Metamath Proof Explorer


Theorem ldual0vcl

Description: The dual zero vector is a functional. (Contributed by NM, 5-Mar-2015)

Ref Expression
Hypotheses ldualv0cl.f 𝐹 = ( LFnl ‘ 𝑊 )
ldualv0cl.d 𝐷 = ( LDual ‘ 𝑊 )
ldualv0cl.o 0 = ( 0g𝐷 )
ldualv0cl.w ( 𝜑𝑊 ∈ LMod )
Assertion ldual0vcl ( 𝜑0𝐹 )

Proof

Step Hyp Ref Expression
1 ldualv0cl.f 𝐹 = ( LFnl ‘ 𝑊 )
2 ldualv0cl.d 𝐷 = ( LDual ‘ 𝑊 )
3 ldualv0cl.o 0 = ( 0g𝐷 )
4 ldualv0cl.w ( 𝜑𝑊 ∈ LMod )
5 eqid ( Base ‘ 𝑊 ) = ( Base ‘ 𝑊 )
6 eqid ( Scalar ‘ 𝑊 ) = ( Scalar ‘ 𝑊 )
7 eqid ( 0g ‘ ( Scalar ‘ 𝑊 ) ) = ( 0g ‘ ( Scalar ‘ 𝑊 ) )
8 5 6 7 2 3 4 ldual0v ( 𝜑0 = ( ( Base ‘ 𝑊 ) × { ( 0g ‘ ( Scalar ‘ 𝑊 ) ) } ) )
9 6 7 5 1 lfl0f ( 𝑊 ∈ LMod → ( ( Base ‘ 𝑊 ) × { ( 0g ‘ ( Scalar ‘ 𝑊 ) ) } ) ∈ 𝐹 )
10 4 9 syl ( 𝜑 → ( ( Base ‘ 𝑊 ) × { ( 0g ‘ ( Scalar ‘ 𝑊 ) ) } ) ∈ 𝐹 )
11 8 10 eqeltrd ( 𝜑0𝐹 )