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