Metamath Proof Explorer


Theorem plycjlem

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

Ref Expression
Hypotheses plycj.1 N=degF
plycj.2 G=*F*
plycjlem.3 A=coeffF
Assertion plycjlem FPolySG=zk=0N*Akzk

Proof

Step Hyp Ref Expression
1 plycj.1 N=degF
2 plycj.2 G=*F*
3 plycjlem.3 A=coeffF
4 cjcl zz
5 4 adantl FPolySzz
6 cjf *:
7 6 a1i FPolyS*:
8 7 feqmptd FPolyS*=zz
9 fzfid FPolySx0NFin
10 3 coef3 FPolySA:0
11 10 adantr FPolySxA:0
12 elfznn0 k0Nk0
13 ffvelcdm A:0k0Ak
14 11 12 13 syl2an FPolySxk0NAk
15 expcl xk0xk
16 12 15 sylan2 xk0Nxk
17 16 adantll FPolySxk0Nxk
18 14 17 mulcld FPolySxk0NAkxk
19 9 18 fsumcl FPolySxk=0NAkxk
20 3 1 coeid FPolySF=xk=0NAkxk
21 fveq2 z=k=0NAkxkz=k=0NAkxk
22 19 20 8 21 fmptco FPolyS*F=xk=0NAkxk
23 oveq1 x=zxk=zk
24 23 oveq2d x=zAkxk=Akzk
25 24 sumeq2sdv x=zk=0NAkxk=k=0NAkzk
26 25 fveq2d x=zk=0NAkxk=k=0NAkzk
27 5 8 22 26 fmptco FPolyS*F*=zk=0NAkzk
28 2 27 eqtrid FPolySG=zk=0NAkzk
29 fzfid FPolySz0NFin
30 10 adantr FPolySzA:0
31 30 12 13 syl2an FPolySzk0NAk
32 expcl zk0zk
33 5 12 32 syl2an FPolySzk0Nzk
34 31 33 mulcld FPolySzk0NAkzk
35 29 34 fsumcj FPolySzk=0NAkzk=k=0NAkzk
36 31 33 cjmuld FPolySzk0NAkzk=Akzk
37 fvco3 A:0k0*Ak=Ak
38 30 12 37 syl2an FPolySzk0N*Ak=Ak
39 cjexp zk0zk=zk
40 5 12 39 syl2an FPolySzk0Nzk=zk
41 cjcj zz=z
42 41 ad2antlr FPolySzk0Nz=z
43 42 oveq1d FPolySzk0Nzk=zk
44 40 43 eqtr2d FPolySzk0Nzk=zk
45 38 44 oveq12d FPolySzk0N*Akzk=Akzk
46 36 45 eqtr4d FPolySzk0NAkzk=*Akzk
47 46 sumeq2dv FPolySzk=0NAkzk=k=0N*Akzk
48 35 47 eqtrd FPolySzk=0NAkzk=k=0N*Akzk
49 48 mpteq2dva FPolySzk=0NAkzk=zk=0N*Akzk
50 28 49 eqtrd FPolySG=zk=0N*Akzk