Metamath Proof Explorer


Theorem pm2mpghmlem1

Description: Lemma 1 for pm2mpghm . (Contributed by AV, 15-Oct-2019) (Revised by AV, 4-Dec-2019)

Ref Expression
Hypotheses pm2mpfo.p P = Poly 1 R
pm2mpfo.c C = N Mat P
pm2mpfo.b B = Base C
pm2mpfo.m ˙ = Q
pm2mpfo.e × ˙ = mulGrp Q
pm2mpfo.x X = var 1 A
pm2mpfo.a A = N Mat R
pm2mpfo.q Q = Poly 1 A
pm2mpfo.l L = Base Q
Assertion pm2mpghmlem1 N Fin R Ring M B K 0 M decompPMat K ˙ K × ˙ X L

Proof

Step Hyp Ref Expression
1 pm2mpfo.p P = Poly 1 R
2 pm2mpfo.c C = N Mat P
3 pm2mpfo.b B = Base C
4 pm2mpfo.m ˙ = Q
5 pm2mpfo.e × ˙ = mulGrp Q
6 pm2mpfo.x X = var 1 A
7 pm2mpfo.a A = N Mat R
8 pm2mpfo.q Q = Poly 1 A
9 pm2mpfo.l L = Base Q
10 7 matring N Fin R Ring A Ring
11 10 3adant3 N Fin R Ring M B A Ring
12 11 adantr N Fin R Ring M B K 0 A Ring
13 simpl2 N Fin R Ring M B K 0 R Ring
14 simpl3 N Fin R Ring M B K 0 M B
15 simpr N Fin R Ring M B K 0 K 0
16 eqid Base A = Base A
17 1 2 3 7 16 decpmatcl R Ring M B K 0 M decompPMat K Base A
18 13 14 15 17 syl3anc N Fin R Ring M B K 0 M decompPMat K Base A
19 eqid mulGrp Q = mulGrp Q
20 16 8 6 4 19 5 9 ply1tmcl A Ring M decompPMat K Base A K 0 M decompPMat K ˙ K × ˙ X L
21 12 18 15 20 syl3anc N Fin R Ring M B K 0 M decompPMat K ˙ K × ˙ X L