Metamath Proof Explorer


Theorem gsummptres

Description: Extend a finite group sum by padding outside with zeroes. Proof generated using OpenAI's proof assistant. (Contributed by Thierry Arnoux, 11-Jul-2020)

Ref Expression
Hypotheses gsummptres.0 ⊢ B = Base G
gsummptres.1 ⊢ 0 ˙ = 0 G
gsummptres.2 ⊢ φ → G ∈ CMnd
gsummptres.3 ⊢ φ → A ∈ Fin
gsummptres.4 ⊢ φ ∧ x ∈ A → C ∈ B
gsummptres.5 ⊢ φ ∧ x ∈ A ∖ D → C = 0 ˙
Assertion gsummptres ⊢ φ → ∑ G x ∈ A C = ∑ G x ∈ A ∩ D C

Proof

Step Hyp Ref Expression
1 gsummptres.0 ⊢ B = Base G
2 gsummptres.1 ⊢ 0 ˙ = 0 G
3 gsummptres.2 ⊢ φ → G ∈ CMnd
4 gsummptres.3 ⊢ φ → A ∈ Fin
5 gsummptres.4 ⊢ φ ∧ x ∈ A → C ∈ B
6 gsummptres.5 ⊢ φ ∧ x ∈ A ∖ D → C = 0 ˙
7 eqid ⊢ + G = + G
8 eqid ⊢ x ∈ A ⟼ C = x ∈ A ⟼ C
9 2 fvexi ⊢ 0 ˙ ∈ V
10 9 a1i ⊢ φ → 0 ˙ ∈ V
11 8 4 5 10 fsuppmptdm ⊢ φ → finSupp 0 ˙⁡ x ∈ A ⟼ C
12 inindif ⊢ A ∩ D ∩ A ∖ D = ∅
13 12 a1i ⊢ φ → A ∩ D ∩ A ∖ D = ∅
14 inundif ⊢ A ∩ D ∪ A ∖ D = A
15 14 eqcomi ⊢ A = A ∩ D ∪ A ∖ D
16 15 a1i ⊢ φ → A = A ∩ D ∪ A ∖ D
17 1 2 7 3 4 5 11 13 16 gsumsplit2 ⊢ φ → ∑ G x ∈ A C = ∑ G x ∈ A ∩ D C + G ∑ G x ∈ A ∖ D C
18 6 mpteq2dva ⊢ φ → x ∈ A ∖ D ⟼ C = x ∈ A ∖ D ⟼ 0 ˙
19 18 oveq2d ⊢ φ → ∑ G x ∈ A ∖ D C = ∑ G x ∈ A ∖ D 0 ˙
20 cmnmnd ⊢ G ∈ CMnd → G ∈ Mnd
21 3 20 syl ⊢ φ → G ∈ Mnd
22 diffi ⊢ A ∈ Fin → A ∖ D ∈ Fin
23 4 22 syl ⊢ φ → A ∖ D ∈ Fin
24 2 gsumz ⊢ G ∈ Mnd ∧ A ∖ D ∈ Fin → ∑ G x ∈ A ∖ D 0 ˙ = 0 ˙
25 21 23 24 syl2anc ⊢ φ → ∑ G x ∈ A ∖ D 0 ˙ = 0 ˙
26 19 25 eqtrd ⊢ φ → ∑ G x ∈ A ∖ D C = 0 ˙
27 26 oveq2d ⊢ φ → ∑ G x ∈ A ∩ D C + G ∑ G x ∈ A ∖ D C = ∑ G x ∈ A ∩ D C + G 0 ˙
28 infi ⊢ A ∈ Fin → A ∩ D ∈ Fin
29 4 28 syl ⊢ φ → A ∩ D ∈ Fin
30 inss1 ⊢ A ∩ D ⊆ A
31 30 sseli ⊢ x ∈ A ∩ D → x ∈ A
32 31 5 sylan2 ⊢ φ ∧ x ∈ A ∩ D → C ∈ B
33 32 ralrimiva ⊢ φ → ∀ x ∈ A ∩ D C ∈ B
34 1 3 29 33 gsummptcl ⊢ φ → ∑ G x ∈ A ∩ D C ∈ B
35 1 7 2 mndrid ⊢ G ∈ Mnd ∧ ∑ G x ∈ A ∩ D C ∈ B → ∑ G x ∈ A ∩ D C + G 0 ˙ = ∑ G x ∈ A ∩ D C
36 21 34 35 syl2anc ⊢ φ → ∑ G x ∈ A ∩ D C + G 0 ˙ = ∑ G x ∈ A ∩ D C
37 27 36 eqtrd ⊢ φ → ∑ G x ∈ A ∩ D C + G ∑ G x ∈ A ∖ D C = ∑ G x ∈ A ∩ D C
38 17 37 eqtrd ⊢ φ → ∑ G x ∈ A C = ∑ G x ∈ A ∩ D C