Metamath Proof Explorer


Theorem gsumsnd

Description: Group sum of a singleton, deduction form. (Contributed by Thierry Arnoux, 30-Jan-2017) (Proof shortened by AV, 11-Dec-2019)

Ref Expression
Hypotheses gsumsnd.b 𝐵 = ( Base ‘ 𝐺 )
gsumsnd.g ( 𝜑𝐺 ∈ Mnd )
gsumsnd.m ( 𝜑𝑀𝑉 )
gsumsnd.c ( 𝜑𝐶𝐵 )
gsumsnd.s ( ( 𝜑𝑘 = 𝑀 ) → 𝐴 = 𝐶 )
Assertion gsumsnd ( 𝜑 → ( 𝐺 Σg ( 𝑘 ∈ { 𝑀 } ↦ 𝐴 ) ) = 𝐶 )

Proof

Step Hyp Ref Expression
1 gsumsnd.b 𝐵 = ( Base ‘ 𝐺 )
2 gsumsnd.g ( 𝜑𝐺 ∈ Mnd )
3 gsumsnd.m ( 𝜑𝑀𝑉 )
4 gsumsnd.c ( 𝜑𝐶𝐵 )
5 gsumsnd.s ( ( 𝜑𝑘 = 𝑀 ) → 𝐴 = 𝐶 )
6 nfv 𝑘 𝜑
7 nfcv 𝑘 𝐶
8 1 2 3 4 5 6 7 gsumsnfd ( 𝜑 → ( 𝐺 Σg ( 𝑘 ∈ { 𝑀 } ↦ 𝐴 ) ) = 𝐶 )