Metamath Proof Explorer


Theorem ldepsprlem

Description: Lemma for ldepspr . (Contributed by AV, 16-Apr-2019)

Ref Expression
Hypotheses snlindsntor.b B=BaseM
snlindsntor.r R=ScalarM
snlindsntor.s S=BaseR
snlindsntor.0 0˙=0R
snlindsntor.z Z=0M
snlindsntor.t ·˙=M
ldepsprlem.1 1˙=1R
ldepsprlem.n N=invgR
Assertion ldepsprlem MLModXBYBASX=A·˙Y1˙·˙X+MNA·˙Y=Z

Proof

Step Hyp Ref Expression
1 snlindsntor.b B=BaseM
2 snlindsntor.r R=ScalarM
3 snlindsntor.s S=BaseR
4 snlindsntor.0 0˙=0R
5 snlindsntor.z Z=0M
6 snlindsntor.t ·˙=M
7 ldepsprlem.1 1˙=1R
8 ldepsprlem.n N=invgR
9 oveq2 X=A·˙Y1˙·˙X=1˙·˙A·˙Y
10 9 oveq1d X=A·˙Y1˙·˙X+MNA·˙Y=1˙·˙A·˙Y+MNA·˙Y
11 simpl MLModXBYBASMLMod
12 2 3 7 lmod1cl MLMod1˙S
13 12 adantr MLModXBYBAS1˙S
14 simpr3 MLModXBYBASAS
15 simpr2 MLModXBYBASYB
16 eqid R=R
17 1 2 6 3 16 lmodvsass MLMod1˙SASYB1˙RA·˙Y=1˙·˙A·˙Y
18 11 13 14 15 17 syl13anc MLModXBYBAS1˙RA·˙Y=1˙·˙A·˙Y
19 18 eqcomd MLModXBYBAS1˙·˙A·˙Y=1˙RA·˙Y
20 19 oveq1d MLModXBYBAS1˙·˙A·˙Y+MNA·˙Y=1˙RA·˙Y+MNA·˙Y
21 2 lmodring MLModRRing
22 simp3 XBYBASAS
23 3 16 7 ringlidm RRingAS1˙RA=A
24 21 22 23 syl2an MLModXBYBAS1˙RA=A
25 24 oveq1d MLModXBYBAS1˙RA·˙Y=A·˙Y
26 25 oveq1d MLModXBYBAS1˙RA·˙Y+MNA·˙Y=A·˙Y+MNA·˙Y
27 2 lmodfgrp MLModRGrp
28 3 8 grpinvcl RGrpASNAS
29 27 22 28 syl2an MLModXBYBASNAS
30 eqid +M=+M
31 eqid +R=+R
32 1 30 2 6 3 31 lmodvsdir MLModASNASYBA+RNA·˙Y=A·˙Y+MNA·˙Y
33 11 14 29 15 32 syl13anc MLModXBYBASA+RNA·˙Y=A·˙Y+MNA·˙Y
34 3 31 4 8 grprinv RGrpASA+RNA=0˙
35 27 22 34 syl2an MLModXBYBASA+RNA=0˙
36 35 oveq1d MLModXBYBASA+RNA·˙Y=0˙·˙Y
37 1 2 6 4 5 lmod0vs MLModYB0˙·˙Y=Z
38 37 3ad2antr2 MLModXBYBAS0˙·˙Y=Z
39 36 38 eqtrd MLModXBYBASA+RNA·˙Y=Z
40 26 33 39 3eqtr2d MLModXBYBAS1˙RA·˙Y+MNA·˙Y=Z
41 20 40 eqtrd MLModXBYBAS1˙·˙A·˙Y+MNA·˙Y=Z
42 10 41 sylan9eqr MLModXBYBASX=A·˙Y1˙·˙X+MNA·˙Y=Z
43 42 ex MLModXBYBASX=A·˙Y1˙·˙X+MNA·˙Y=Z