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 = ( 0g ‘ 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 Σg 𝐹 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 gsummpt1n0.0 ⊢ 0 = ( 0g ‘ 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 Σg 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 Σg 𝐹 ) = 𝐴 )