ
38 37 com13 ⊢ n ∈ ℕ 0 → N ∈ ℤ ≥ 2 → 2 ⁢ n + 1 = N → n ∈ ℕ
39 38 impcom ⊢ N ∈ ℤ ≥ 2 ∧ n ∈ ℕ 0 → 2 ⁢ n + 1 = N → n ∈ ℕ
40 39 pm4.71rd ⊢ N ∈ ℤ ≥ 2 ∧ n ∈ ℕ 0 → 2 ⁢ n + 1 = N ↔ n ∈ ℕ ∧ 2 ⁢ n + 1 = N
41 40 bicomd ⊢ N ∈ ℤ ≥ 2 ∧ n ∈ ℕ 0 → n ∈ ℕ ∧ 2 ⁢ n + 1 = N ↔ 2 ⁢ n + 1 = N
42 41 rexbidva ⊢ N ∈ ℤ ≥ 2 → ∃ n ∈ ℕ 0 n ∈ ℕ ∧ 2 ⁢ n + 1 = N ↔ ∃ n ∈ ℕ 0 2 ⁢ n + 1 = N
43 nnssnn0 ⊢ ℕ ⊆ ℕ 0
44 rexss ⊢ ℕ ⊆ ℕ 0 → ∃ n ∈ ℕ 2 ⁢ n + 1 = N ↔ ∃ n ∈ ℕ 0 n ∈ ℕ ∧ 2 ⁢ n + 1 = N
45 43 44 mp1i ⊢ N ∈ ℤ ≥ 2 → ∃ n ∈ ℕ 2 ⁢ n + 1 = N ↔ ∃ n ∈ ℕ 0 n ∈ ℕ ∧ 2 ⁢ n + 1 = N
46 eluzge2nn0 ⊢ N ∈ ℤ ≥ 2 → N ∈ ℕ 0
47 oddnn02np1 ⊢ N ∈ ℕ 0 → ¬ 2 ∥ N ↔ ∃ n ∈ ℕ 0 2 ⁢ n + 1 = N
48 46 47 syl ⊢ N ∈ ℤ ≥ 2 → ¬ 2 ∥ N ↔ ∃ n ∈ ℕ 0 2 ⁢ n + 1 = N
49 42 45 48 3bitr4rd ⊢ N ∈ ℤ ≥ 2 → ¬ 2 ∥ N ↔ ∃ n ∈ ℕ 2 ⁢ n + 1 = N