Metamath Proof Explorer


Theorem rmynn

Description: rmY is positive for positive arguments. (Contributed by Stefan O'Rear, 16-Oct-2014)

Ref Expression
Assertion rmynn ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → A Y rm N ∈ ℕ

Proof

Step Hyp Ref Expression
1 nnz ⊢ N ∈ ℕ → N ∈ ℤ
2 frmy ⊢ Y rm : ℤ ≥ 2 × ℤ ⟶ ℤ
3 2 fovcl ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A Y rm N ∈ ℤ
4 1 3 sylan2 ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → A Y rm N ∈ ℤ
5 rmy0 ⊢ A ∈ ℤ ≥ 2 → A Y rm 0 = 0
6 5 adantr ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → A Y rm 0 = 0
7 nngt0 ⊢ N ∈ ℕ → 0 < N
8 7 adantl ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → 0 < N
9 simpl ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → A ∈ ℤ ≥ 2
10 0zd ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → 0 ∈ ℤ
11 1 adantl ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → N ∈ ℤ
12 ltrmy ⊢ A ∈ ℤ ≥ 2 ∧ 0 ∈ ℤ ∧ N ∈ ℤ → 0 < N ↔ A Y rm 0 < A Y rm N
13 9 10 11 12 syl3anc ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → 0 < N ↔ A Y rm 0 < A Y rm N
14 8 13 mpbid ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → A Y rm 0 < A Y rm N
15 6 14 eqbrtrrd ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → 0 < A Y rm N
16 elnnz ⊢ A Y rm N ∈ ℕ ↔ A Y rm N ∈ ℤ ∧ 0 < A Y rm N
17 4 15 16 sylanbrc ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℕ → A Y rm N ∈ ℕ