Metamath Proof Explorer


Theorem dvhvscaval

Description: The scalar product operation for the constructed full vector space H. (Contributed by NM, 20-Nov-2013)

Ref Expression
Hypothesis dvhvscaval.s ⊢ · ˙ = s ∈ E , f ∈ T × E ⟼ s ⁡ 1 st ⁡ f s ∘ 2 nd ⁡ f
Assertion dvhvscaval ⊢ U ∈ E ∧ F ∈ T × E → U · ˙ F = U ⁡ 1 st ⁡ F U ∘ 2 nd ⁡ F

Proof

Step Hyp Ref Expression
1 dvhvscaval.s ⊢ · ˙ = s ∈ E , f ∈ T × E ⟼ s ⁡ 1 st ⁡ f s ∘ 2 nd ⁡ f
2 fveq1 ⊢ t = U → t ⁡ 1 st ⁡ g = U ⁡ 1 st ⁡ g
3 coeq1 ⊢ t = U → t ∘ 2 nd ⁡ g = U ∘ 2 nd ⁡ g
4 2 3 opeq12d ⊢ t = U → t ⁡ 1 st ⁡ g t ∘ 2 nd ⁡ g = U ⁡ 1 st ⁡ g U ∘ 2 nd ⁡ g
5 2fveq3 ⊢ g = F → U ⁡ 1 st ⁡ g = U ⁡ 1 st ⁡ F
6 fveq2 ⊢ g = F → 2 nd ⁡ g = 2 nd ⁡ F
7 6 coeq2d ⊢ g = F → U ∘ 2 nd ⁡ g = U ∘ 2 nd ⁡ F
8 5 7 opeq12d ⊢ g = F → U ⁡ 1 st ⁡ g U ∘ 2 nd ⁡ g = U ⁡ 1 st ⁡ F U ∘ 2 nd ⁡ F
9 1 dvhvscacbv ⊢ · ˙ = t ∈ E , g ∈ T × E ⟼ t ⁡ 1 st ⁡ g t ∘ 2 nd ⁡ g
10 opex ⊢ U ⁡ 1 st ⁡ F U ∘ 2 nd ⁡ F ∈ V
11 4 8 9 10 ovmpo ⊢ U ∈ E ∧ F ∈ T × E → U · ˙ F = U ⁡ 1 st ⁡ F U ∘ 2 nd ⁡ F