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Metamath Proof Explorer

Theorem erngring-rN

Description: An endomorphism ring is a ring. TODO: fix comment. (Contributed by NM, 4-Aug-2013) (New usage is discouraged.)

Ref Expression
Hypotheses ernggrp.h-r 
|- H = ( LHyp ` K )
ernggrp.d-r 
|- D = ( ( EDRingR ` K ) ` W )
Assertion erngring-rN 
|- ( ( K e. HL /\ W e. H ) -> D e. Ring )

Proof

Step Hyp Ref Expression
1 ernggrp.h-r 
 |-  H = ( LHyp ` K )
2 ernggrp.d-r 
 |-  D = ( ( EDRingR ` K ) ` W )
3 eqid 
 |-  ( Base ` K ) = ( Base ` K )
4 eqid 
 |-  ( ( LTrn ` K ) ` W ) = ( ( LTrn ` K ) ` W )
5 eqid 
 |-  ( ( TEndo ` K ) ` W ) = ( ( TEndo ` K ) ` W )
6 eqid 
 |-  ( a e. ( ( TEndo ` K ) ` W ) , b e. ( ( TEndo ` K ) ` W ) |-> ( f e. ( ( LTrn ` K ) ` W ) |-> ( ( a ` f ) o. ( b ` f ) ) ) ) = ( a e. ( ( TEndo ` K ) ` W ) , b e. ( ( TEndo ` K ) ` W ) |-> ( f e. ( ( LTrn ` K ) ` W ) |-> ( ( a ` f ) o. ( b ` f ) ) ) )
7 eqid 
 |-  ( f e. ( ( LTrn ` K ) ` W ) |-> ( _I |` ( Base ` K ) ) ) = ( f e. ( ( LTrn ` K ) ` W ) |-> ( _I |` ( Base ` K ) ) )
8 eqid 
 |-  ( a e. ( ( TEndo ` K ) ` W ) |-> ( f e. ( ( LTrn ` K ) ` W ) |-> `' ( a ` f ) ) ) = ( a e. ( ( TEndo ` K ) ` W ) |-> ( f e. ( ( LTrn ` K ) ` W ) |-> `' ( a ` f ) ) )
9 eqid 
 |-  ( a e. ( ( TEndo ` K ) ` W ) , b e. ( ( TEndo ` K ) ` W ) |-> ( b o. a ) ) = ( a e. ( ( TEndo ` K ) ` W ) , b e. ( ( TEndo ` K ) ` W ) |-> ( b o. a ) )
10 1 2 3 4 5 6 7 8 9 erngdvlem3-rN 
 |-  ( ( K e. HL /\ W e. H ) -> D e. Ring )