Metamath Proof Explorer


Theorem znlidl

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 ` ZZring )
Assertion znlidl
|- ( N e. ZZ -> ( S ` { N } ) e. ( LIdeal ` ZZring ) )

Proof

Step Hyp Ref Expression
1 znval.s
 |-  S = ( RSpan ` ZZring )
2 zringring
 |-  ZZring e. Ring
3 snssi
 |-  ( N e. ZZ -> { N } C_ ZZ )
4 zringbas
 |-  ZZ = ( Base ` ZZring )
5 eqid
 |-  ( LIdeal ` ZZring ) = ( LIdeal ` ZZring )
6 1 4 5 rspcl
 |-  ( ( ZZring e. Ring /\ { N } C_ ZZ ) -> ( S ` { N } ) e. ( LIdeal ` ZZring ) )
7 2 3 6 sylancr
 |-  ( N e. ZZ -> ( S ` { N } ) e. ( LIdeal ` ZZring ) )