Metamath Proof Explorer


Theorem plyrecj

Description: A polynomial with real coefficients distributes under conjugation. (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Assertion plyrecj F Poly A F A = F A

Proof

Step Hyp Ref Expression
1 fzfid F Poly A 0 deg F Fin
2 0re 0
3 eqid coeff F = coeff F
4 3 coef2 F Poly 0 coeff F : 0
5 2 4 mpan2 F Poly coeff F : 0
6 5 adantr F Poly A coeff F : 0
7 elfznn0 x 0 deg F x 0
8 ffvelrn coeff F : 0 x 0 coeff F x
9 6 7 8 syl2an F Poly A x 0 deg F coeff F x
10 9 recnd F Poly A x 0 deg F coeff F x
11 simpr F Poly A A
12 expcl A x 0 A x
13 11 7 12 syl2an F Poly A x 0 deg F A x
14 10 13 mulcld F Poly A x 0 deg F coeff F x A x
15 1 14 fsumcj F Poly A x = 0 deg F coeff F x A x = x = 0 deg F coeff F x A x
16 10 13 cjmuld F Poly A x 0 deg F coeff F x A x = coeff F x A x
17 9 cjred F Poly A x 0 deg F coeff F x = coeff F x
18 cjexp A x 0 A x = A x
19 11 7 18 syl2an F Poly A x 0 deg F A x = A x
20 17 19 oveq12d F Poly A x 0 deg F coeff F x A x = coeff F x A x
21 16 20 eqtrd F Poly A x 0 deg F coeff F x A x = coeff F x A x
22 21 sumeq2dv F Poly A x = 0 deg F coeff F x A x = x = 0 deg F coeff F x A x
23 15 22 eqtrd F Poly A x = 0 deg F coeff F x A x = x = 0 deg F coeff F x A x
24 eqid deg F = deg F
25 3 24 coeid2 F Poly A F A = x = 0 deg F coeff F x A x
26 25 fveq2d F Poly A F A = x = 0 deg F coeff F x A x
27 cjcl A A
28 3 24 coeid2 F Poly A F A = x = 0 deg F coeff F x A x
29 27 28 sylan2 F Poly A F A = x = 0 deg F coeff F x A x
30 23 26 29 3eqtr4d F Poly A F A = F A