Metamath Proof Explorer


Theorem dgreq0

Description: The leading coefficient of a polynomial is nonzero, unless the entire polynomial is zero. (Contributed by Mario Carneiro, 22-Jul-2014) (Proof shortened by Fan Zheng, 21-Jun-2016)

Ref Expression
Hypotheses dgreq0.1 N = deg F
dgreq0.2 A = coeff F
Assertion dgreq0 F Poly S F = 0 𝑝 A N = 0

Proof

Step Hyp Ref Expression
1 dgreq0.1 N = deg F
2 dgreq0.2 A = coeff F
3 fveq2 F = 0 𝑝 coeff F = coeff 0 𝑝
4 2 3 eqtrid F = 0 𝑝 A = coeff 0 𝑝
5 coe0 coeff 0 𝑝 = 0 × 0
6 4 5 eqtrdi F = 0 𝑝 A = 0 × 0
7 fveq2 F = 0 𝑝 deg F = deg 0 𝑝
8 1 7 eqtrid F = 0 𝑝 N = deg 0 𝑝
9 dgr0 deg 0 𝑝 = 0
10 8 9 eqtrdi F = 0 𝑝 N = 0
11 6 10 fveq12d F = 0 𝑝 A N = 0 × 0 0
12 0nn0 0 0
13 fvconst2g 0 0 0 0 0 × 0 0 = 0
14 12 12 13 mp2an 0 × 0 0 = 0
15 11 14 eqtrdi F = 0 𝑝 A N = 0
16 2 coefv0 F Poly S F 0 = A 0
17 16 adantr F Poly S A N = 0 F 0 = A 0
18 simpr F Poly S A N = 0 N N
19 18 nnred F Poly S A N = 0 N N
20 19 ltm1d F Poly S A N = 0 N N 1 < N
21 nnre N N
22 21 adantl F Poly S A N = 0 N N
23 peano2rem N N 1
24 22 23 syl F Poly S A N = 0 N N 1
25 simpll F Poly S A N = 0 N F Poly S
26 nnm1nn0 N N 1 0
27 26 adantl F Poly S A N = 0 N N 1 0
28 2 1 dgrub F Poly S k 0 A k 0 k N
29 28 3expia F Poly S k 0 A k 0 k N
30 29 ad2ant2rl F Poly S A N = 0 N k 0 A k 0 k N
31 simplr F Poly S A N = 0 N k 0 A N = 0
32 fveqeq2 N = k A N = 0 A k = 0
33 31 32 syl5ibcom F Poly S A N = 0 N k 0 N = k A k = 0
34 33 necon3d F Poly S A N = 0 N k 0 A k 0 N k
35 30 34 jcad F Poly S A N = 0 N k 0 A k 0 k N N k
36 nn0re k 0 k
37 36 ad2antll F Poly S A N = 0 N k 0 k
38 21 ad2antrl F Poly S A N = 0 N k 0 N
39 37 38 ltlend F Poly S A N = 0 N k 0 k < N k N N k
40 nn0z k 0 k
41 40 ad2antll F Poly S A N = 0 N k 0 k
42 nnz N N
43 42 ad2antrl F Poly S A N = 0 N k 0 N
44 zltlem1 k N k < N k N 1
45 41 43 44 syl2anc F Poly S A N = 0 N k 0 k < N k N 1
46 39 45 bitr3d F Poly S A N = 0 N k 0 k N N k k N 1
47 35 46 sylibd F Poly S A N = 0 N k 0 A k 0 k N 1
48 47 expr F Poly S A N = 0 N k 0 A k 0 k N 1
49 48 ralrimiv F Poly S A N = 0 N k 0 A k 0 k N 1
50 2 coef3 F Poly S A : 0
51 50 ad2antrr F Poly S A N = 0 N A : 0
52 plyco0 N 1 0 A : 0 A N - 1 + 1 = 0 k 0 A k 0 k N 1
53 27 51 52 syl2anc F Poly S A N = 0 N A N - 1 + 1 = 0 k 0 A k 0 k N 1
54 49 53 mpbird F Poly S A N = 0 N A N - 1 + 1 = 0
55 2 1 dgrlb F Poly S N 1 0 A N - 1 + 1 = 0 N N 1
56 25 27 54 55 syl3anc F Poly S A N = 0 N N N 1
57 22 24 56 lensymd F Poly S A N = 0 N ¬ N 1 < N
58 20 57 pm2.65da F Poly S A N = 0 ¬ N
59 dgrcl F Poly S deg F 0
60 1 59 eqeltrid F Poly S N 0
61 60 adantr F Poly S A N = 0 N 0
62 elnn0 N 0 N N = 0
63 61 62 sylib F Poly S A N = 0 N N = 0
64 63 ord F Poly S A N = 0 ¬ N N = 0
65 58 64 mpd F Poly S A N = 0 N = 0
66 65 fveq2d F Poly S A N = 0 A N = A 0
67 simpr F Poly S A N = 0 A N = 0
68 17 66 67 3eqtr2d F Poly S A N = 0 F 0 = 0
69 68 sneqd F Poly S A N = 0 F 0 = 0
70 69 xpeq2d F Poly S A N = 0 × F 0 = × 0
71 1 65 eqtr3id F Poly S A N = 0 deg F = 0
72 0dgrb F Poly S deg F = 0 F = × F 0
73 72 adantr F Poly S A N = 0 deg F = 0 F = × F 0
74 71 73 mpbid F Poly S A N = 0 F = × F 0
75 df-0p 0 𝑝 = × 0
76 75 a1i F Poly S A N = 0 0 𝑝 = × 0
77 70 74 76 3eqtr4d F Poly S A N = 0 F = 0 𝑝
78 77 ex F Poly S A N = 0 F = 0 𝑝
79 15 78 impbid2 F Poly S F = 0 𝑝 A N = 0