Metamath Proof Explorer


Theorem dgrlb

Description: If all the coefficients above M are zero, then the degree of F is at most M . (Contributed by Mario Carneiro, 22-Jul-2014)

Ref Expression
Hypotheses dgrub.1 A = coeff F
dgrub.2 N = deg F
Assertion dgrlb F Poly S M 0 A M + 1 = 0 N M

Proof

Step Hyp Ref Expression
1 dgrub.1 A = coeff F
2 dgrub.2 N = deg F
3 dgrcl F Poly S deg F 0
4 2 3 eqeltrid F Poly S N 0
5 4 3ad2ant1 F Poly S M 0 A M + 1 = 0 N 0
6 5 nn0red F Poly S M 0 A M + 1 = 0 N
7 simp2 F Poly S M 0 A M + 1 = 0 M 0
8 7 nn0red F Poly S M 0 A M + 1 = 0 M
9 1 dgrlem F Poly S A : 0 S 0 n x A -1 0 x n
10 9 simpld F Poly S A : 0 S 0
11 10 3ad2ant1 F Poly S M 0 A M + 1 = 0 A : 0 S 0
12 ffn A : 0 S 0 A Fn 0
13 elpreima A Fn 0 y A -1 0 y 0 A y 0
14 11 12 13 3syl F Poly S M 0 A M + 1 = 0 y A -1 0 y 0 A y 0
15 14 biimpa F Poly S M 0 A M + 1 = 0 y A -1 0 y 0 A y 0
16 15 simpld F Poly S M 0 A M + 1 = 0 y A -1 0 y 0
17 16 nn0red F Poly S M 0 A M + 1 = 0 y A -1 0 y
18 8 adantr F Poly S M 0 A M + 1 = 0 y A -1 0 M
19 eldifsni A y 0 A y 0
20 15 19 simpl2im F Poly S M 0 A M + 1 = 0 y A -1 0 A y 0
21 simp3 F Poly S M 0 A M + 1 = 0 A M + 1 = 0
22 1 coef3 F Poly S A : 0
23 22 3ad2ant1 F Poly S M 0 A M + 1 = 0 A : 0
24 plyco0 M 0 A : 0 A M + 1 = 0 y 0 A y 0 y M
25 7 23 24 syl2anc F Poly S M 0 A M + 1 = 0 A M + 1 = 0 y 0 A y 0 y M
26 21 25 mpbid F Poly S M 0 A M + 1 = 0 y 0 A y 0 y M
27 26 r19.21bi F Poly S M 0 A M + 1 = 0 y 0 A y 0 y M
28 16 27 syldan F Poly S M 0 A M + 1 = 0 y A -1 0 A y 0 y M
29 20 28 mpd F Poly S M 0 A M + 1 = 0 y A -1 0 y M
30 17 18 29 lensymd F Poly S M 0 A M + 1 = 0 y A -1 0 ¬ M < y
31 30 ralrimiva F Poly S M 0 A M + 1 = 0 y A -1 0 ¬ M < y
32 nn0ssre 0
33 ltso < Or
34 soss 0 < Or < Or 0
35 32 33 34 mp2 < Or 0
36 35 a1i F Poly S M 0 A M + 1 = 0 < Or 0
37 0zd F Poly S 0
38 cnvimass A -1 0 dom A
39 38 10 fssdm F Poly S A -1 0 0
40 9 simprd F Poly S n x A -1 0 x n
41 nn0uz 0 = 0
42 41 uzsupss 0 A -1 0 0 n x A -1 0 x n n 0 x A -1 0 ¬ n < x x 0 x < n y A -1 0 x < y
43 37 39 40 42 syl3anc F Poly S n 0 x A -1 0 ¬ n < x x 0 x < n y A -1 0 x < y
44 43 3ad2ant1 F Poly S M 0 A M + 1 = 0 n 0 x A -1 0 ¬ n < x x 0 x < n y A -1 0 x < y
45 36 44 supnub F Poly S M 0 A M + 1 = 0 M 0 y A -1 0 ¬ M < y ¬ M < sup A -1 0 0 <
46 7 31 45 mp2and F Poly S M 0 A M + 1 = 0 ¬ M < sup A -1 0 0 <
47 1 dgrval F Poly S deg F = sup A -1 0 0 <
48 2 47 eqtrid F Poly S N = sup A -1 0 0 <
49 48 3ad2ant1 F Poly S M 0 A M + 1 = 0 N = sup A -1 0 0 <
50 49 breq2d F Poly S M 0 A M + 1 = 0 M < N M < sup A -1 0 0 <
51 46 50 mtbird F Poly S M 0 A M + 1 = 0 ¬ M < N
52 6 8 51 nltled F Poly S M 0 A M + 1 = 0 N M