Metamath Proof Explorer


Theorem proththdlem

Description: Lemma for proththd . (Contributed by AV, 4-Jul-2020)

Ref Expression
Hypotheses proththd.n ⊢ φ → N ∈ ℕ
proththd.k ⊢ φ → K ∈ ℕ
proththd.p ⊢ φ → P = K ⁢ 2 N + 1
Assertion proththdlem ⊢ φ → P ∈ ℕ ∧ 1 < P ∧ P − 1 2 ∈ ℕ

Proof

Step Hyp Ref Expression
1 proththd.n ⊢ φ → N ∈ ℕ
2 proththd.k ⊢ φ → K ∈ ℕ
3 proththd.p ⊢ φ → P = K ⁢ 2 N + 1
4 2nn ⊢ 2 ∈ ℕ
5 4 a1i ⊢ φ → 2 ∈ ℕ
6 1 nnnn0d ⊢ φ → N ∈ ℕ 0
7 5 6 nnexpcld ⊢ φ → 2 N ∈ ℕ
8 2 7 nnmulcld ⊢ φ → K ⁢ 2 N ∈ ℕ
9 8 peano2nnd ⊢ φ → K ⁢ 2 N + 1 ∈ ℕ
10 1m1e0 ⊢ 1 − 1 = 0
11 8 nngt0d ⊢ φ → 0 < K ⁢ 2 N
12 10 11 eqbrtrid ⊢ φ → 1 − 1 < K ⁢ 2 N
13 1red ⊢ φ → 1 ∈ ℝ
14 8 nnred ⊢ φ → K ⁢ 2 N ∈ ℝ
15 13 13 14 ltsubaddd ⊢ φ → 1 − 1 < K ⁢ 2 N ↔ 1 < K ⁢ 2 N + 1
16 12 15 mpbid ⊢ φ → 1 < K ⁢ 2 N + 1
17 8 nncnd ⊢ φ → K ⁢ 2 N ∈ ℂ
18 pncan1 ⊢ K ⁢ 2 N ∈ ℂ → K ⁢ 2 N + 1 - 1 = K ⁢ 2 N
19 17 18 syl ⊢ φ → K ⁢ 2 N + 1 - 1 = K ⁢ 2 N
20 19 oveq1d ⊢ φ → K ⁢ 2 N + 1 - 1 2 = K ⁢ 2 N 2
21 2z ⊢ 2 ∈ ℤ
22 21 a1i ⊢ φ → 2 ∈ ℤ
23 2 nnzd ⊢ φ → K ∈ ℤ
24 7 nnzd ⊢ φ → 2 N ∈ ℤ
25 22 23 24 3jca ⊢ φ → 2 ∈ ℤ ∧ K ∈ ℤ ∧ 2 N ∈ ℤ
26 iddvdsexp ⊢ 2 ∈ ℤ ∧ N ∈ ℕ → 2 ∥ 2 N
27 22 1 26 syl2anc ⊢ φ → 2 ∥ 2 N
28 dvdsmultr2 ⊢ 2 ∈ ℤ ∧ K ∈ ℤ ∧ 2 N ∈ ℤ → 2 ∥ 2 N → 2 ∥ K ⁢ 2 N
29 25 27 28 sylc ⊢ φ → 2 ∥ K ⁢ 2 N
30 nndivdvds ⊢ K ⁢ 2 N ∈ ℕ ∧ 2 ∈ ℕ → 2 ∥ K ⁢ 2 N ↔ K ⁢ 2 N 2 ∈ ℕ
31 8 5 30 syl2anc ⊢ φ → 2 ∥ K ⁢ 2 N ↔ K ⁢ 2 N 2 ∈ ℕ
32 29 31 mpbid ⊢ φ → K ⁢ 2 N 2 ∈ ℕ
33 20 32 eqeltrd ⊢ φ → K ⁢ 2 N + 1 - 1 2 ∈ ℕ
34 9 16 33 3jca ⊢ φ → K ⁢ 2 N + 1 ∈ ℕ ∧ 1 < K ⁢ 2 N + 1 ∧ K ⁢ 2 N + 1 - 1 2 ∈ ℕ
35 eleq1 ⊢ P = K ⁢ 2 N + 1 → P ∈ ℕ ↔ K ⁢ 2 N + 1 ∈ ℕ
36 breq2 ⊢ P = K ⁢ 2 N + 1 → 1 < P ↔ 1 < K ⁢ 2 N + 1
37 oveq1 ⊢ P = K ⁢ 2 N + 1 → P − 1 = K ⁢ 2 N + 1 - 1
38 37 oveq1d ⊢ P = K ⁢ 2 N + 1 → P − 1 2 = K ⁢ 2 N + 1 - 1 2
39 38 eleq1d ⊢ P = K ⁢ 2 N + 1 → P − 1 2 ∈ ℕ ↔ K ⁢ 2 N + 1 - 1 2 ∈ ℕ
40 35 36 39 3anbi123d ⊢ P = K ⁢ 2 N + 1 → P ∈ ℕ ∧ 1 < P ∧ P − 1 2 ∈ ℕ ↔ K ⁢ 2 N + 1 ∈ ℕ ∧ 1 < K ⁢ 2 N + 1 ∧ K ⁢ 2 N + 1 - 1 2 ∈ ℕ
41 34 40 syl5ibrcom ⊢ φ → P = K ⁢ 2 N + 1 → P ∈ ℕ ∧ 1 < P ∧ P − 1 2 ∈ ℕ
42 3 41 mpd ⊢ φ → P ∈ ℕ ∧ 1 < P ∧ P − 1 2 ∈ ℕ