Metamath Proof Explorer


Theorem plymul02

Description: Product of a polynomial with the zero polynomial. (Contributed by Thierry Arnoux, 26-Sep-2018)

Ref Expression
Assertion plymul02
|- ( F e. ( Poly ` S ) -> ( 0p oF x. F ) = 0p )

Proof

Step Hyp Ref Expression
1 plyf
 |-  ( F e. ( Poly ` S ) -> F : CC --> CC )
2 1 ffvelrnda
 |-  ( ( F e. ( Poly ` S ) /\ x e. CC ) -> ( F ` x ) e. CC )
3 2 mul02d
 |-  ( ( F e. ( Poly ` S ) /\ x e. CC ) -> ( 0 x. ( F ` x ) ) = 0 )
4 3 mpteq2dva
 |-  ( F e. ( Poly ` S ) -> ( x e. CC |-> ( 0 x. ( F ` x ) ) ) = ( x e. CC |-> 0 ) )
5 c0ex
 |-  0 e. _V
6 5 fconst
 |-  ( CC X. { 0 } ) : CC --> { 0 }
7 df-0p
 |-  0p = ( CC X. { 0 } )
8 7 feq1i
 |-  ( 0p : CC --> { 0 } <-> ( CC X. { 0 } ) : CC --> { 0 } )
9 6 8 mpbir
 |-  0p : CC --> { 0 }
10 ffn
 |-  ( 0p : CC --> { 0 } -> 0p Fn CC )
11 9 10 mp1i
 |-  ( F e. ( Poly ` S ) -> 0p Fn CC )
12 1 ffnd
 |-  ( F e. ( Poly ` S ) -> F Fn CC )
13 cnex
 |-  CC e. _V
14 13 a1i
 |-  ( F e. ( Poly ` S ) -> CC e. _V )
15 inidm
 |-  ( CC i^i CC ) = CC
16 0pval
 |-  ( x e. CC -> ( 0p ` x ) = 0 )
17 16 adantl
 |-  ( ( F e. ( Poly ` S ) /\ x e. CC ) -> ( 0p ` x ) = 0 )
18 eqidd
 |-  ( ( F e. ( Poly ` S ) /\ x e. CC ) -> ( F ` x ) = ( F ` x ) )
19 11 12 14 14 15 17 18 offval
 |-  ( F e. ( Poly ` S ) -> ( 0p oF x. F ) = ( x e. CC |-> ( 0 x. ( F ` x ) ) ) )
20 fconstmpt
 |-  ( CC X. { 0 } ) = ( x e. CC |-> 0 )
21 7 20 eqtri
 |-  0p = ( x e. CC |-> 0 )
22 21 a1i
 |-  ( F e. ( Poly ` S ) -> 0p = ( x e. CC |-> 0 ) )
23 4 19 22 3eqtr4d
 |-  ( F e. ( Poly ` S ) -> ( 0p oF x. F ) = 0p )