Metamath Proof Explorer


Theorem gsumvsmul1

Description: Pull a scalar multiplication out of a sum of vectors. This theorem properly generalizes gsummulc1 , since every ring is a left module over itself. (Contributed by Thierry Arnoux, 12-Jun-2023)

Ref Expression
Hypotheses gsumvsmul1.b ⊢ B = Base R
gsumvsmul1.s ⊢ S = Scalar ⁡ R
gsumvsmul1.k ⊢ K = Base S
gsumvsmul1.z ⊢ 0 ˙ = 0 S
gsumvsmul1.t ⊢ · ˙ = ⋅ R
gsumvsmul1.r ⊢ φ → R ∈ LMod
gsumvsmul1.1 ⊢ φ → S ∈ CMnd
gsumvsmul1.a ⊢ φ → A ∈ V
gsumvsmul1.x ⊢ φ → Y ∈ B
gsumvsmul1.y ⊢ φ ∧ k ∈ A → X ∈ K
gsumvsmul1.n ⊢ φ → finSupp 0 ˙⁡ k ∈ A ⟼ X
Assertion gsumvsmul1 ⊢ φ → ∑ R k ∈ A X · ˙ Y = ∑ S k ∈ A X · ˙ Y

Proof

Step Hyp Ref Expression
1 gsumvsmul1.b ⊢ B = Base R
2 gsumvsmul1.s ⊢ S = Scalar ⁡ R
3 gsumvsmul1.k ⊢ K = Base S
4 gsumvsmul1.z ⊢ 0 ˙ = 0 S
5 gsumvsmul1.t ⊢ · ˙ = ⋅ R
6 gsumvsmul1.r ⊢ φ → R ∈ LMod
7 gsumvsmul1.1 ⊢ φ → S ∈ CMnd
8 gsumvsmul1.a ⊢ φ → A ∈ V
9 gsumvsmul1.x ⊢ φ → Y ∈ B
10 gsumvsmul1.y ⊢ φ ∧ k ∈ A → X ∈ K
11 gsumvsmul1.n ⊢ φ → finSupp 0 ˙⁡ k ∈ A ⟼ X
12 lmodcmn ⊢ R ∈ LMod → R ∈ CMnd
13 cmnmnd ⊢ R ∈ CMnd → R ∈ Mnd
14 6 12 13 3syl ⊢ φ → R ∈ Mnd
15 1 2 5 3 lmodvslmhm ⊢ R ∈ LMod ∧ Y ∈ B → x ∈ K ⟼ x · ˙ Y ∈ S GrpHom R
16 6 9 15 syl2anc ⊢ φ → x ∈ K ⟼ x · ˙ Y ∈ S GrpHom R
17 ghmmhm ⊢ x ∈ K ⟼ x · ˙ Y ∈ S GrpHom R → x ∈ K ⟼ x · ˙ Y ∈ S MndHom R
18 16 17 syl ⊢ φ → x ∈ K ⟼ x · ˙ Y ∈ S MndHom R
19 oveq1 ⊢ x = X → x · ˙ Y = X · ˙ Y
20 oveq1 ⊢ x = ∑ S k ∈ A X → x · ˙ Y = ∑ S k ∈ A X · ˙ Y
21 3 4 7 14 8 18 10 11 19 20 gsummhm2 ⊢ φ → ∑ R k ∈ A X · ˙ Y = ∑ S k ∈ A X · ˙ Y