Metamath Proof Explorer


Theorem ldualsmul

Description: Scalar multiplication for the dual of a vector space. (Contributed by NM, 19-Oct-2014) (Revised by Mario Carneiro, 22-Sep-2015)

Ref Expression
Hypotheses ldualsmul.f 𝐹 = ( Scalar ‘ 𝑊 )
ldualsmul.k 𝐾 = ( Base ‘ 𝐹 )
ldualsmul.t · = ( .r𝐹 )
ldualsmul.d 𝐷 = ( LDual ‘ 𝑊 )
ldualsmul.r 𝑅 = ( Scalar ‘ 𝐷 )
ldualsmul.m = ( .r𝑅 )
ldualsmul.w ( 𝜑𝑊𝑉 )
ldualsmul.x ( 𝜑𝑋𝐾 )
ldualsmul.y ( 𝜑𝑌𝐾 )
Assertion ldualsmul ( 𝜑 → ( 𝑋 𝑌 ) = ( 𝑌 · 𝑋 ) )

Proof

Step Hyp Ref Expression
1 ldualsmul.f 𝐹 = ( Scalar ‘ 𝑊 )
2 ldualsmul.k 𝐾 = ( Base ‘ 𝐹 )
3 ldualsmul.t · = ( .r𝐹 )
4 ldualsmul.d 𝐷 = ( LDual ‘ 𝑊 )
5 ldualsmul.r 𝑅 = ( Scalar ‘ 𝐷 )
6 ldualsmul.m = ( .r𝑅 )
7 ldualsmul.w ( 𝜑𝑊𝑉 )
8 ldualsmul.x ( 𝜑𝑋𝐾 )
9 ldualsmul.y ( 𝜑𝑌𝐾 )
10 eqid ( oppr𝐹 ) = ( oppr𝐹 )
11 1 10 4 5 7 ldualsca ( 𝜑𝑅 = ( oppr𝐹 ) )
12 11 fveq2d ( 𝜑 → ( .r𝑅 ) = ( .r ‘ ( oppr𝐹 ) ) )
13 6 12 eqtrid ( 𝜑 = ( .r ‘ ( oppr𝐹 ) ) )
14 13 oveqd ( 𝜑 → ( 𝑋 𝑌 ) = ( 𝑋 ( .r ‘ ( oppr𝐹 ) ) 𝑌 ) )
15 eqid ( .r ‘ ( oppr𝐹 ) ) = ( .r ‘ ( oppr𝐹 ) )
16 2 3 10 15 opprmul ( 𝑋 ( .r ‘ ( oppr𝐹 ) ) 𝑌 ) = ( 𝑌 · 𝑋 )
17 14 16 eqtrdi ( 𝜑 → ( 𝑋 𝑌 ) = ( 𝑌 · 𝑋 ) )