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 ) )