Metamath Proof Explorer


Theorem isumnn0nn

Description: Sum from 0 to infinity in terms of sum from 1 to infinity. (Contributed by NM, 2-Jan-2006) (Revised by Mario Carneiro, 24-Apr-2014)

Ref Expression
Hypotheses isumnn0nn.1 ⊢ k = 0 → A = B
isumnn0nn.2 ⊢ φ ∧ k ∈ ℕ 0 → F ⁡ k = A
isumnn0nn.3 ⊢ φ ∧ k ∈ ℕ 0 → A ∈ ℂ
isumnn0nn.4 ⊢ φ → seq 0 + F ∈ dom ⁡ ⇝
Assertion isumnn0nn ⊢ φ → ∑ k ∈ ℕ 0 A = B + ∑ k ∈ ℕ A

Proof

Step Hyp Ref Expression
1 isumnn0nn.1 ⊢ k = 0 → A = B
2 isumnn0nn.2 ⊢ φ ∧ k ∈ ℕ 0 → F ⁡ k = A
3 isumnn0nn.3 ⊢ φ ∧ k ∈ ℕ 0 → A ∈ ℂ
4 isumnn0nn.4 ⊢ φ → seq 0 + F ∈ dom ⁡ ⇝
5 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
6 0zd ⊢ φ → 0 ∈ ℤ
7 5 6 2 3 4 isum1p ⊢ φ → ∑ k ∈ ℕ 0 A = F ⁡ 0 + ∑ k ∈ ℤ ≥ 0 + 1 A
8 fveq2 ⊢ k = 0 → F ⁡ k = F ⁡ 0
9 8 1 eqeq12d ⊢ k = 0 → F ⁡ k = A ↔ F ⁡ 0 = B
10 2 ralrimiva ⊢ φ → ∀ k ∈ ℕ 0 F ⁡ k = A
11 0nn0 ⊢ 0 ∈ ℕ 0
12 11 a1i ⊢ φ → 0 ∈ ℕ 0
13 9 10 12 rspcdva ⊢ φ → F ⁡ 0 = B
14 0p1e1 ⊢ 0 + 1 = 1
15 14 fveq2i ⊢ ℤ ≥ 0 + 1 = ℤ ≥ 1
16 nnuz ⊢ ℕ = ℤ ≥ 1
17 15 16 eqtr4i ⊢ ℤ ≥ 0 + 1 = ℕ
18 17 sumeq1i ⊢ ∑ k ∈ ℤ ≥ 0 + 1 A = ∑ k ∈ ℕ A
19 18 a1i ⊢ φ → ∑ k ∈ ℤ ≥ 0 + 1 A = ∑ k ∈ ℕ A
20 13 19 oveq12d ⊢ φ → F ⁡ 0 + ∑ k ∈ ℤ ≥ 0 + 1 A = B + ∑ k ∈ ℕ A
21 7 20 eqtrd ⊢ φ → ∑ k ∈ ℕ 0 A = B + ∑ k ∈ ℕ A