Metamath Proof Explorer


Theorem fsumrev2

Description: Reversal of a finite sum. (Contributed by NM, 27-Nov-2005) (Revised by Mario Carneiro, 13-Apr-2016)

Ref Expression
Hypotheses fsumrev2.1 ⊢ φ ∧ j ∈ M … N → A ∈ ℂ
fsumrev2.2 ⊢ j = M + N - k → A = B
Assertion fsumrev2 ⊢ φ → ∑ j = M N A = ∑ k = M N B

Proof

Step Hyp Ref Expression
1 fsumrev2.1 ⊢ φ ∧ j ∈ M … N → A ∈ ℂ
2 fsumrev2.2 ⊢ j = M + N - k → A = B
3 sum0 ⊢ ∑ j ∈ ∅ A = 0
4 sum0 ⊢ ∑ k ∈ ∅ B = 0
5 3 4 eqtr4i ⊢ ∑ j ∈ ∅ A = ∑ k ∈ ∅ B
6 sumeq1 ⊢ M … N = ∅ → ∑ j = M N A = ∑ j ∈ ∅ A
7 sumeq1 ⊢ M … N = ∅ → ∑ k = M N B = ∑ k ∈ ∅ B
8 5 6 7 3eqtr4a ⊢ M … N = ∅ → ∑ j = M N A = ∑ k = M N B
9 8 adantl ⊢ φ ∧ M … N = ∅ → ∑ j = M N A = ∑ k = M N B
10 fzn0 ⊢ M … N ≠ ∅ ↔ N ∈ ℤ ≥ M
11 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
12 11 adantl ⊢ φ ∧ N ∈ ℤ ≥ M → M ∈ ℤ
13 eluzelz ⊢ N ∈ ℤ ≥ M → N ∈ ℤ
14 13 adantl ⊢ φ ∧ N ∈ ℤ ≥ M → N ∈ ℤ
15 12 14 zaddcld ⊢ φ ∧ N ∈ ℤ ≥ M → M + N ∈ ℤ
16 1 adantlr ⊢ φ ∧ N ∈ ℤ ≥ M ∧ j ∈ M … N → A ∈ ℂ
17 15 12 14 16 2 fsumrev ⊢ φ ∧ N ∈ ℤ ≥ M → ∑ j = M N A = ∑ k = M + N - N M + N - M B
18 12 zcnd ⊢ φ ∧ N ∈ ℤ ≥ M → M ∈ ℂ
19 14 zcnd ⊢ φ ∧ N ∈ ℤ ≥ M → N ∈ ℂ
20 18 19 pncand ⊢ φ ∧ N ∈ ℤ ≥ M → M + N - N = M
21 18 19 pncan2d ⊢ φ ∧ N ∈ ℤ ≥ M → M + N - M = N
22 20 21 oveq12d ⊢ φ ∧ N ∈ ℤ ≥ M → M + N - N … M + N - M = M … N
23 22 sumeq1d ⊢ φ ∧ N ∈ ℤ ≥ M → ∑ k = M + N - N M + N - M B = ∑ k = M N B
24 17 23 eqtrd ⊢ φ ∧ N ∈ ℤ ≥ M → ∑ j = M N A = ∑ k = M N B
25 10 24 sylan2b ⊢ φ ∧ M … N ≠ ∅ → ∑ j = M N A = ∑ k = M N B
26 9 25 pm2.61dane ⊢ φ → ∑ j = M N A = ∑ k = M N B