Metamath Proof Explorer


Theorem bernneq3

Description: A corollary of bernneq . (Contributed by Mario Carneiro, 11-Mar-2014)

Ref Expression
Assertion bernneq3 ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N < P N

Proof

Step Hyp Ref Expression
1 nn0re ⊢ N ∈ ℕ 0 → N ∈ ℝ
2 1 adantl ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N ∈ ℝ
3 peano2re ⊢ N ∈ ℝ → N + 1 ∈ ℝ
4 2 3 syl ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N + 1 ∈ ℝ
5 eluzelre ⊢ P ∈ ℤ ≥ 2 → P ∈ ℝ
6 reexpcl ⊢ P ∈ ℝ ∧ N ∈ ℕ 0 → P N ∈ ℝ
7 5 6 sylan ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → P N ∈ ℝ
8 2 ltp1d ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N < N + 1
9 uz2m1nn ⊢ P ∈ ℤ ≥ 2 → P − 1 ∈ ℕ
10 9 adantr ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → P − 1 ∈ ℕ
11 10 nnred ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → P − 1 ∈ ℝ
12 11 2 remulcld ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → P − 1 ⋅ N ∈ ℝ
13 peano2re ⊢ P − 1 ⋅ N ∈ ℝ → P − 1 ⋅ N + 1 ∈ ℝ
14 12 13 syl ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → P − 1 ⋅ N + 1 ∈ ℝ
15 1red ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → 1 ∈ ℝ
16 nn0ge0 ⊢ N ∈ ℕ 0 → 0 ≤ N
17 16 adantl ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → 0 ≤ N
18 10 nnge1d ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → 1 ≤ P − 1
19 2 11 17 18 lemulge12d ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N ≤ P − 1 ⋅ N
20 2 12 15 19 leadd1dd ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N + 1 ≤ P − 1 ⋅ N + 1
21 5 adantr ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → P ∈ ℝ
22 simpr ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N ∈ ℕ 0
23 eluzge2nn0 ⊢ P ∈ ℤ ≥ 2 → P ∈ ℕ 0
24 nn0ge0 ⊢ P ∈ ℕ 0 → 0 ≤ P
25 23 24 syl ⊢ P ∈ ℤ ≥ 2 → 0 ≤ P
26 25 adantr ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → 0 ≤ P
27 bernneq2 ⊢ P ∈ ℝ ∧ N ∈ ℕ 0 ∧ 0 ≤ P → P − 1 ⋅ N + 1 ≤ P N
28 21 22 26 27 syl3anc ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → P − 1 ⋅ N + 1 ≤ P N
29 4 14 7 20 28 letrd ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N + 1 ≤ P N
30 2 4 7 8 29 ltletrd ⊢ P ∈ ℤ ≥ 2 ∧ N ∈ ℕ 0 → N < P N