Metamath Proof Explorer


Theorem lermxnn0

Description: The X-sequence is monotonic on NN0 . (Contributed by Stefan O'Rear, 4-Oct-2014)

Ref Expression
Assertion lermxnn0 ⊢ A ∈ ℤ ≥ 2 ∧ M ∈ ℕ 0 ∧ N ∈ ℕ 0 → M ≤ N ↔ A X rm M ≤ A X rm N

Proof

Step Hyp Ref Expression
1 oveq2 ⊢ a = b → A X rm a = A X rm b
2 oveq2 ⊢ a = M → A X rm a = A X rm M
3 oveq2 ⊢ a = N → A X rm a = A X rm N
4 nn0ssre ⊢ ℕ 0 ⊆ ℝ
5 nn0z ⊢ a ∈ ℕ 0 → a ∈ ℤ
6 frmx ⊢ X rm : ℤ ≥ 2 × ℤ ⟶ ℕ 0
7 6 fovcl ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℤ → A X rm a ∈ ℕ 0
8 5 7 sylan2 ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℕ 0 → A X rm a ∈ ℕ 0
9 8 nn0red ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℕ 0 → A X rm a ∈ ℝ
10 ltrmxnn0 ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → a < b ↔ A X rm a < A X rm b
11 10 biimpd ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → a < b → A X rm a < A X rm b
12 11 3expb ⊢ A ∈ ℤ ≥ 2 ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → a < b → A X rm a < A X rm b
13 1 2 3 4 9 12 leord1 ⊢ A ∈ ℤ ≥ 2 ∧ M ∈ ℕ 0 ∧ N ∈ ℕ 0 → M ≤ N ↔ A X rm M ≤ A X rm N
14 13 3impb ⊢ A ∈ ℤ ≥ 2 ∧ M ∈ ℕ 0 ∧ N ∈ ℕ 0 → M ≤ N ↔ A X rm M ≤ A X rm N