Metamath Proof Explorer
Description: The set n ZZ is an ideal in ZZ . (Contributed by Mario
Carneiro, 14-Jun-2015) (Revised by AV, 13-Jun-2019)
|
|
Ref |
Expression |
|
Hypothesis |
znval.s |
⊢ 𝑆 = ( RSpan ‘ ℤring ) |
|
Assertion |
znlidl |
⊢ ( 𝑁 ∈ ℤ → ( 𝑆 ‘ { 𝑁 } ) ∈ ( LIdeal ‘ ℤring ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
znval.s |
⊢ 𝑆 = ( RSpan ‘ ℤring ) |
2 |
|
zringring |
⊢ ℤring ∈ Ring |
3 |
|
snssi |
⊢ ( 𝑁 ∈ ℤ → { 𝑁 } ⊆ ℤ ) |
4 |
|
zringbas |
⊢ ℤ = ( Base ‘ ℤring ) |
5 |
|
eqid |
⊢ ( LIdeal ‘ ℤring ) = ( LIdeal ‘ ℤring ) |
6 |
1 4 5
|
rspcl |
⊢ ( ( ℤring ∈ Ring ∧ { 𝑁 } ⊆ ℤ ) → ( 𝑆 ‘ { 𝑁 } ) ∈ ( LIdeal ‘ ℤring ) ) |
7 |
2 3 6
|
sylancr |
⊢ ( 𝑁 ∈ ℤ → ( 𝑆 ‘ { 𝑁 } ) ∈ ( LIdeal ‘ ℤring ) ) |