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=deg1R
deg1tm.k K=BaseR
deg1tm.p P=Poly1R
deg1tm.x X=var1R
deg1tm.m ·˙=P
deg1tm.n N=mulGrpP
deg1tm.e ×˙=N
deg1tm.z 0˙=0R
Assertion deg1tm RRingCKC0˙F0DC·˙F×˙X=F

Proof

Step Hyp Ref Expression
1 deg1tm.d D=deg1R
2 deg1tm.k K=BaseR
3 deg1tm.p P=Poly1R
4 deg1tm.x X=var1R
5 deg1tm.m ·˙=P
6 deg1tm.n N=mulGrpP
7 deg1tm.e ×˙=N
8 deg1tm.z 0˙=0R
9 eqid BaseP=BaseP
10 2 3 4 5 6 7 9 ply1tmcl RRingCKF0C·˙F×˙XBaseP
11 10 3adant2r RRingCKC0˙F0C·˙F×˙XBaseP
12 1 3 9 deg1xrcl C·˙F×˙XBasePDC·˙F×˙X*
13 11 12 syl RRingCKC0˙F0DC·˙F×˙X*
14 simp3 RRingCKC0˙F0F0
15 14 nn0red RRingCKC0˙F0F
16 15 rexrd RRingCKC0˙F0F*
17 1 2 3 4 5 6 7 deg1tmle RRingCKF0DC·˙F×˙XF
18 17 3adant2r RRingCKC0˙F0DC·˙F×˙XF
19 8 2 3 4 5 6 7 coe1tmfv1 RRingCKF0coe1C·˙F×˙XF=C
20 19 3adant2r RRingCKC0˙F0coe1C·˙F×˙XF=C
21 simp2r RRingCKC0˙F0C0˙
22 20 21 eqnetrd RRingCKC0˙F0coe1C·˙F×˙XF0˙
23 eqid coe1C·˙F×˙X=coe1C·˙F×˙X
24 1 3 9 8 23 deg1ge C·˙F×˙XBasePF0coe1C·˙F×˙XF0˙FDC·˙F×˙X
25 11 14 22 24 syl3anc RRingCKC0˙F0FDC·˙F×˙X
26 13 16 18 25 xrletrid RRingCKC0˙F0DC·˙F×˙X=F