Metamath Proof Explorer


Theorem islvec

Description: The predicate "is a left vector space". (Contributed by NM, 11-Nov-2013)

Ref Expression
Hypothesis islvec.1
|- F = ( Scalar ` W )
Assertion islvec
|- ( W e. LVec <-> ( W e. LMod /\ F e. DivRing ) )

Proof

Step Hyp Ref Expression
1 islvec.1
 |-  F = ( Scalar ` W )
2 fveq2
 |-  ( f = W -> ( Scalar ` f ) = ( Scalar ` W ) )
3 2 1 eqtr4di
 |-  ( f = W -> ( Scalar ` f ) = F )
4 3 eleq1d
 |-  ( f = W -> ( ( Scalar ` f ) e. DivRing <-> F e. DivRing ) )
5 df-lvec
 |-  LVec = { f e. LMod | ( Scalar ` f ) e. DivRing }
6 4 5 elrab2
 |-  ( W e. LVec <-> ( W e. LMod /\ F e. DivRing ) )