Metamath Proof Explorer


Theorem plycjlem

Description: Lemma for plycj and coecj . (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Hypotheses plycj.1 N = deg F
plycj.2 G = * F *
plycjlem.3 A = coeff F
Assertion plycjlem F Poly S G = z k = 0 N * A k z k

Proof

Step Hyp Ref Expression
1 plycj.1 N = deg F
2 plycj.2 G = * F *
3 plycjlem.3 A = coeff F
4 cjcl z z
5 4 adantl F Poly S z z
6 cjf * :
7 6 a1i F Poly S * :
8 7 feqmptd F Poly S * = z z
9 fzfid F Poly S x 0 N Fin
10 3 coef3 F Poly S A : 0
11 10 adantr F Poly S x A : 0
12 elfznn0 k 0 N k 0
13 ffvelrn A : 0 k 0 A k
14 11 12 13 syl2an F Poly S x k 0 N A k
15 expcl x k 0 x k
16 12 15 sylan2 x k 0 N x k
17 16 adantll F Poly S x k 0 N x k
18 14 17 mulcld F Poly S x k 0 N A k x k
19 9 18 fsumcl F Poly S x k = 0 N A k x k
20 3 1 coeid F Poly S F = x k = 0 N A k x k
21 fveq2 z = k = 0 N A k x k z = k = 0 N A k x k
22 19 20 8 21 fmptco F Poly S * F = x k = 0 N A k x k
23 oveq1 x = z x k = z k
24 23 oveq2d x = z A k x k = A k z k
25 24 sumeq2sdv x = z k = 0 N A k x k = k = 0 N A k z k
26 25 fveq2d x = z k = 0 N A k x k = k = 0 N A k z k
27 5 8 22 26 fmptco F Poly S * F * = z k = 0 N A k z k
28 2 27 eqtrid F Poly S G = z k = 0 N A k z k
29 fzfid F Poly S z 0 N Fin
30 10 adantr F Poly S z A : 0
31 30 12 13 syl2an F Poly S z k 0 N A k
32 expcl z k 0 z k
33 5 12 32 syl2an F Poly S z k 0 N z k
34 31 33 mulcld F Poly S z k 0 N A k z k
35 29 34 fsumcj F Poly S z k = 0 N A k z k = k = 0 N A k z k
36 31 33 cjmuld F Poly S z k 0 N A k z k = A k z k
37 fvco3 A : 0 k 0 * A k = A k
38 30 12 37 syl2an F Poly S z k 0 N * A k = A k
39 cjexp z k 0 z k = z k
40 5 12 39 syl2an F Poly S z k 0 N z k = z k
41 cjcj z z = z
42 41 ad2antlr F Poly S z k 0 N z = z
43 42 oveq1d F Poly S z k 0 N z k = z k
44 40 43 eqtr2d F Poly S z k 0 N z k = z k
45 38 44 oveq12d F Poly S z k 0 N * A k z k = A k z k
46 36 45 eqtr4d F Poly S z k 0 N A k z k = * A k z k
47 46 sumeq2dv F Poly S z k = 0 N A k z k = k = 0 N * A k z k
48 35 47 eqtrd F Poly S z k = 0 N A k z k = k = 0 N * A k z k
49 48 mpteq2dva F Poly S z k = 0 N A k z k = z k = 0 N * A k z k
50 28 49 eqtrd F Poly S G = z k = 0 N * A k z k