Metamath Proof Explorer


Theorem nprmdvdsfacm1lem1

Description: Lemma 1 for nprmdvdsfacm1 . (Contributed by AV, 7-Apr-2026)

Ref Expression
Assertion nprmdvdsfacm1lem1 ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → N ∥ A ⁢ 2 ⁢ A

Proof

Step Hyp Ref Expression
1 2z ⊢ 2 ∈ ℤ
2 eluzelz ⊢ N ∈ ℤ ≥ 6 → N ∈ ℤ
3 dvdsmul2 ⊢ 2 ∈ ℤ ∧ N ∈ ℤ → N ∥ 2 ⋅ N
4 1 2 3 sylancr ⊢ N ∈ ℤ ≥ 6 → N ∥ 2 ⋅ N
5 4 3ad2ant1 ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → N ∥ 2 ⋅ N
6 elfzoelz ⊢ A ∈ 2 ..^ N → A ∈ ℤ
7 6 zcnd ⊢ A ∈ 2 ..^ N → A ∈ ℂ
8 2cnd ⊢ A ∈ 2 ..^ N → 2 ∈ ℂ
9 7 8 7 mul12d ⊢ A ∈ 2 ..^ N → A ⁢ 2 ⁢ A = 2 ⁢ A ⁢ A
10 9 3ad2ant2 ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → A ⁢ 2 ⁢ A = 2 ⁢ A ⁢ A
11 simp3 ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → N = A 2
12 7 sqvald ⊢ A ∈ 2 ..^ N → A 2 = A ⁢ A
13 12 3ad2ant2 ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → A 2 = A ⁢ A
14 11 13 eqtr2d ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → A ⁢ A = N
15 14 oveq2d ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → 2 ⁢ A ⁢ A = 2 ⋅ N
16 10 15 eqtrd ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → A ⁢ 2 ⁢ A = 2 ⋅ N
17 5 16 breqtrrd ⊢ N ∈ ℤ ≥ 6 ∧ A ∈ 2 ..^ N ∧ N = A 2 → N ∥ A ⁢ 2 ⁢ A