Metamath Proof Explorer


Theorem rmyeq0

Description: Y is zero only at zero. (Contributed by Stefan O'Rear, 26-Sep-2014)

Ref Expression
Assertion rmyeq0 ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℤ ) → ( 𝑁 = 0 ↔ ( 𝐴 Yrm 𝑁 ) = 0 ) )

Proof

Step Hyp Ref Expression
1 0z 0 ∈ ℤ
2 oveq2 ( 𝑎 = 𝑏 → ( 𝐴 Yrm 𝑎 ) = ( 𝐴 Yrm 𝑏 ) )
3 oveq2 ( 𝑎 = 𝑁 → ( 𝐴 Yrm 𝑎 ) = ( 𝐴 Yrm 𝑁 ) )
4 oveq2 ( 𝑎 = 0 → ( 𝐴 Yrm 𝑎 ) = ( 𝐴 Yrm 0 ) )
5 zssre ℤ ⊆ ℝ
6 frmy Yrm : ( ( ℤ ‘ 2 ) × ℤ ) ⟶ ℤ
7 6 fovcl ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑎 ∈ ℤ ) → ( 𝐴 Yrm 𝑎 ) ∈ ℤ )
8 7 zred ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑎 ∈ ℤ ) → ( 𝐴 Yrm 𝑎 ) ∈ ℝ )
9 ltrmy ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑎 ∈ ℤ ∧ 𝑏 ∈ ℤ ) → ( 𝑎 < 𝑏 ↔ ( 𝐴 Yrm 𝑎 ) < ( 𝐴 Yrm 𝑏 ) ) )
10 9 biimpd ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑎 ∈ ℤ ∧ 𝑏 ∈ ℤ ) → ( 𝑎 < 𝑏 → ( 𝐴 Yrm 𝑎 ) < ( 𝐴 Yrm 𝑏 ) ) )
11 10 3expb ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ ( 𝑎 ∈ ℤ ∧ 𝑏 ∈ ℤ ) ) → ( 𝑎 < 𝑏 → ( 𝐴 Yrm 𝑎 ) < ( 𝐴 Yrm 𝑏 ) ) )
12 2 3 4 5 8 11 eqord1 ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ ( 𝑁 ∈ ℤ ∧ 0 ∈ ℤ ) ) → ( 𝑁 = 0 ↔ ( 𝐴 Yrm 𝑁 ) = ( 𝐴 Yrm 0 ) ) )
13 1 12 mpanr2 ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℤ ) → ( 𝑁 = 0 ↔ ( 𝐴 Yrm 𝑁 ) = ( 𝐴 Yrm 0 ) ) )
14 rmy0 ( 𝐴 ∈ ( ℤ ‘ 2 ) → ( 𝐴 Yrm 0 ) = 0 )
15 14 adantr ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℤ ) → ( 𝐴 Yrm 0 ) = 0 )
16 15 eqeq2d ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℤ ) → ( ( 𝐴 Yrm 𝑁 ) = ( 𝐴 Yrm 0 ) ↔ ( 𝐴 Yrm 𝑁 ) = 0 ) )
17 13 16 bitrd ( ( 𝐴 ∈ ( ℤ ‘ 2 ) ∧ 𝑁 ∈ ℤ ) → ( 𝑁 = 0 ↔ ( 𝐴 Yrm 𝑁 ) = 0 ) )