Metamath Proof Explorer


Theorem ldualsca

Description: The ring of scalars of the dual of a vector space. (Contributed by NM, 18-Oct-2014)

Ref Expression
Hypotheses ldualsca.f ⊢ F = Scalar ⁡ W
ldualsca.o ⊢ O = opp r ⁡ F
ldualsca.d ⊢ D = LDual ⁡ W
ldualsca.r ⊢ R = Scalar ⁡ D
ldualsca.w ⊢ φ → W ∈ X
Assertion ldualsca ⊢ φ → R = O

Proof

Step Hyp Ref Expression
1 ldualsca.f ⊢ F = Scalar ⁡ W
2 ldualsca.o ⊢ O = opp r ⁡ F
3 ldualsca.d ⊢ D = LDual ⁡ W
4 ldualsca.r ⊢ R = Scalar ⁡ D
5 ldualsca.w ⊢ φ → W ∈ X
6 eqid ⊢ Base W = Base W
7 eqid ⊢ + F = + F
8 eqid ⊢ ∘ f ⁡ + F ↾ LFnl ⁡ W × LFnl ⁡ W = ∘ f ⁡ + F ↾ LFnl ⁡ W × LFnl ⁡ W
9 eqid ⊢ LFnl ⁡ W = LFnl ⁡ W
10 eqid ⊢ Base F = Base F
11 eqid ⊢ ⋅ F = ⋅ F
12 eqid ⊢ k ∈ Base F , f ∈ LFnl ⁡ W ⟼ f ⋅ F f Base W × k = k ∈ Base F , f ∈ LFnl ⁡ W ⟼ f ⋅ F f Base W × k
13 6 7 8 9 3 1 10 11 2 12 5 ldualset ⊢ φ → D = Base ndx LFnl ⁡ W + ndx ∘ f ⁡ + F ↾ LFnl ⁡ W × LFnl ⁡ W Scalar ⁡ ndx O ∪ ⋅ ndx k ∈ Base F , f ∈ LFnl ⁡ W ⟼ f ⋅ F f Base W × k
14 13 fveq2d ⊢ φ → Scalar ⁡ D = Scalar ⁡ Base ndx LFnl ⁡ W + ndx ∘ f ⁡ + F ↾ LFnl ⁡ W × LFnl ⁡ W Scalar ⁡ ndx O ∪ ⋅ ndx k ∈ Base F , f ∈ LFnl ⁡ W ⟼ f ⋅ F f Base W × k
15 2 fvexi ⊢ O ∈ V
16 eqid ⊢ Base ndx LFnl ⁡ W + ndx ∘ f ⁡ + F ↾ LFnl ⁡ W × LFnl ⁡ W Scalar ⁡ ndx O ∪ ⋅ ndx k ∈ Base F , f ∈ LFnl ⁡ W ⟼ f ⋅ F f Base W × k = Base ndx LFnl ⁡ W + ndx ∘ f ⁡ + F ↾ LFnl ⁡ W × LFnl ⁡ W Scalar ⁡ ndx O ∪ ⋅ ndx k ∈ Base F , f ∈ LFnl ⁡ W ⟼ f ⋅ F f Base W × k
17 16 lmodsca ⊢ O ∈ V → O = Scalar ⁡ Base ndx LFnl ⁡ W + ndx ∘ f ⁡ + F ↾ LFnl ⁡ W × LFnl ⁡ W Scalar ⁡ ndx O ∪ ⋅ ndx k ∈ Base F , f ∈ LFnl ⁡ W ⟼ f ⋅ F f Base W × k
18 15 17 ax-mp ⊢ O = Scalar ⁡ Base ndx LFnl ⁡ W + ndx ∘ f ⁡ + F ↾ LFnl ⁡ W × LFnl ⁡ W Scalar ⁡ ndx O ∪ ⋅ ndx k ∈ Base F , f ∈ LFnl ⁡ W ⟼ f ⋅ F f Base W × k
19 14 4 18 3eqtr4g ⊢ φ → R = O