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 ⊢ × ˙ = ⋅ R
ldualfvs.d ⊢ D = LDual ⁡ W
ldualfvs.s ⊢ ∙ ˙ = ⋅ D
ldualfvs.w ⊢ φ → W ∈ Y
ldualvs.x ⊢ φ → X ∈ K
ldualvs.g ⊢ φ → G ∈ F
Assertion ldualvs ⊢ φ → X ∙ ˙ G = G × ˙ f 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 ⊢ × ˙ = ⋅ R
6 ldualfvs.d ⊢ D = LDual ⁡ W
7 ldualfvs.s ⊢ ∙ ˙ = ⋅ D
8 ldualfvs.w ⊢ φ → W ∈ Y
9 ldualvs.x ⊢ φ → X ∈ K
10 ldualvs.g ⊢ φ → G ∈ F
11 eqid ⊢ k ∈ K , f ∈ F ⟼ f × ˙ f V × k = k ∈ K , f ∈ F ⟼ f × ˙ f V × k
12 1 2 3 4 5 6 7 8 11 ldualfvs ⊢ φ → ∙ ˙ = k ∈ K , f ∈ F ⟼ f × ˙ f V × k
13 12 oveqd ⊢ φ → X ∙ ˙ G = X k ∈ K , f ∈ F ⟼ f × ˙ f V × k G
14 sneq ⊢ k = X → k = X
15 14 xpeq2d ⊢ k = X → V × k = V × X
16 15 oveq2d ⊢ k = X → f × ˙ f V × k = f × ˙ f V × X
17 oveq1 ⊢ f = G → f × ˙ f V × X = G × ˙ f V × X
18 ovex ⊢ G × ˙ f V × X ∈ V
19 16 17 11 18 ovmpo ⊢ X ∈ K ∧ G ∈ F → X k ∈ K , f ∈ F ⟼ f × ˙ f V × k G = G × ˙ f V × X
20 9 10 19 syl2anc ⊢ φ → X k ∈ K , f ∈ F ⟼ f × ˙ f V × k G = G × ˙ f V × X
21 13 20 eqtrd ⊢ φ → X ∙ ˙ G = G × ˙ f V × X