Metamath Proof Explorer


Theorem deg1tm

Description: Exact degree of a polynomial term. (Contributed by Stefan O'Rear, 27-Mar-2015)

Ref Expression
Hypotheses deg1tm.d D = deg 1 R
deg1tm.k K = Base R
deg1tm.p P = Poly 1 R
deg1tm.x X = var 1 R
deg1tm.m · ˙ = P
deg1tm.n N = mulGrp P
deg1tm.e × ˙ = N
deg1tm.z 0 ˙ = 0 R
Assertion deg1tm R Ring C K C 0 ˙ F 0 D C · ˙ F × ˙ X = F

Proof

Step Hyp Ref Expression
1 deg1tm.d D = deg 1 R
2 deg1tm.k K = Base R
3 deg1tm.p P = Poly 1 R
4 deg1tm.x X = var 1 R
5 deg1tm.m · ˙ = P
6 deg1tm.n N = mulGrp P
7 deg1tm.e × ˙ = N
8 deg1tm.z 0 ˙ = 0 R
9 eqid Base P = Base P
10 2 3 4 5 6 7 9 ply1tmcl R Ring C K F 0 C · ˙ F × ˙ X Base P
11 10 3adant2r R Ring C K C 0 ˙ F 0 C · ˙ F × ˙ X Base P
12 1 3 9 deg1xrcl C · ˙ F × ˙ X Base P D C · ˙ F × ˙ X *
13 11 12 syl R Ring C K C 0 ˙ F 0 D C · ˙ F × ˙ X *
14 simp3 R Ring C K C 0 ˙ F 0 F 0
15 14 nn0red R Ring C K C 0 ˙ F 0 F
16 15 rexrd R Ring C K C 0 ˙ F 0 F *
17 1 2 3 4 5 6 7 deg1tmle R Ring C K F 0 D C · ˙ F × ˙ X F
18 17 3adant2r R Ring C K C 0 ˙ F 0 D C · ˙ F × ˙ X F
19 8 2 3 4 5 6 7 coe1tmfv1 R Ring C K F 0 coe 1 C · ˙ F × ˙ X F = C
20 19 3adant2r R Ring C K C 0 ˙ F 0 coe 1 C · ˙ F × ˙ X F = C
21 simp2r R Ring C K C 0 ˙ F 0 C 0 ˙
22 20 21 eqnetrd R Ring C K C 0 ˙ F 0 coe 1 C · ˙ F × ˙ X F 0 ˙
23 eqid coe 1 C · ˙ F × ˙ X = coe 1 C · ˙ F × ˙ X
24 1 3 9 8 23 deg1ge C · ˙ F × ˙ X Base P F 0 coe 1 C · ˙ F × ˙ X F 0 ˙ F D C · ˙ F × ˙ X
25 11 14 22 24 syl3anc R Ring C K C 0 ˙ F 0 F D C · ˙ F × ˙ X
26 13 16 18 25 xrletrid R Ring C K C 0 ˙ F 0 D C · ˙ F × ˙ X = F