Metamath Proof Explorer


Theorem rmxyval

Description: Main definition of the X and Y sequences. Compare definition 2.3 of JonesMatijasevic p. 694. (Contributed by Stefan O'Rear, 19-Oct-2014)

Ref Expression
Assertion rmxyval ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A X rm N + A 2 − 1 ⁢ A Y rm N = A + A 2 − 1 N

Proof

Step Hyp Ref Expression
1 rmxfval ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A X rm N = 1 st ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
2 rmyfval ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A Y rm N = 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
3 2 oveq2d ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A 2 − 1 ⁢ A Y rm N = A 2 − 1 ⁢ 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
4 1 3 oveq12d ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A X rm N + A 2 − 1 ⁢ A Y rm N = 1 st ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N + A 2 − 1 ⁢ 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
5 rmxyelxp ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N ∈ ℕ 0 × ℤ
6 fveq2 ⊢ a = b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N → 1 st ⁡ a = 1 st ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
7 fveq2 ⊢ a = b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N → 2 nd ⁡ a = 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
8 7 oveq2d ⊢ a = b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N → A 2 − 1 ⁢ 2 nd ⁡ a = A 2 − 1 ⁢ 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
9 6 8 oveq12d ⊢ a = b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N → 1 st ⁡ a + A 2 − 1 ⁢ 2 nd ⁡ a = 1 st ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N + A 2 − 1 ⁢ 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
10 fveq2 ⊢ b = a → 1 st ⁡ b = 1 st ⁡ a
11 fveq2 ⊢ b = a → 2 nd ⁡ b = 2 nd ⁡ a
12 11 oveq2d ⊢ b = a → A 2 − 1 ⁢ 2 nd ⁡ b = A 2 − 1 ⁢ 2 nd ⁡ a
13 10 12 oveq12d ⊢ b = a → 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b = 1 st ⁡ a + A 2 − 1 ⁢ 2 nd ⁡ a
14 13 cbvmptv ⊢ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b = a ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ a + A 2 − 1 ⁢ 2 nd ⁡ a
15 ovex ⊢ 1 st ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N + A 2 − 1 ⁢ 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N ∈ V
16 9 14 15 fvmpt ⊢ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N ∈ ℕ 0 × ℤ → b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N = 1 st ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N + A 2 − 1 ⁢ 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
17 5 16 syl ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N = 1 st ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N + A 2 − 1 ⁢ 2 nd ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N
18 rmxypairf1o ⊢ A ∈ ℤ ≥ 2 → b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b : ℕ 0 × ℤ ⟶ 1-1 onto a | ∃ c ∈ ℕ 0 ∃ d ∈ ℤ a = c + A 2 − 1 ⁢ d
19 18 adantr ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b : ℕ 0 × ℤ ⟶ 1-1 onto a | ∃ c ∈ ℕ 0 ∃ d ∈ ℤ a = c + A 2 − 1 ⁢ d
20 rmxyelqirr ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A + A 2 − 1 N ∈ a | ∃ c ∈ ℕ 0 ∃ d ∈ ℤ a = c + A 2 − 1 ⁢ d
21 f1ocnvfv2 ⊢ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b : ℕ 0 × ℤ ⟶ 1-1 onto a | ∃ c ∈ ℕ 0 ∃ d ∈ ℤ a = c + A 2 − 1 ⁢ d ∧ A + A 2 − 1 N ∈ a | ∃ c ∈ ℕ 0 ∃ d ∈ ℤ a = c + A 2 − 1 ⁢ d → b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N = A + A 2 − 1 N
22 19 20 21 syl2anc ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b ⁡ b ∈ ℕ 0 × ℤ ⟼ 1 st ⁡ b + A 2 − 1 ⁢ 2 nd ⁡ b -1 ⁡ A + A 2 − 1 N = A + A 2 − 1 N
23 4 17 22 3eqtr2d ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A X rm N + A 2 − 1 ⁢ A Y rm N = A + A 2 − 1 N