Metamath Proof Explorer


Theorem ldualvs

Description: Scalar product operation value (which is a functional) for the dual of a vector space. (Contributed by NM, 18-Oct-2014)

Ref Expression
Hypotheses ldualfvs.f 𝐹 = ( LFnl ‘ 𝑊 )
ldualfvs.v 𝑉 = ( Base ‘ 𝑊 )
ldualfvs.r 𝑅 = ( Scalar ‘ 𝑊 )
ldualfvs.k 𝐾 = ( Base ‘ 𝑅 )
ldualfvs.t × = ( .r𝑅 )
ldualfvs.d 𝐷 = ( LDual ‘ 𝑊 )
ldualfvs.s = ( ·𝑠𝐷 )
ldualfvs.w ( 𝜑𝑊𝑌 )
ldualvs.x ( 𝜑𝑋𝐾 )
ldualvs.g ( 𝜑𝐺𝐹 )
Assertion ldualvs ( 𝜑 → ( 𝑋 𝐺 ) = ( 𝐺f × ( 𝑉 × { 𝑋 } ) ) )

Proof

Step Hyp Ref Expression
1 ldualfvs.f 𝐹 = ( LFnl ‘ 𝑊 )
2 ldualfvs.v 𝑉 = ( Base ‘ 𝑊 )
3 ldualfvs.r 𝑅 = ( Scalar ‘ 𝑊 )
4 ldualfvs.k 𝐾 = ( Base ‘ 𝑅 )
5 ldualfvs.t × = ( .r𝑅 )
6 ldualfvs.d 𝐷 = ( LDual ‘ 𝑊 )
7 ldualfvs.s = ( ·𝑠𝐷 )
8 ldualfvs.w ( 𝜑𝑊𝑌 )
9 ldualvs.x ( 𝜑𝑋𝐾 )
10 ldualvs.g ( 𝜑𝐺𝐹 )
11 eqid ( 𝑘𝐾 , 𝑓𝐹 ↦ ( 𝑓f × ( 𝑉 × { 𝑘 } ) ) ) = ( 𝑘𝐾 , 𝑓𝐹 ↦ ( 𝑓f × ( 𝑉 × { 𝑘 } ) ) )
12 1 2 3 4 5 6 7 8 11 ldualfvs ( 𝜑 = ( 𝑘𝐾 , 𝑓𝐹 ↦ ( 𝑓f × ( 𝑉 × { 𝑘 } ) ) ) )
13 12 oveqd ( 𝜑 → ( 𝑋 𝐺 ) = ( 𝑋 ( 𝑘𝐾 , 𝑓𝐹 ↦ ( 𝑓f × ( 𝑉 × { 𝑘 } ) ) ) 𝐺 ) )
14 sneq ( 𝑘 = 𝑋 → { 𝑘 } = { 𝑋 } )
15 14 xpeq2d ( 𝑘 = 𝑋 → ( 𝑉 × { 𝑘 } ) = ( 𝑉 × { 𝑋 } ) )
16 15 oveq2d ( 𝑘 = 𝑋 → ( 𝑓f × ( 𝑉 × { 𝑘 } ) ) = ( 𝑓f × ( 𝑉 × { 𝑋 } ) ) )
17 oveq1 ( 𝑓 = 𝐺 → ( 𝑓f × ( 𝑉 × { 𝑋 } ) ) = ( 𝐺f × ( 𝑉 × { 𝑋 } ) ) )
18 ovex ( 𝐺f × ( 𝑉 × { 𝑋 } ) ) ∈ V
19 16 17 11 18 ovmpo ( ( 𝑋𝐾𝐺𝐹 ) → ( 𝑋 ( 𝑘𝐾 , 𝑓𝐹 ↦ ( 𝑓f × ( 𝑉 × { 𝑘 } ) ) ) 𝐺 ) = ( 𝐺f × ( 𝑉 × { 𝑋 } ) ) )
20 9 10 19 syl2anc ( 𝜑 → ( 𝑋 ( 𝑘𝐾 , 𝑓𝐹 ↦ ( 𝑓f × ( 𝑉 × { 𝑘 } ) ) ) 𝐺 ) = ( 𝐺f × ( 𝑉 × { 𝑋 } ) ) )
21 13 20 eqtrd ( 𝜑 → ( 𝑋 𝐺 ) = ( 𝐺f × ( 𝑉 × { 𝑋 } ) ) )