Metamath Proof Explorer


Theorem gsumsnf

Description: Group sum of a singleton, using bound-variable hypotheses instead of distinct variable conditions. (Contributed by Mario Carneiro, 19-Dec-2014) (Revised by Thierry Arnoux, 28-Mar-2018) (Proof shortened by AV, 11-Dec-2019)

Ref Expression
Hypotheses gsumsnf.c ⊢ Ⅎ 𝑘 𝐶
gsumsnf.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
gsumsnf.s ⊢ ( 𝑘 = 𝑀 → 𝐴 = 𝐶 )
Assertion gsumsnf ( ( 𝐺 ∈ Mnd ∧ 𝑀 ∈ 𝑉 ∧ 𝐶 ∈ 𝐵 ) → ( 𝐺 Σg ( 𝑘 ∈ { 𝑀 } ↦ 𝐴 ) ) = 𝐶 )

Proof

Step Hyp Ref Expression
1 gsumsnf.c ⊢ Ⅎ 𝑘 𝐶
2 gsumsnf.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
3 gsumsnf.s ⊢ ( 𝑘 = 𝑀 → 𝐴 = 𝐶 )
4 simp1 ⊢ ( ( 𝐺 ∈ Mnd ∧ 𝑀 ∈ 𝑉 ∧ 𝐶 ∈ 𝐵 ) → 𝐺 ∈ Mnd )
5 simp2 ⊢ ( ( 𝐺 ∈ Mnd ∧ 𝑀 ∈ 𝑉 ∧ 𝐶 ∈ 𝐵 ) → 𝑀 ∈ 𝑉 )
6 simp3 ⊢ ( ( 𝐺 ∈ Mnd ∧ 𝑀 ∈ 𝑉 ∧ 𝐶 ∈ 𝐵 ) → 𝐶 ∈ 𝐵 )
7 3 adantl ⊢ ( ( ( 𝐺 ∈ Mnd ∧ 𝑀 ∈ 𝑉 ∧ 𝐶 ∈ 𝐵 ) ∧ 𝑘 = 𝑀 ) → 𝐴 = 𝐶 )
8 nfv ⊢ Ⅎ 𝑘 𝐺 ∈ Mnd
9 nfv ⊢ Ⅎ 𝑘 𝑀 ∈ 𝑉
10 1 nfel1 ⊢ Ⅎ 𝑘 𝐶 ∈ 𝐵
11 8 9 10 nf3an ⊢ Ⅎ 𝑘 ( 𝐺 ∈ Mnd ∧ 𝑀 ∈ 𝑉 ∧ 𝐶 ∈ 𝐵 )
12 2 4 5 6 7 11 1 gsumsnfd ⊢ ( ( 𝐺 ∈ Mnd ∧ 𝑀 ∈ 𝑉 ∧ 𝐶 ∈ 𝐵 ) → ( 𝐺 Σg ( 𝑘 ∈ { 𝑀 } ↦ 𝐴 ) ) = 𝐶 )