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 ⊢ · ˙ = ⋅ F
ldualsmul.d ⊢ D = LDual ⁡ W
ldualsmul.r ⊢ R = Scalar ⁡ D
ldualsmul.m ⊢ ∙ ˙ = ⋅ R
ldualsmul.w ⊢ φ → W ∈ V
ldualsmul.x ⊢ φ → X ∈ K
ldualsmul.y ⊢ φ → Y ∈ K
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 ⊢ · ˙ = ⋅ F
4 ldualsmul.d ⊢ D = LDual ⁡ W
5 ldualsmul.r ⊢ R = Scalar ⁡ D
6 ldualsmul.m ⊢ ∙ ˙ = ⋅ R
7 ldualsmul.w ⊢ φ → W ∈ V
8 ldualsmul.x ⊢ φ → X ∈ K
9 ldualsmul.y ⊢ φ → Y ∈ K
10 eqid ⊢ opp r ⁡ F = opp r ⁡ F
11 1 10 4 5 7 ldualsca ⊢ φ → R = opp r ⁡ F
12 11 fveq2d ⊢ φ → ⋅ R = ⋅ opp r ⁡ F
13 6 12 eqtrid ⊢ φ → ∙ ˙ = ⋅ opp r ⁡ F
14 13 oveqd ⊢ φ → X ∙ ˙ Y = X ⋅ opp r ⁡ F Y
15 eqid ⊢ ⋅ opp r ⁡ F = ⋅ opp r ⁡ F
16 2 3 10 15 opprmul ⊢ X ⋅ opp r ⁡ F Y = Y · ˙ X
17 14 16 eqtrdi ⊢ φ → X ∙ ˙ Y = Y · ˙ X