Metamath Proof Explorer


Theorem gsumdifsndf

Description: Extract a summand from a finitely supported group sum. (Contributed by AV, 4-Sep-2019)

Ref Expression
Hypotheses gsumdifsndf.k ⊢ Ⅎ _ k Y
gsumdifsndf.n ⊢ Ⅎ k φ
gsumdifsndf.b ⊢ B = Base G
gsumdifsndf.p ⊢ + ˙ = + G
gsumdifsndf.g ⊢ φ → G ∈ CMnd
gsumdifsndf.a ⊢ φ → A ∈ W
gsumdifsndf.f ⊢ φ → finSupp 0 G⁡ k ∈ A ⟼ X
gsumdifsndf.e ⊢ φ ∧ k ∈ A → X ∈ B
gsumdifsndf.m ⊢ φ → M ∈ A
gsumdifsndf.y ⊢ φ → Y ∈ B
gsumdifsndf.s ⊢ φ ∧ k = M → X = Y
Assertion gsumdifsndf ⊢ φ → ∑ G k ∈ A X = ∑ G k ∈ A ∖ M X + ˙ Y

Proof

Step Hyp Ref Expression
1 gsumdifsndf.k ⊢ Ⅎ _ k Y
2 gsumdifsndf.n ⊢ Ⅎ k φ
3 gsumdifsndf.b ⊢ B = Base G
4 gsumdifsndf.p ⊢ + ˙ = + G
5 gsumdifsndf.g ⊢ φ → G ∈ CMnd
6 gsumdifsndf.a ⊢ φ → A ∈ W
7 gsumdifsndf.f ⊢ φ → finSupp 0 G⁡ k ∈ A ⟼ X
8 gsumdifsndf.e ⊢ φ ∧ k ∈ A → X ∈ B
9 gsumdifsndf.m ⊢ φ → M ∈ A
10 gsumdifsndf.y ⊢ φ → Y ∈ B
11 gsumdifsndf.s ⊢ φ ∧ k = M → X = Y
12 eqid ⊢ 0 G = 0 G
13 9 snssd ⊢ φ → M ⊆ A
14 difin2 ⊢ M ⊆ A → M ∖ M = A ∖ M ∩ M
15 13 14 syl ⊢ φ → M ∖ M = A ∖ M ∩ M
16 difid ⊢ M ∖ M = ∅
17 15 16 eqtr3di ⊢ φ → A ∖ M ∩ M = ∅
18 difsnid ⊢ M ∈ A → A ∖ M ∪ M = A
19 9 18 syl ⊢ φ → A ∖ M ∪ M = A
20 19 eqcomd ⊢ φ → A = A ∖ M ∪ M
21 2 3 12 4 5 6 8 7 17 20 gsumsplit2f ⊢ φ → ∑ G k ∈ A X = ∑ G k ∈ A ∖ M X + ˙ ∑ G k ∈ M X
22 cmnmnd ⊢ G ∈ CMnd → G ∈ Mnd
23 5 22 syl ⊢ φ → G ∈ Mnd
24 3 23 9 10 11 2 1 gsumsnfd ⊢ φ → ∑ G k ∈ M X = Y
25 24 oveq2d ⊢ φ → ∑ G k ∈ A ∖ M X + ˙ ∑ G k ∈ M X = ∑ G k ∈ A ∖ M X + ˙ Y
26 21 25 eqtrd ⊢ φ → ∑ G k ∈ A X = ∑ G k ∈ A ∖ M X + ˙ Y