Metamath Proof Explorer


Theorem rmyabs

Description: rmY commutes with abs . (Contributed by Stefan O'Rear, 26-Sep-2014)

Ref Expression
Assertion rmyabs ⊢ A ∈ ℤ ≥ 2 ∧ B ∈ ℤ → A Y rm B = A Y rm B

Proof

Step Hyp Ref Expression
1 frmy ⊢ Y rm : ℤ ≥ 2 × ℤ ⟶ ℤ
2 1 fovcl ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℤ → A Y rm a ∈ ℤ
3 2 zred ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℤ → A Y rm a ∈ ℝ
4 simp1 ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℤ ∧ 0 ≤ a → A ∈ ℤ ≥ 2
5 elnn0z ⊢ a ∈ ℕ 0 ↔ a ∈ ℤ ∧ 0 ≤ a
6 5 biimpri ⊢ a ∈ ℤ ∧ 0 ≤ a → a ∈ ℕ 0
7 6 3adant1 ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℤ ∧ 0 ≤ a → a ∈ ℕ 0
8 rmxypos ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℕ 0 → 0 < A X rm a ∧ 0 ≤ A Y rm a
9 4 7 8 syl2anc ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℤ ∧ 0 ≤ a → 0 < A X rm a ∧ 0 ≤ A Y rm a
10 9 simprd ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℤ ∧ 0 ≤ a → 0 ≤ A Y rm a
11 rmyneg ⊢ A ∈ ℤ ≥ 2 ∧ b ∈ ℤ → A Y rm − b = − A Y rm b
12 oveq2 ⊢ a = b → A Y rm a = A Y rm b
13 oveq2 ⊢ a = − b → A Y rm a = A Y rm − b
14 oveq2 ⊢ a = B → A Y rm a = A Y rm B
15 oveq2 ⊢ a = B → A Y rm a = A Y rm B
16 3 10 11 12 13 14 15 oddcomabszz ⊢ A ∈ ℤ ≥ 2 ∧ B ∈ ℤ → A Y rm B = A Y rm B