Metamath Proof Explorer


Theorem sumz

Description: Any sum of zero over a summable set is zero. (Contributed by Mario Carneiro, 12-Aug-2013) (Revised by Mario Carneiro, 20-Apr-2014)

Ref Expression
Assertion sumz ⊢ A ⊆ ℤ ≥ M ∨ A ∈ Fin → ∑ k ∈ A 0 = 0

Proof

Step Hyp Ref Expression
1 eqid ⊢ ℤ ≥ M = ℤ ≥ M
2 simpr ⊢ A ⊆ ℤ ≥ M ∧ M ∈ ℤ → M ∈ ℤ
3 simpl ⊢ A ⊆ ℤ ≥ M ∧ M ∈ ℤ → A ⊆ ℤ ≥ M
4 c0ex ⊢ 0 ∈ V
5 4 fvconst2 ⊢ k ∈ ℤ ≥ M → ℤ ≥ M × 0 ⁡ k = 0
6 ifid ⊢ if k ∈ A 0 0 = 0
7 5 6 eqtr4di ⊢ k ∈ ℤ ≥ M → ℤ ≥ M × 0 ⁡ k = if k ∈ A 0 0
8 7 adantl ⊢ A ⊆ ℤ ≥ M ∧ M ∈ ℤ ∧ k ∈ ℤ ≥ M → ℤ ≥ M × 0 ⁡ k = if k ∈ A 0 0
9 0cnd ⊢ A ⊆ ℤ ≥ M ∧ M ∈ ℤ ∧ k ∈ A → 0 ∈ ℂ
10 1 2 3 8 9 zsum ⊢ A ⊆ ℤ ≥ M ∧ M ∈ ℤ → ∑ k ∈ A 0 = ⇝ ⁡ seq M + ℤ ≥ M × 0
11 fclim ⊢ ⇝ : dom ⁡ ⇝ ⟶ ℂ
12 ffun ⊢ ⇝ : dom ⁡ ⇝ ⟶ ℂ → Fun ⁡ ⇝
13 11 12 ax-mp ⊢ Fun ⁡ ⇝
14 serclim0 ⊢ M ∈ ℤ → seq M + ℤ ≥ M × 0 ⇝ 0
15 14 adantl ⊢ A ⊆ ℤ ≥ M ∧ M ∈ ℤ → seq M + ℤ ≥ M × 0 ⇝ 0
16 funbrfv ⊢ Fun ⁡ ⇝ → seq M + ℤ ≥ M × 0 ⇝ 0 → ⇝ ⁡ seq M + ℤ ≥ M × 0 = 0
17 13 15 16 mpsyl ⊢ A ⊆ ℤ ≥ M ∧ M ∈ ℤ → ⇝ ⁡ seq M + ℤ ≥ M × 0 = 0
18 10 17 eqtrd ⊢ A ⊆ ℤ ≥ M ∧ M ∈ ℤ → ∑ k ∈ A 0 = 0
19 uzf ⊢ ℤ ≥ : ℤ ⟶ 𝒫 ℤ
20 19 fdmi ⊢ dom ⁡ ℤ ≥ = ℤ
21 20 eleq2i ⊢ M ∈ dom ⁡ ℤ ≥ ↔ M ∈ ℤ
22 ndmfv ⊢ ¬ M ∈ dom ⁡ ℤ ≥ → ℤ ≥ M = ∅
23 21 22 sylnbir ⊢ ¬ M ∈ ℤ → ℤ ≥ M = ∅
24 23 sseq2d ⊢ ¬ M ∈ ℤ → A ⊆ ℤ ≥ M ↔ A ⊆ ∅
25 24 biimpac ⊢ A ⊆ ℤ ≥ M ∧ ¬ M ∈ ℤ → A ⊆ ∅
26 ss0 ⊢ A ⊆ ∅ → A = ∅
27 sumeq1 ⊢ A = ∅ → ∑ k ∈ A 0 = ∑ k ∈ ∅ 0
28 sum0 ⊢ ∑ k ∈ ∅ 0 = 0
29 27 28 eqtrdi ⊢ A = ∅ → ∑ k ∈ A 0 = 0
30 25 26 29 3syl ⊢ A ⊆ ℤ ≥ M ∧ ¬ M ∈ ℤ → ∑ k ∈ A 0 = 0
31 18 30 pm2.61dan ⊢ A ⊆ ℤ ≥ M → ∑ k ∈ A 0 = 0
32 fz1f1o ⊢ A ∈ Fin → A = ∅ ∨ A ∈ ℕ ∧ ∃ f f : 1 … A ⟶ 1-1 onto A
33 eqidd ⊢ k = f ⁡ n → 0 = 0
34 simpl ⊢ A ∈ ℕ ∧ f : 1 … A ⟶ 1-1 onto A → A ∈ ℕ
35 simpr ⊢ A ∈ ℕ ∧ f : 1 … A ⟶ 1-1 onto A → f : 1 … A ⟶ 1-1 onto A
36 0cnd ⊢ A ∈ ℕ ∧ f : 1 … A ⟶ 1-1 onto A ∧ k ∈ A → 0 ∈ ℂ
37 elfznn ⊢ n ∈ 1 … A → n ∈ ℕ
38 4 fvconst2 ⊢ n ∈ ℕ → ℕ × 0 ⁡ n = 0
39 37 38 syl ⊢ n ∈ 1 … A → ℕ × 0 ⁡ n = 0
40 39 adantl ⊢ A ∈ ℕ ∧ f : 1 … A ⟶ 1-1 onto A ∧ n ∈ 1 … A → ℕ × 0 ⁡ n = 0
41 33 34 35 36 40 fsum ⊢ A ∈ ℕ ∧ f : 1 … A ⟶ 1-1 onto A → ∑ k ∈ A 0 = seq 1 + ℕ × 0 ⁡ A
42 nnuz ⊢ ℕ = ℤ ≥ 1
43 42 ser0 ⊢ A ∈ ℕ → seq 1 + ℕ × 0 ⁡ A = 0
44 43 adantr ⊢ A ∈ ℕ ∧ f : 1 … A ⟶ 1-1 onto A → seq 1 + ℕ × 0 ⁡ A = 0
45 41 44 eqtrd ⊢ A ∈ ℕ ∧ f : 1 … A ⟶ 1-1 onto A → ∑ k ∈ A 0 = 0
46 45 ex ⊢ A ∈ ℕ → f : 1 … A ⟶ 1-1 onto A → ∑ k ∈ A 0 = 0
47 46 exlimdv ⊢ A ∈ ℕ → ∃ f f : 1 … A ⟶ 1-1 onto A → ∑ k ∈ A 0 = 0
48 47 imp ⊢ A ∈ ℕ ∧ ∃ f f : 1 … A ⟶ 1-1 onto A → ∑ k ∈ A 0 = 0
49 29 48 jaoi ⊢ A = ∅ ∨ A ∈ ℕ ∧ ∃ f f : 1 … A ⟶ 1-1 onto A → ∑ k ∈ A 0 = 0
50 32 49 syl ⊢ A ∈ Fin → ∑ k ∈ A 0 = 0
51 31 50 jaoi ⊢ A ⊆ ℤ ≥ M ∨ A ∈ Fin → ∑ k ∈ A 0 = 0