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 = ( Poly1 ` R )
ply1scl.a
|- A = ( algSc ` P )
coe1scl.k
|- K = ( Base ` R )
ply1sclf.b
|- B = ( Base ` P )
Assertion ply1sclf
|- ( R e. Ring -> A : K --> B )

Proof

Step Hyp Ref Expression
1 ply1scl.p
 |-  P = ( Poly1 ` 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 e. Ring -> P e. Ring )
7 1 ply1lmod
 |-  ( R e. Ring -> P e. LMod )
8 baseid
 |-  Base = Slot ( Base ` ndx )
9 8 3 strfvi
 |-  K = ( Base ` ( _I ` R ) )
10 2 5 6 7 9 4 asclf
 |-  ( R e. Ring -> A : K --> B )