Metamath Proof Explorer


Theorem fsumsplitsndif

Description: Separate out a term in a finite sum by splitting the sum into two parts. (Contributed by Alexander van der Vekens, 31-Aug-2018)

Ref Expression
Assertion fsumsplitsndif ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ∑ k ∈ A B = ∑ k ∈ A ∖ X B + ⦋ X / k⦌ B

Proof

Step Hyp Ref Expression
1 neldifsnd ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ¬ X ∈ A ∖ X
2 disjsn ⊢ A ∖ X ∩ X = ∅ ↔ ¬ X ∈ A ∖ X
3 1 2 sylibr ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → A ∖ X ∩ X = ∅
4 uncom ⊢ A ∖ X ∪ X = X ∪ A ∖ X
5 simp2 ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → X ∈ A
6 5 snssd ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → X ⊆ A
7 undif ⊢ X ⊆ A ↔ X ∪ A ∖ X = A
8 6 7 sylib ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → X ∪ A ∖ X = A
9 4 8 eqtr2id ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → A = A ∖ X ∪ X
10 simp1 ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → A ∈ Fin
11 rspcsbela ⊢ x ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ⦋ x / k⦌ B ∈ ℤ
12 11 zcnd ⊢ x ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ⦋ x / k⦌ B ∈ ℂ
13 12 expcom ⊢ ∀ k ∈ A B ∈ ℤ → x ∈ A → ⦋ x / k⦌ B ∈ ℂ
14 13 3ad2ant3 ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → x ∈ A → ⦋ x / k⦌ B ∈ ℂ
15 14 imp ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ ∧ x ∈ A → ⦋ x / k⦌ B ∈ ℂ
16 3 9 10 15 fsumsplit ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ∑ x ∈ A ⦋ x / k⦌ B = ∑ x ∈ A ∖ X ⦋ x / k⦌ B + ∑ x ∈ X ⦋ x / k⦌ B
17 csbeq1a ⊢ k = x → B = ⦋ x / k⦌ B
18 nfcv ⊢ Ⅎ _ x B
19 nfcsb1v ⊢ Ⅎ _ k ⦋ x / k⦌ B
20 17 18 19 cbvsum ⊢ ∑ k ∈ A B = ∑ x ∈ A ⦋ x / k⦌ B
21 17 18 19 cbvsum ⊢ ∑ k ∈ A ∖ X B = ∑ x ∈ A ∖ X ⦋ x / k⦌ B
22 17 18 19 cbvsum ⊢ ∑ k ∈ X B = ∑ x ∈ X ⦋ x / k⦌ B
23 21 22 oveq12i ⊢ ∑ k ∈ A ∖ X B + ∑ k ∈ X B = ∑ x ∈ A ∖ X ⦋ x / k⦌ B + ∑ x ∈ X ⦋ x / k⦌ B
24 16 20 23 3eqtr4g ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ∑ k ∈ A B = ∑ k ∈ A ∖ X B + ∑ k ∈ X B
25 rspcsbela ⊢ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ⦋ X / k⦌ B ∈ ℤ
26 25 zcnd ⊢ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ⦋ X / k⦌ B ∈ ℂ
27 26 3adant1 ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ⦋ X / k⦌ B ∈ ℂ
28 sumsns ⊢ X ∈ A ∧ ⦋ X / k⦌ B ∈ ℂ → ∑ k ∈ X B = ⦋ X / k⦌ B
29 5 27 28 syl2anc ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ∑ k ∈ X B = ⦋ X / k⦌ B
30 29 oveq2d ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ∑ k ∈ A ∖ X B + ∑ k ∈ X B = ∑ k ∈ A ∖ X B + ⦋ X / k⦌ B
31 24 30 eqtrd ⊢ A ∈ Fin ∧ X ∈ A ∧ ∀ k ∈ A B ∈ ℤ → ∑ k ∈ A B = ∑ k ∈ A ∖ X B + ⦋ X / k⦌ B