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 ∈ LVec ↔ ( 𝑊 ∈ LMod ∧ 𝐹 ∈ 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 ) ∈ DivRing ↔ 𝐹 ∈ DivRing ) )
5 df-lvec LVec = { 𝑓 ∈ LMod ∣ ( Scalar ‘ 𝑓 ) ∈ DivRing }
6 4 5 elrab2 ( 𝑊 ∈ LVec ↔ ( 𝑊 ∈ LMod ∧ 𝐹 ∈ DivRing ) )