Metamath Proof Explorer


Theorem fsum1p

Description: Separate out the first term in a finite sum. (Contributed by NM, 3-Jan-2006) (Revised by Mario Carneiro, 23-Apr-2014)

Ref Expression
Hypotheses fsumm1.1 ⊢ φ → N ∈ ℤ ≥ M
fsumm1.2 ⊢ φ ∧ k ∈ M … N → A ∈ ℂ
fsum1p.3 ⊢ k = M → A = B
Assertion fsum1p ⊢ φ → ∑ k = M N A = B + ∑ k = M + 1 N A

Proof

Step Hyp Ref Expression
1 fsumm1.1 ⊢ φ → N ∈ ℤ ≥ M
2 fsumm1.2 ⊢ φ ∧ k ∈ M … N → A ∈ ℂ
3 fsum1p.3 ⊢ k = M → A = B
4 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
5 1 4 syl ⊢ φ → M ∈ ℤ
6 fzsn ⊢ M ∈ ℤ → M … M = M
7 5 6 syl ⊢ φ → M … M = M
8 7 ineq1d ⊢ φ → M … M ∩ M + 1 … N = M ∩ M + 1 … N
9 5 zred ⊢ φ → M ∈ ℝ
10 9 ltp1d ⊢ φ → M < M + 1
11 fzdisj ⊢ M < M + 1 → M … M ∩ M + 1 … N = ∅
12 10 11 syl ⊢ φ → M … M ∩ M + 1 … N = ∅
13 8 12 eqtr3d ⊢ φ → M ∩ M + 1 … N = ∅
14 eluzfz1 ⊢ N ∈ ℤ ≥ M → M ∈ M … N
15 1 14 syl ⊢ φ → M ∈ M … N
16 fzsplit ⊢ M ∈ M … N → M … N = M … M ∪ M + 1 … N
17 15 16 syl ⊢ φ → M … N = M … M ∪ M + 1 … N
18 7 uneq1d ⊢ φ → M … M ∪ M + 1 … N = M ∪ M + 1 … N
19 17 18 eqtrd ⊢ φ → M … N = M ∪ M + 1 … N
20 fzfid ⊢ φ → M … N ∈ Fin
21 13 19 20 2 fsumsplit ⊢ φ → ∑ k = M N A = ∑ k ∈ M A + ∑ k = M + 1 N A
22 3 eleq1d ⊢ k = M → A ∈ ℂ ↔ B ∈ ℂ
23 2 ralrimiva ⊢ φ → ∀ k ∈ M … N A ∈ ℂ
24 22 23 15 rspcdva ⊢ φ → B ∈ ℂ
25 3 sumsn ⊢ M ∈ ℤ ∧ B ∈ ℂ → ∑ k ∈ M A = B
26 5 24 25 syl2anc ⊢ φ → ∑ k ∈ M A = B
27 26 oveq1d ⊢ φ → ∑ k ∈ M A + ∑ k = M + 1 N A = B + ∑ k = M + 1 N A
28 21 27 eqtrd ⊢ φ → ∑ k = M N A = B + ∑ k = M + 1 N A