Metamath Proof Explorer


Theorem chpp1

Description: The second Chebyshev function at a successor. (Contributed by Mario Carneiro, 11-Apr-2016)

Ref Expression
Assertion chpp1 ⊢ A ∈ ℕ 0 → ψ ⁡ A + 1 = ψ ⁡ A + Λ ⁡ A + 1

Proof

Step Hyp Ref Expression
1 nn0p1nn ⊢ A ∈ ℕ 0 → A + 1 ∈ ℕ
2 nnuz ⊢ ℕ = ℤ ≥ 1
3 1 2 eleqtrdi ⊢ A ∈ ℕ 0 → A + 1 ∈ ℤ ≥ 1
4 elfznn ⊢ n ∈ 1 … A + 1 → n ∈ ℕ
5 4 adantl ⊢ A ∈ ℕ 0 ∧ n ∈ 1 … A + 1 → n ∈ ℕ
6 vmacl ⊢ n ∈ ℕ → Λ ⁡ n ∈ ℝ
7 5 6 syl ⊢ A ∈ ℕ 0 ∧ n ∈ 1 … A + 1 → Λ ⁡ n ∈ ℝ
8 7 recnd ⊢ A ∈ ℕ 0 ∧ n ∈ 1 … A + 1 → Λ ⁡ n ∈ ℂ
9 fveq2 ⊢ n = A + 1 → Λ ⁡ n = Λ ⁡ A + 1
10 3 8 9 fsumm1 ⊢ A ∈ ℕ 0 → ∑ n = 1 A + 1 Λ ⁡ n = ∑ n = 1 A + 1 - 1 Λ ⁡ n + Λ ⁡ A + 1
11 nn0re ⊢ A ∈ ℕ 0 → A ∈ ℝ
12 peano2re ⊢ A ∈ ℝ → A + 1 ∈ ℝ
13 chpval ⊢ A + 1 ∈ ℝ → ψ ⁡ A + 1 = ∑ n = 1 A + 1 Λ ⁡ n
14 11 12 13 3syl ⊢ A ∈ ℕ 0 → ψ ⁡ A + 1 = ∑ n = 1 A + 1 Λ ⁡ n
15 nn0z ⊢ A ∈ ℕ 0 → A ∈ ℤ
16 15 peano2zd ⊢ A ∈ ℕ 0 → A + 1 ∈ ℤ
17 flid ⊢ A + 1 ∈ ℤ → A + 1 = A + 1
18 16 17 syl ⊢ A ∈ ℕ 0 → A + 1 = A + 1
19 18 oveq2d ⊢ A ∈ ℕ 0 → 1 … A + 1 = 1 … A + 1
20 19 sumeq1d ⊢ A ∈ ℕ 0 → ∑ n = 1 A + 1 Λ ⁡ n = ∑ n = 1 A + 1 Λ ⁡ n
21 14 20 eqtrd ⊢ A ∈ ℕ 0 → ψ ⁡ A + 1 = ∑ n = 1 A + 1 Λ ⁡ n
22 chpval ⊢ A ∈ ℝ → ψ ⁡ A = ∑ n = 1 A Λ ⁡ n
23 11 22 syl ⊢ A ∈ ℕ 0 → ψ ⁡ A = ∑ n = 1 A Λ ⁡ n
24 flid ⊢ A ∈ ℤ → A = A
25 15 24 syl ⊢ A ∈ ℕ 0 → A = A
26 nn0cn ⊢ A ∈ ℕ 0 → A ∈ ℂ
27 ax-1cn ⊢ 1 ∈ ℂ
28 pncan ⊢ A ∈ ℂ ∧ 1 ∈ ℂ → A + 1 - 1 = A
29 26 27 28 sylancl ⊢ A ∈ ℕ 0 → A + 1 - 1 = A
30 25 29 eqtr4d ⊢ A ∈ ℕ 0 → A = A + 1 - 1
31 30 oveq2d ⊢ A ∈ ℕ 0 → 1 … A = 1 … A + 1 - 1
32 31 sumeq1d ⊢ A ∈ ℕ 0 → ∑ n = 1 A Λ ⁡ n = ∑ n = 1 A + 1 - 1 Λ ⁡ n
33 23 32 eqtrd ⊢ A ∈ ℕ 0 → ψ ⁡ A = ∑ n = 1 A + 1 - 1 Λ ⁡ n
34 33 oveq1d ⊢ A ∈ ℕ 0 → ψ ⁡ A + Λ ⁡ A + 1 = ∑ n = 1 A + 1 - 1 Λ ⁡ n + Λ ⁡ A + 1
35 10 21 34 3eqtr4d ⊢ A ∈ ℕ 0 → ψ ⁡ A + 1 = ψ ⁡ A + Λ ⁡ A + 1