Metamath Proof Explorer


Theorem nmmulg

Description: The norm of a group product, provided the ZZ -module is normed. (Contributed by Thierry Arnoux, 8-Nov-2017)

Ref Expression
Hypotheses nmmulg.x B = Base R
nmmulg.n N = norm R
nmmulg.z Z = ℤMod R
nmmulg.t · ˙ = R
Assertion nmmulg Z NrmMod M X B N M · ˙ X = M N X

Proof

Step Hyp Ref Expression
1 nmmulg.x B = Base R
2 nmmulg.n N = norm R
3 nmmulg.z Z = ℤMod R
4 nmmulg.t · ˙ = R
5 simp2 Z NrmMod M X B M
6 zringbas = Base ring
7 nlmlmod Z NrmMod Z LMod
8 3 zlmlmod R Abel Z LMod
9 7 8 sylibr Z NrmMod R Abel
10 9 3ad2ant1 Z NrmMod M X B R Abel
11 3 zlmsca R Abel ring = Scalar Z
12 10 11 syl Z NrmMod M X B ring = Scalar Z
13 12 fveq2d Z NrmMod M X B Base ring = Base Scalar Z
14 6 13 syl5eq Z NrmMod M X B = Base Scalar Z
15 5 14 eleqtrd Z NrmMod M X B M Base Scalar Z
16 3 1 zlmbas B = Base Z
17 eqid norm Z = norm Z
18 3 4 zlmvsca · ˙ = Z
19 eqid Scalar Z = Scalar Z
20 eqid Base Scalar Z = Base Scalar Z
21 eqid norm Scalar Z = norm Scalar Z
22 16 17 18 19 20 21 nmvs Z NrmMod M Base Scalar Z X B norm Z M · ˙ X = norm Scalar Z M norm Z X
23 15 22 syld3an2 Z NrmMod M X B norm Z M · ˙ X = norm Scalar Z M norm Z X
24 3 2 zlmnm R Abel N = norm Z
25 10 24 syl Z NrmMod M X B N = norm Z
26 25 fveq1d Z NrmMod M X B N M · ˙ X = norm Z M · ˙ X
27 zzsnm M M = norm ring M
28 27 3ad2ant2 Z NrmMod M X B M = norm ring M
29 12 fveq2d Z NrmMod M X B norm ring = norm Scalar Z
30 29 fveq1d Z NrmMod M X B norm ring M = norm Scalar Z M
31 28 30 eqtrd Z NrmMod M X B M = norm Scalar Z M
32 25 fveq1d Z NrmMod M X B N X = norm Z X
33 31 32 oveq12d Z NrmMod M X B M N X = norm Scalar Z M norm Z X
34 23 26 33 3eqtr4d Z NrmMod M X B N M · ˙ X = M N X