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Metamath Proof Explorer

Theorem maxidln1

Description: One is not contained in any maximal ideal. (Contributed by Jeff Madsen, 17-Jun-2011)

Ref Expression
Hypotheses maxidln1.1 
|- H = ( 2nd ` R )
maxidln1.2 
|- U = ( GId ` H )
Assertion maxidln1 
|- ( ( R e. RingOps /\ M e. ( MaxIdl ` R ) ) -> -. U e. M )

Proof

Step Hyp Ref Expression
1 maxidln1.1 
 |-  H = ( 2nd ` R )
2 maxidln1.2 
 |-  U = ( GId ` H )
3 eqid 
 |-  ( 1st ` R ) = ( 1st ` R )
4 eqid 
 |-  ran ( 1st ` R ) = ran ( 1st ` R )
5 3 4 maxidlnr 
 |-  ( ( R e. RingOps /\ M e. ( MaxIdl ` R ) ) -> M =/= ran ( 1st ` R ) )
6 maxidlidl 
 |-  ( ( R e. RingOps /\ M e. ( MaxIdl ` R ) ) -> M e. ( Idl ` R ) )
7 3 1 4 2 1idl 
 |-  ( ( R e. RingOps /\ M e. ( Idl ` R ) ) -> ( U e. M <-> M = ran ( 1st ` R ) ) )
8 7 necon3bbid 
 |-  ( ( R e. RingOps /\ M e. ( Idl ` R ) ) -> ( -. U e. M <-> M =/= ran ( 1st ` R ) ) )
9 6 8 syldan 
 |-  ( ( R e. RingOps /\ M e. ( MaxIdl ` R ) ) -> ( -. U e. M <-> M =/= ran ( 1st ` R ) ) )
10 5 9 mpbird 
 |-  ( ( R e. RingOps /\ M e. ( MaxIdl ` R ) ) -> -. U e. M )