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Metamath Proof Explorer

Theorem ringinvdv

Description: Write the inverse function in terms of division. (Contributed by Mario Carneiro, 2-Jul-2014)

Ref Expression
Hypotheses ringinvdv.b 
|- B = ( Base ` R )
ringinvdv.u 
|- U = ( Unit ` R )
ringinvdv.d 
|- ./ = ( /r ` R )
ringinvdv.o 
|- .1. = ( 1r ` R )
ringinvdv.i 
|- I = ( invr ` R )
Assertion ringinvdv 
|- ( ( R e. Ring /\ X e. U ) -> ( I ` X ) = ( .1. ./ X ) )

Proof

Step Hyp Ref Expression
1 ringinvdv.b 
 |-  B = ( Base ` R )
2 ringinvdv.u 
 |-  U = ( Unit ` R )
3 ringinvdv.d 
 |-  ./ = ( /r ` R )
4 ringinvdv.o 
 |-  .1. = ( 1r ` R )
5 ringinvdv.i 
 |-  I = ( invr ` R )
6 1 4 ringidcl 
 |-  ( R e. Ring -> .1. e. B )
7 eqid 
 |-  ( .r ` R ) = ( .r ` R )
8 1 7 2 5 3 dvrval 
 |-  ( ( .1. e. B /\ X e. U ) -> ( .1. ./ X ) = ( .1. ( .r ` R ) ( I ` X ) ) )
9 6 8 sylan 
 |-  ( ( R e. Ring /\ X e. U ) -> ( .1. ./ X ) = ( .1. ( .r ` R ) ( I ` X ) ) )
10 2 5 1 ringinvcl 
 |-  ( ( R e. Ring /\ X e. U ) -> ( I ` X ) e. B )
11 1 7 4 ringlidm 
 |-  ( ( R e. Ring /\ ( I ` X ) e. B ) -> ( .1. ( .r ` R ) ( I ` X ) ) = ( I ` X ) )
12 10 11 syldan 
 |-  ( ( R e. Ring /\ X e. U ) -> ( .1. ( .r ` R ) ( I ` X ) ) = ( I ` X ) )
13 9 12 eqtr2d 
 |-  ( ( R e. Ring /\ X e. U ) -> ( I ` X ) = ( .1. ./ X ) )