Metamath Proof Explorer


Theorem ply1sclf

Description: A scalar polynomial is a polynomial. (Contributed by Stefan O'Rear, 28-Mar-2015)

Ref Expression
Hypotheses ply1scl.p ⊢ P = Poly 1 ⁡ R
ply1scl.a ⊢ A = algSc ⁡ P
coe1scl.k ⊢ K = Base R
ply1sclf.b ⊢ B = Base P
Assertion ply1sclf ⊢ R ∈ Ring → A : K ⟶ B

Proof

Step Hyp Ref Expression
1 ply1scl.p ⊢ P = Poly 1 ⁡ R
2 ply1scl.a ⊢ A = algSc ⁡ P
3 coe1scl.k ⊢ K = Base R
4 ply1sclf.b ⊢ B = Base P
5 1 ply1sca2 ⊢ I ⁡ R = Scalar ⁡ P
6 1 ply1ring ⊢ R ∈ Ring → P ∈ Ring
7 1 ply1lmod ⊢ R ∈ Ring → P ∈ LMod
8 baseid ⊢ Base = Slot Base ndx
9 8 3 strfvi ⊢ K = Base I ⁡ R
10 2 5 6 7 9 4 asclf ⊢ R ∈ Ring → A : K ⟶ B