Metamath Proof Explorer


Theorem ceilhalfnn

Description: The ceiling of half of a positive integer is a positive integer. (Contributed by AV, 2-Nov-2025)

Ref Expression
Assertion ceilhalfnn ⊢ N ∈ ℕ → N 2 ∈ ℕ

Proof

Step Hyp Ref Expression
1 nnre ⊢ N ∈ ℕ → N ∈ ℝ
2 1 rehalfcld ⊢ N ∈ ℕ → N 2 ∈ ℝ
3 2 ceilcld ⊢ N ∈ ℕ → N 2 ∈ ℤ
4 elnn1uz2 ⊢ N ∈ ℕ ↔ N = 1 ∨ N ∈ ℤ ≥ 2
5 1le1 ⊢ 1 ≤ 1
6 fvoveq1 ⊢ N = 1 → N 2 = 1 2
7 ceilhalf1 ⊢ 1 2 = 1
8 6 7 eqtrdi ⊢ N = 1 → N 2 = 1
9 5 8 breqtrrid ⊢ N = 1 → 1 ≤ N 2
10 1red ⊢ N ∈ ℤ ≥ 2 → 1 ∈ ℝ
11 eluzelre ⊢ N ∈ ℤ ≥ 2 → N ∈ ℝ
12 11 rehalfcld ⊢ N ∈ ℤ ≥ 2 → N 2 ∈ ℝ
13 12 ceilcld ⊢ N ∈ ℤ ≥ 2 → N 2 ∈ ℤ
14 13 zred ⊢ N ∈ ℤ ≥ 2 → N 2 ∈ ℝ
15 eluzle ⊢ N ∈ ℤ ≥ 2 → 2 ≤ N
16 2re ⊢ 2 ∈ ℝ
17 elicopnf ⊢ 2 ∈ ℝ → N ∈ 2 +∞ ↔ N ∈ ℝ ∧ 2 ≤ N
18 16 17 ax-mp ⊢ N ∈ 2 +∞ ↔ N ∈ ℝ ∧ 2 ≤ N
19 11 15 18 sylanbrc ⊢ N ∈ ℤ ≥ 2 → N ∈ 2 +∞
20 rehalfge1 ⊢ N ∈ 2 +∞ → 1 ≤ N 2
21 19 20 syl ⊢ N ∈ ℤ ≥ 2 → 1 ≤ N 2
22 12 ceilged ⊢ N ∈ ℤ ≥ 2 → N 2 ≤ N 2
23 10 12 14 21 22 letrd ⊢ N ∈ ℤ ≥ 2 → 1 ≤ N 2
24 9 23 jaoi ⊢ N = 1 ∨ N ∈ ℤ ≥ 2 → 1 ≤ N 2
25 4 24 sylbi ⊢ N ∈ ℕ → 1 ≤ N 2
26 elnnz1 ⊢ N 2 ∈ ℕ ↔ N 2 ∈ ℤ ∧ 1 ≤ N 2
27 3 25 26 sylanbrc ⊢ N ∈ ℕ → N 2 ∈ ℕ