Metamath Proof Explorer
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 |
⊢ ( 𝜑 → 𝐺 ∈ Mnd ) |
|
|
gsumsnd.m |
⊢ ( 𝜑 → 𝑀 ∈ 𝑉 ) |
|
|
gsumsnd.c |
⊢ ( 𝜑 → 𝐶 ∈ 𝐵 ) |
|
|
gsumsnd.s |
⊢ ( ( 𝜑 ∧ 𝑘 = 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 |
⊢ ( 𝜑 → 𝐺 ∈ Mnd ) |
3 |
|
gsumsnd.m |
⊢ ( 𝜑 → 𝑀 ∈ 𝑉 ) |
4 |
|
gsumsnd.c |
⊢ ( 𝜑 → 𝐶 ∈ 𝐵 ) |
5 |
|
gsumsnd.s |
⊢ ( ( 𝜑 ∧ 𝑘 = 𝑀 ) → 𝐴 = 𝐶 ) |
6 |
|
nfv |
⊢ Ⅎ 𝑘 𝜑 |
7 |
|
nfcv |
⊢ Ⅎ 𝑘 𝐶 |
8 |
1 2 3 4 5 6 7
|
gsumsnfd |
⊢ ( 𝜑 → ( 𝐺 Σg ( 𝑘 ∈ { 𝑀 } ↦ 𝐴 ) ) = 𝐶 ) |