Metamath Proof Explorer


Theorem pwsvscafval

Description: Scalar multiplication in a structure power is pointwise. (Contributed by Mario Carneiro, 11-Jan-2015)

Ref Expression
Hypotheses pwsvscaval.y ⊢ Y = R ↑ 𝑠 I
pwsvscaval.b ⊢ B = Base Y
pwsvscaval.s ⊢ · ˙ = ⋅ R
pwsvscaval.t ⊢ ∙ ˙ = ⋅ Y
pwsvscaval.f ⊢ F = Scalar ⁡ R
pwsvscaval.k ⊢ K = Base F
pwsvscaval.r ⊢ φ → R ∈ V
pwsvscaval.i ⊢ φ → I ∈ W
pwsvscaval.a ⊢ φ → A ∈ K
pwsvscaval.x ⊢ φ → X ∈ B
Assertion pwsvscafval ⊢ φ → A ∙ ˙ X = I × A · ˙ f X

Proof

Step Hyp Ref Expression
1 pwsvscaval.y ⊢ Y = R ↑ 𝑠 I
2 pwsvscaval.b ⊢ B = Base Y
3 pwsvscaval.s ⊢ · ˙ = ⋅ R
4 pwsvscaval.t ⊢ ∙ ˙ = ⋅ Y
5 pwsvscaval.f ⊢ F = Scalar ⁡ R
6 pwsvscaval.k ⊢ K = Base F
7 pwsvscaval.r ⊢ φ → R ∈ V
8 pwsvscaval.i ⊢ φ → I ∈ W
9 pwsvscaval.a ⊢ φ → A ∈ K
10 pwsvscaval.x ⊢ φ → X ∈ B
11 1 5 pwsval ⊢ R ∈ V ∧ I ∈ W → Y = F ⨉ 𝑠 I × R
12 7 8 11 syl2anc ⊢ φ → Y = F ⨉ 𝑠 I × R
13 12 fveq2d ⊢ φ → ⋅ Y = ⋅ F ⨉ 𝑠 I × R
14 4 13 eqtrid ⊢ φ → ∙ ˙ = ⋅ F ⨉ 𝑠 I × R
15 14 oveqd ⊢ φ → A ∙ ˙ X = A ⋅ F ⨉ 𝑠 I × R X
16 eqid ⊢ F ⨉ 𝑠 I × R = F ⨉ 𝑠 I × R
17 eqid ⊢ Base F ⨉ 𝑠 I × R = Base F ⨉ 𝑠 I × R
18 eqid ⊢ ⋅ F ⨉ 𝑠 I × R = ⋅ F ⨉ 𝑠 I × R
19 5 fvexi ⊢ F ∈ V
20 19 a1i ⊢ φ → F ∈ V
21 fnconstg ⊢ R ∈ V → I × R Fn I
22 7 21 syl ⊢ φ → I × R Fn I
23 12 fveq2d ⊢ φ → Base Y = Base F ⨉ 𝑠 I × R
24 2 23 eqtrid ⊢ φ → B = Base F ⨉ 𝑠 I × R
25 10 24 eleqtrd ⊢ φ → X ∈ Base F ⨉ 𝑠 I × R
26 16 17 18 6 20 8 22 9 25 prdsvscaval ⊢ φ → A ⋅ F ⨉ 𝑠 I × R X = x ∈ I ⟼ A ⋅ I × R ⁡ x X ⁡ x
27 fvconst2g ⊢ R ∈ V ∧ x ∈ I → I × R ⁡ x = R
28 7 27 sylan ⊢ φ ∧ x ∈ I → I × R ⁡ x = R
29 28 fveq2d ⊢ φ ∧ x ∈ I → ⋅ I × R ⁡ x = ⋅ R
30 29 3 eqtr4di ⊢ φ ∧ x ∈ I → ⋅ I × R ⁡ x = · ˙
31 30 oveqd ⊢ φ ∧ x ∈ I → A ⋅ I × R ⁡ x X ⁡ x = A · ˙ X ⁡ x
32 31 mpteq2dva ⊢ φ → x ∈ I ⟼ A ⋅ I × R ⁡ x X ⁡ x = x ∈ I ⟼ A · ˙ X ⁡ x
33 9 adantr ⊢ φ ∧ x ∈ I → A ∈ K
34 fvexd ⊢ φ ∧ x ∈ I → X ⁡ x ∈ V
35 fconstmpt ⊢ I × A = x ∈ I ⟼ A
36 35 a1i ⊢ φ → I × A = x ∈ I ⟼ A
37 eqid ⊢ Base R = Base R
38 1 37 2 7 8 10 pwselbas ⊢ φ → X : I ⟶ Base R
39 38 feqmptd ⊢ φ → X = x ∈ I ⟼ X ⁡ x
40 8 33 34 36 39 offval2 ⊢ φ → I × A · ˙ f X = x ∈ I ⟼ A · ˙ X ⁡ x
41 32 40 eqtr4d ⊢ φ → x ∈ I ⟼ A ⋅ I × R ⁡ x X ⁡ x = I × A · ˙ f X
42 15 26 41 3eqtrd ⊢ φ → A ∙ ˙ X = I × A · ˙ f X