Metamath Proof Explorer


Theorem lighneallem1

Description: Lemma 1 for lighneal . (Contributed by AV, 11-Aug-2021)

Ref Expression
Assertion lighneallem1 ⊢ P = 2 ∧ M ∈ ℕ ∧ N ∈ ℕ → 2 N − 1 ≠ P M

Proof

Step Hyp Ref Expression
1 2z ⊢ 2 ∈ ℤ
2 simp2 ⊢ P = 2 ∧ M ∈ ℕ ∧ N ∈ ℕ → M ∈ ℕ
3 iddvdsexp ⊢ 2 ∈ ℤ ∧ M ∈ ℕ → 2 ∥ 2 M
4 1 2 3 sylancr ⊢ P = 2 ∧ M ∈ ℕ ∧ N ∈ ℕ → 2 ∥ 2 M
5 oveq1 ⊢ P = 2 → P M = 2 M
6 5 breq2d ⊢ P = 2 → 2 ∥ P M ↔ 2 ∥ 2 M
7 6 3ad2ant1 ⊢ P = 2 ∧ M ∈ ℕ ∧ N ∈ ℕ → 2 ∥ P M ↔ 2 ∥ 2 M
8 4 7 mpbird ⊢ P = 2 ∧ M ∈ ℕ ∧ N ∈ ℕ → 2 ∥ P M
9 iddvdsexp ⊢ 2 ∈ ℤ ∧ N ∈ ℕ → 2 ∥ 2 N
10 1 9 mpan ⊢ N ∈ ℕ → 2 ∥ 2 N
11 10 notnotd ⊢ N ∈ ℕ → ¬ ¬ 2 ∥ 2 N
12 2nn ⊢ 2 ∈ ℕ
13 12 a1i ⊢ N ∈ ℕ → 2 ∈ ℕ
14 nnnn0 ⊢ N ∈ ℕ → N ∈ ℕ 0
15 13 14 nnexpcld ⊢ N ∈ ℕ → 2 N ∈ ℕ
16 15 nnzd ⊢ N ∈ ℕ → 2 N ∈ ℤ
17 oddm1even ⊢ 2 N ∈ ℤ → ¬ 2 ∥ 2 N ↔ 2 ∥ 2 N − 1
18 16 17 syl ⊢ N ∈ ℕ → ¬ 2 ∥ 2 N ↔ 2 ∥ 2 N − 1
19 11 18 mtbid ⊢ N ∈ ℕ → ¬ 2 ∥ 2 N − 1
20 19 3ad2ant3 ⊢ P = 2 ∧ M ∈ ℕ ∧ N ∈ ℕ → ¬ 2 ∥ 2 N − 1
21 nbrne1 ⊢ 2 ∥ P M ∧ ¬ 2 ∥ 2 N − 1 → P M ≠ 2 N − 1
22 8 20 21 syl2anc ⊢ P = 2 ∧ M ∈ ℕ ∧ N ∈ ℕ → P M ≠ 2 N − 1
23 22 necomd ⊢ P = 2 ∧ M ∈ ℕ ∧ N ∈ ℕ → 2 N − 1 ≠ P M