Metamath Proof Explorer


Theorem fsumsers

Description: Special case of series sum over a finite upper integer index set. (Contributed by Mario Carneiro, 26-Jul-2013) (Revised by Mario Carneiro, 21-Apr-2014)

Ref Expression
Hypotheses fsumsers.1 ⊢ φ ∧ k ∈ ℤ ≥ M → F ⁡ k = if k ∈ A B 0
fsumsers.2 ⊢ φ → N ∈ ℤ ≥ M
fsumsers.3 ⊢ φ ∧ k ∈ A → B ∈ ℂ
fsumsers.4 ⊢ φ → A ⊆ M … N
Assertion fsumsers ⊢ φ → ∑ k ∈ A B = seq M + F ⁡ N

Proof

Step Hyp Ref Expression
1 fsumsers.1 ⊢ φ ∧ k ∈ ℤ ≥ M → F ⁡ k = if k ∈ A B 0
2 fsumsers.2 ⊢ φ → N ∈ ℤ ≥ M
3 fsumsers.3 ⊢ φ ∧ k ∈ A → B ∈ ℂ
4 fsumsers.4 ⊢ φ → A ⊆ M … N
5 eqid ⊢ ℤ ≥ M = ℤ ≥ M
6 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
7 2 6 syl ⊢ φ → M ∈ ℤ
8 fzssuz ⊢ M … N ⊆ ℤ ≥ M
9 4 8 sstrdi ⊢ φ → A ⊆ ℤ ≥ M
10 5 7 9 1 3 zsum ⊢ φ → ∑ k ∈ A B = ⇝ ⁡ seq M + F
11 fclim ⊢ ⇝ : dom ⁡ ⇝ ⟶ ℂ
12 ffun ⊢ ⇝ : dom ⁡ ⇝ ⟶ ℂ → Fun ⁡ ⇝
13 11 12 ax-mp ⊢ Fun ⁡ ⇝
14 1 2 3 4 fsumcvg2 ⊢ φ → seq M + F ⇝ seq M + F ⁡ N
15 funbrfv ⊢ Fun ⁡ ⇝ → seq M + F ⇝ seq M + F ⁡ N → ⇝ ⁡ seq M + F = seq M + F ⁡ N
16 13 14 15 mpsyl ⊢ φ → ⇝ ⁡ seq M + F = seq M + F ⁡ N
17 10 16 eqtrd ⊢ φ → ∑ k ∈ A B = seq M + F ⁡ N