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Metamath Proof Explorer

Theorem subrgid

Description: Every ring is a subring of itself. (Contributed by Stefan O'Rear, 30-Nov-2014)

Ref Expression
Hypothesis subrgss.1 
|- B = ( Base ` R )
Assertion subrgid 
|- ( R e. Ring -> B e. ( SubRing ` R ) )

Proof

Step Hyp Ref Expression
1 subrgss.1 
 |-  B = ( Base ` R )
2 id 
 |-  ( R e. Ring -> R e. Ring )
3 1 ressid 
 |-  ( R e. Ring -> ( R |`s B ) = R )
4 3 2 eqeltrd 
 |-  ( R e. Ring -> ( R |`s B ) e. Ring )
5 eqid 
 |-  ( 1r ` R ) = ( 1r ` R )
6 1 5 ringidcl 
 |-  ( R e. Ring -> ( 1r ` R ) e. B )
7 ssid 
 |-  B C_ B
8 6 7 jctil 
 |-  ( R e. Ring -> ( B C_ B /\ ( 1r ` R ) e. B ) )
9 1 5 issubrg 
 |-  ( B e. ( SubRing ` R ) <-> ( ( R e. Ring /\ ( R |`s B ) e. Ring ) /\ ( B C_ B /\ ( 1r ` R ) e. B ) ) )
10 2 4 8 9 syl21anbrc 
 |-  ( R e. Ring -> B e. ( SubRing ` R ) )