Metamath Proof Explorer


Theorem coecj

Description: Double conjugation of a polynomial causes the coefficients to be conjugated. (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Hypotheses plycj.2 G = * F *
coecj.3 A = coeff F
Assertion coecj F Poly S coeff G = * A

Proof

Step Hyp Ref Expression
1 plycj.2 G = * F *
2 coecj.3 A = coeff F
3 cjcl x x
4 3 adantl F Poly S x x
5 plyssc Poly S Poly
6 5 sseli F Poly S F Poly
7 1 4 6 plycj F Poly S G Poly
8 dgrcl F Poly S deg F 0
9 cjf * :
10 2 coef3 F Poly S A : 0
11 fco * : A : 0 * A : 0
12 9 10 11 sylancr F Poly S * A : 0
13 fvco3 A : 0 k 0 * A k = A k
14 10 13 sylan F Poly S k 0 * A k = A k
15 cj0 0 = 0
16 15 eqcomi 0 = 0
17 16 a1i F Poly S k 0 0 = 0
18 14 17 eqeq12d F Poly S k 0 * A k = 0 A k = 0
19 10 ffvelcdmda F Poly S k 0 A k
20 0cnd F Poly S k 0 0
21 cj11 A k 0 A k = 0 A k = 0
22 19 20 21 syl2anc F Poly S k 0 A k = 0 A k = 0
23 18 22 bitrd F Poly S k 0 * A k = 0 A k = 0
24 23 necon3bid F Poly S k 0 * A k 0 A k 0
25 eqid deg F = deg F
26 2 25 dgrub2 F Poly S A deg F + 1 = 0
27 plyco0 deg F 0 A : 0 A deg F + 1 = 0 k 0 A k 0 k deg F
28 8 10 27 syl2anc F Poly S A deg F + 1 = 0 k 0 A k 0 k deg F
29 26 28 mpbid F Poly S k 0 A k 0 k deg F
30 29 r19.21bi F Poly S k 0 A k 0 k deg F
31 24 30 sylbid F Poly S k 0 * A k 0 k deg F
32 31 ralrimiva F Poly S k 0 * A k 0 k deg F
33 plyco0 deg F 0 * A : 0 * A deg F + 1 = 0 k 0 * A k 0 k deg F
34 8 12 33 syl2anc F Poly S * A deg F + 1 = 0 k 0 * A k 0 k deg F
35 32 34 mpbird F Poly S * A deg F + 1 = 0
36 25 1 2 plycjlem F Poly S G = y z = 0 deg F * A z y z
37 7 8 12 35 36 coeeq F Poly S coeff G = * A