Step |
Hyp |
Ref |
Expression |
1 |
|
zcn |
⊢ ( 𝐴 ∈ ℤ → 𝐴 ∈ ℂ ) |
2 |
|
sgmval |
⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℕ ) → ( 𝐴 σ 𝐵 ) = Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑𝑐 𝐴 ) ) |
3 |
1 2
|
sylan |
⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) → ( 𝐴 σ 𝐵 ) = Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑𝑐 𝐴 ) ) |
4 |
|
ssrab2 |
⊢ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ⊆ ℕ |
5 |
|
simpr |
⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) |
6 |
4 5
|
sselid |
⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → 𝑘 ∈ ℕ ) |
7 |
6
|
nncnd |
⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → 𝑘 ∈ ℂ ) |
8 |
6
|
nnne0d |
⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → 𝑘 ≠ 0 ) |
9 |
|
simpll |
⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → 𝐴 ∈ ℤ ) |
10 |
7 8 9
|
cxpexpzd |
⊢ ( ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) ∧ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ) → ( 𝑘 ↑𝑐 𝐴 ) = ( 𝑘 ↑ 𝐴 ) ) |
11 |
10
|
sumeq2dv |
⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) → Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑𝑐 𝐴 ) = Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ) |
12 |
3 11
|
eqtrd |
⊢ ( ( 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ ) → ( 𝐴 σ 𝐵 ) = Σ 𝑘 ∈ { 𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐵 } ( 𝑘 ↑ 𝐴 ) ) |