Metamath Proof Explorer


Theorem gsummptif1n0

Description: If only one summand in a finite group sum is not zero, the whole sum equals this summand. (Contributed by AV, 17-Feb-2019) (Proof shortened by AV, 11-Oct-2019)

Ref Expression
Hypotheses gsummpt1n0.0 ⊢ 0 ˙ = 0 G
gsummpt1n0.g ⊢ φ → G ∈ Mnd
gsummpt1n0.i ⊢ φ → I ∈ W
gsummpt1n0.x ⊢ φ → X ∈ I
gsummpt1n0.f ⊢ F = n ∈ I ⟼ if n = X A 0 ˙
gsummptif1n0.a ⊢ φ → A ∈ Base G
Assertion gsummptif1n0 ⊢ φ → ∑ G F = A

Proof

Step Hyp Ref Expression
1 gsummpt1n0.0 ⊢ 0 ˙ = 0 G
2 gsummpt1n0.g ⊢ φ → G ∈ Mnd
3 gsummpt1n0.i ⊢ φ → I ∈ W
4 gsummpt1n0.x ⊢ φ → X ∈ I
5 gsummpt1n0.f ⊢ F = n ∈ I ⟼ if n = X A 0 ˙
6 gsummptif1n0.a ⊢ φ → A ∈ Base G
7 6 ralrimivw ⊢ φ → ∀ n ∈ I A ∈ Base G
8 1 2 3 4 5 7 gsummpt1n0 ⊢ φ → ∑ G F = ⦋ X / n⦌ A
9 csbconstg ⊢ X ∈ I → ⦋ X / n⦌ A = A
10 4 9 syl ⊢ φ → ⦋ X / n⦌ A = A
11 8 10 eqtrd ⊢ φ → ∑ G F = A