Metamath Proof Explorer


Theorem fsumm1

Description: Separate out the last term in a finite sum. (Contributed by Mario Carneiro, 26-Apr-2014)

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

Proof

Step Hyp Ref Expression
1 fsumm1.1 ⊢ φ → N ∈ ℤ ≥ M
2 fsumm1.2 ⊢ φ ∧ k ∈ M … N → A ∈ ℂ
3 fsumm1.3 ⊢ k = N → A = B
4 eluzelz ⊢ N ∈ ℤ ≥ M → N ∈ ℤ
5 1 4 syl ⊢ φ → N ∈ ℤ
6 fzsn ⊢ N ∈ ℤ → N … N = N
7 5 6 syl ⊢ φ → N … N = N
8 7 ineq2d ⊢ φ → M … N − 1 ∩ N … N = M … N − 1 ∩ N
9 5 zred ⊢ φ → N ∈ ℝ
10 9 ltm1d ⊢ φ → N − 1 < N
11 fzdisj ⊢ N − 1 < N → M … N − 1 ∩ N … N = ∅
12 10 11 syl ⊢ φ → M … N − 1 ∩ N … N = ∅
13 8 12 eqtr3d ⊢ φ → M … N − 1 ∩ N = ∅
14 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
15 1 14 syl ⊢ φ → M ∈ ℤ
16 peano2zm ⊢ M ∈ ℤ → M − 1 ∈ ℤ
17 15 16 syl ⊢ φ → M − 1 ∈ ℤ
18 15 zcnd ⊢ φ → M ∈ ℂ
19 ax-1cn ⊢ 1 ∈ ℂ
20 npcan ⊢ M ∈ ℂ ∧ 1 ∈ ℂ → M - 1 + 1 = M
21 18 19 20 sylancl ⊢ φ → M - 1 + 1 = M
22 21 fveq2d ⊢ φ → ℤ ≥ M - 1 + 1 = ℤ ≥ M
23 1 22 eleqtrrd ⊢ φ → N ∈ ℤ ≥ M - 1 + 1
24 eluzp1m1 ⊢ M − 1 ∈ ℤ ∧ N ∈ ℤ ≥ M - 1 + 1 → N − 1 ∈ ℤ ≥ M − 1
25 17 23 24 syl2anc ⊢ φ → N − 1 ∈ ℤ ≥ M − 1
26 fzsuc2 ⊢ M ∈ ℤ ∧ N − 1 ∈ ℤ ≥ M − 1 → M … N - 1 + 1 = M … N − 1 ∪ N - 1 + 1
27 15 25 26 syl2anc ⊢ φ → M … N - 1 + 1 = M … N − 1 ∪ N - 1 + 1
28 5 zcnd ⊢ φ → N ∈ ℂ
29 npcan ⊢ N ∈ ℂ ∧ 1 ∈ ℂ → N - 1 + 1 = N
30 28 19 29 sylancl ⊢ φ → N - 1 + 1 = N
31 30 oveq2d ⊢ φ → M … N - 1 + 1 = M … N
32 27 31 eqtr3d ⊢ φ → M … N − 1 ∪ N - 1 + 1 = M … N
33 30 sneqd ⊢ φ → N - 1 + 1 = N
34 33 uneq2d ⊢ φ → M … N − 1 ∪ N - 1 + 1 = M … N − 1 ∪ N
35 32 34 eqtr3d ⊢ φ → M … N = M … N − 1 ∪ N
36 fzfid ⊢ φ → M … N ∈ Fin
37 13 35 36 2 fsumsplit ⊢ φ → ∑ k = M N A = ∑ k = M N − 1 A + ∑ k ∈ N A
38 3 eleq1d ⊢ k = N → A ∈ ℂ ↔ B ∈ ℂ
39 2 ralrimiva ⊢ φ → ∀ k ∈ M … N A ∈ ℂ
40 eluzfz2 ⊢ N ∈ ℤ ≥ M → N ∈ M … N
41 1 40 syl ⊢ φ → N ∈ M … N
42 38 39 41 rspcdva ⊢ φ → B ∈ ℂ
43 3 sumsn ⊢ N ∈ ℤ ≥ M ∧ B ∈ ℂ → ∑ k ∈ N A = B
44 1 42 43 syl2anc ⊢ φ → ∑ k ∈ N A = B
45 44 oveq2d ⊢ φ → ∑ k = M N − 1 A + ∑ k ∈ N A = ∑ k = M N − 1 A + B
46 37 45 eqtrd ⊢ φ → ∑ k = M N A = ∑ k = M N − 1 A + B