Metamath Proof Explorer

Theorem plypow

Description: A power is a polynomial. (Contributed by Mario Carneiro, 17-Jul-2014)

Ref Expression
Assertion plypow 
|- ( ( S C_ CC /\ 1 e. S /\ N e. NN0 ) -> ( z e. CC |-> ( z ^ N ) ) e. ( Poly ` S ) )

Proof

Step Hyp Ref Expression
1 id 
 |-  ( z e. CC -> z e. CC )
2 simp3 
 |-  ( ( S C_ CC /\ 1 e. S /\ N e. NN0 ) -> N e. NN0 )
3 expcl 
 |-  ( ( z e. CC /\ N e. NN0 ) -> ( z ^ N ) e. CC )
4 1 2 3 syl2anr 
 |-  ( ( ( S C_ CC /\ 1 e. S /\ N e. NN0 ) /\ z e. CC ) -> ( z ^ N ) e. CC )
5 4 mullidd 
 |-  ( ( ( S C_ CC /\ 1 e. S /\ N e. NN0 ) /\ z e. CC ) -> ( 1 x. ( z ^ N ) ) = ( z ^ N ) )
6 5 mpteq2dva 
 |-  ( ( S C_ CC /\ 1 e. S /\ N e. NN0 ) -> ( z e. CC |-> ( 1 x. ( z ^ N ) ) ) = ( z e. CC |-> ( z ^ N ) ) )
7 eqid 
 |-  ( z e. CC |-> ( 1 x. ( z ^ N ) ) ) = ( z e. CC |-> ( 1 x. ( z ^ N ) ) )
8 7 ply1term 
 |-  ( ( S C_ CC /\ 1 e. S /\ N e. NN0 ) -> ( z e. CC |-> ( 1 x. ( z ^ N ) ) ) e. ( Poly ` S ) )
9 6 8 eqeltrrd 
 |-  ( ( S C_ CC /\ 1 e. S /\ N e. NN0 ) -> ( z e. CC |-> ( z ^ N ) ) e. ( Poly ` S ) )