Metamath Proof Explorer


Theorem fsfnn0gsumfsffz

Description: Replacing a finitely supported function over the nonnegative integers by a function over a finite set of sequential integers in a finite group sum. (Contributed by AV, 9-Oct-2019)

Ref Expression
Hypotheses nn0gsumfz.b ⊢ B = Base G
nn0gsumfz.0 ⊢ 0 ˙ = 0 G
nn0gsumfz.g ⊢ φ → G ∈ CMnd
nn0gsumfz.f ⊢ φ → F ∈ B ℕ 0
fsfnn0gsumfsffz.s ⊢ φ → S ∈ ℕ 0
fsfnn0gsumfsffz.h ⊢ H = F ↾ 0 … S
Assertion fsfnn0gsumfsffz ⊢ φ → ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → ∑ G F = ∑ G H

Proof

Step Hyp Ref Expression
1 nn0gsumfz.b ⊢ B = Base G
2 nn0gsumfz.0 ⊢ 0 ˙ = 0 G
3 nn0gsumfz.g ⊢ φ → G ∈ CMnd
4 nn0gsumfz.f ⊢ φ → F ∈ B ℕ 0
5 fsfnn0gsumfsffz.s ⊢ φ → S ∈ ℕ 0
6 fsfnn0gsumfsffz.h ⊢ H = F ↾ 0 … S
7 6 oveq2i ⊢ ∑ G H = ∑ G F ↾ 0 … S
8 3 adantr ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → G ∈ CMnd
9 nn0ex ⊢ ℕ 0 ∈ V
10 9 a1i ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → ℕ 0 ∈ V
11 elmapi ⊢ F ∈ B ℕ 0 → F : ℕ 0 ⟶ B
12 4 11 syl ⊢ φ → F : ℕ 0 ⟶ B
13 12 adantr ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → F : ℕ 0 ⟶ B
14 2 fvexi ⊢ 0 ˙ ∈ V
15 14 a1i ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → 0 ˙ ∈ V
16 4 adantr ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → F ∈ B ℕ 0
17 5 adantr ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → S ∈ ℕ 0
18 simpr ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙
19 15 16 17 18 suppssfz ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → F supp 0 ˙ ⊆ 0 … S
20 elmapfun ⊢ F ∈ B ℕ 0 → Fun ⁡ F
21 4 20 syl ⊢ φ → Fun ⁡ F
22 14 a1i ⊢ φ → 0 ˙ ∈ V
23 4 21 22 3jca ⊢ φ → F ∈ B ℕ 0 ∧ Fun ⁡ F ∧ 0 ˙ ∈ V
24 fzfid ⊢ φ → 0 … S ∈ Fin
25 24 anim1i ⊢ φ ∧ F supp 0 ˙ ⊆ 0 … S → 0 … S ∈ Fin ∧ F supp 0 ˙ ⊆ 0 … S
26 suppssfifsupp ⊢ F ∈ B ℕ 0 ∧ Fun ⁡ F ∧ 0 ˙ ∈ V ∧ 0 … S ∈ Fin ∧ F supp 0 ˙ ⊆ 0 … S → finSupp 0 ˙⁡ F
27 23 25 26 syl2an2r ⊢ φ ∧ F supp 0 ˙ ⊆ 0 … S → finSupp 0 ˙⁡ F
28 19 27 syldan ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → finSupp 0 ˙⁡ F
29 1 2 8 10 13 19 28 gsumres ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → ∑ G F ↾ 0 … S = ∑ G F
30 7 29 eqtr2id ⊢ φ ∧ ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → ∑ G F = ∑ G H
31 30 ex ⊢ φ → ∀ x ∈ ℕ 0 S < x → F ⁡ x = 0 ˙ → ∑ G F = ∑ G H