Theorem opprringb

Description: Bidirectional form of opprring . (Contributed by Mario Carneiro, 6-Dec-2014)

	Ref	Expression
Hypothesis	opprbas.1	|- O = ( oppR ` R )
Assertion	opprringb	|- ( R e. Ring <-> O e. Ring )

Proof

Step	Hyp	Ref	Expression
1		opprbas.1	|- O = ( oppR ` R )
2	1	opprring	|- ( R e. Ring -> O e. Ring )
3		eqid	|- ( oppR ` O ) = ( oppR ` O )
4	3	opprring	|- ( O e. Ring -> ( oppR ` O ) e. Ring )
5		eqidd	|- ( T. -> ( Base ` R ) = ( Base ` R ) )
6		eqid	|- ( Base ` R ) = ( Base ` R )
7	1 6	opprbas	|- ( Base ` R ) = ( Base ` O )
8	3 7	opprbas	|- ( Base ` R ) = ( Base ` ( oppR ` O ) )
9	8	a1i	|- ( T. -> ( Base ` R ) = ( Base ` ( oppR ` O ) ) )
10		eqid	|- ( +g ` R ) = ( +g ` R )
11	1 10	oppradd	|- ( +g ` R ) = ( +g ` O )
12	3 11	oppradd	|- ( +g ` R ) = ( +g ` ( oppR ` O ) )
13	12	oveqi	|- ( x ( +g ` R ) y ) = ( x ( +g ` ( oppR ` O ) ) y )
14	13	a1i	|- ( ( T. /\ ( x e. ( Base ` R ) /\ y e. ( Base ` R ) ) ) -> ( x ( +g ` R ) y ) = ( x ( +g ` ( oppR ` O ) ) y ) )
15		eqid	|- ( .r ` O ) = ( .r ` O )
16		eqid	|- ( .r ` ( oppR ` O ) ) = ( .r ` ( oppR ` O ) )
17	7 15 3 16	opprmul	|- ( x ( .r ` ( oppR ` O ) ) y ) = ( y ( .r ` O ) x )
18		eqid	|- ( .r ` R ) = ( .r ` R )
19	6 18 1 15	opprmul	|- ( y ( .r ` O ) x ) = ( x ( .r ` R ) y )
20	17 19	eqtr2i	|- ( x ( .r ` R ) y ) = ( x ( .r ` ( oppR ` O ) ) y )
21	20	a1i	|- ( ( T. /\ ( x e. ( Base ` R ) /\ y e. ( Base ` R ) ) ) -> ( x ( .r ` R ) y ) = ( x ( .r ` ( oppR ` O ) ) y ) )
22	5 9 14 21	ringpropd	|- ( T. -> ( R e. Ring <-> ( oppR ` O ) e. Ring ) )
23	22	mptru	|- ( R e. Ring <-> ( oppR ` O ) e. Ring )
24	4 23	sylibr	|- ( O e. Ring -> R e. Ring )
25	2 24	impbii	|- ( R e. Ring <-> O e. Ring )