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 Yrm 𝑁 ) ∈ ℕ )

Proof

Step Hyp Ref Expression
1 nnz ( 𝑁 ∈ ℕ → 𝑁 ∈ ℤ )
2 frmy Yrm : ( ( ℤ ‘ 2 ) × ℤ ) ⟶ ℤ
3 2 fovcl ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℤ ) → ( 𝐴 Yrm 𝑁 ) ∈ ℤ )
4 1 3 sylan2 ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℕ ) → ( 𝐴 Yrm 𝑁 ) ∈ ℤ )
5 rmy0 ( 𝐴 ∈ ( ℤ ‘ 2 ) → ( 𝐴 Yrm 0 ) = 0 )
6 5 adantr ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℕ ) → ( 𝐴 Yrm 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 Yrm 0 ) < ( 𝐴 Yrm 𝑁 ) ) )
13 9 10 11 12 syl3anc ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℕ ) → ( 0 < 𝑁 ↔ ( 𝐴 Yrm 0 ) < ( 𝐴 Yrm 𝑁 ) ) )
14 8 13 mpbid ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℕ ) → ( 𝐴 Yrm 0 ) < ( 𝐴 Yrm 𝑁 ) )
15 6 14 eqbrtrrd ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℕ ) → 0 < ( 𝐴 Yrm 𝑁 ) )
16 elnnz ( ( 𝐴 Yrm 𝑁 ) ∈ ℕ ↔ ( ( 𝐴 Yrm 𝑁 ) ∈ ℤ ∧ 0 < ( 𝐴 Yrm 𝑁 ) ) )
17 4 15 16 sylanbrc ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℕ ) → ( 𝐴 Yrm 𝑁 ) ∈ ℕ )