Metamath Proof Explorer


Theorem frlmsubgval

Description: Subtraction in a free module. (Contributed by Thierry Arnoux, 30-Jun-2019)

Ref Expression
Hypotheses frlmsubval.y Y=RfreeLModI
frlmsubval.b B=BaseY
frlmsubval.r φRRing
frlmsubval.i φIW
frlmsubval.f φFB
frlmsubval.g φGB
frlmsubval.a -˙=-R
frlmsubval.p M=-Y
Assertion frlmsubgval φFMG=F-˙fG

Proof

Step Hyp Ref Expression
1 frlmsubval.y Y=RfreeLModI
2 frlmsubval.b B=BaseY
3 frlmsubval.r φRRing
4 frlmsubval.i φIW
5 frlmsubval.f φFB
6 frlmsubval.g φGB
7 frlmsubval.a -˙=-R
8 frlmsubval.p M=-Y
9 1 2 frlmpws RRingIWY=ringLModR𝑠I𝑠B
10 3 4 9 syl2anc φY=ringLModR𝑠I𝑠B
11 10 fveq2d φ-Y=-ringLModR𝑠I𝑠B
12 8 11 eqtrid φM=-ringLModR𝑠I𝑠B
13 12 oveqd φFMG=F-ringLModR𝑠I𝑠BG
14 rlmlmod RRingringLModRLMod
15 3 14 syl φringLModRLMod
16 eqid ringLModR𝑠I=ringLModR𝑠I
17 16 pwslmod ringLModRLModIWringLModR𝑠ILMod
18 15 4 17 syl2anc φringLModR𝑠ILMod
19 eqid LSubSpringLModR𝑠I=LSubSpringLModR𝑠I
20 1 2 19 frlmlss RRingIWBLSubSpringLModR𝑠I
21 3 4 20 syl2anc φBLSubSpringLModR𝑠I
22 19 lsssubg ringLModR𝑠ILModBLSubSpringLModR𝑠IBSubGrpringLModR𝑠I
23 18 21 22 syl2anc φBSubGrpringLModR𝑠I
24 eqid -ringLModR𝑠I=-ringLModR𝑠I
25 eqid ringLModR𝑠I𝑠B=ringLModR𝑠I𝑠B
26 eqid -ringLModR𝑠I𝑠B=-ringLModR𝑠I𝑠B
27 24 25 26 subgsub BSubGrpringLModR𝑠IFBGBF-ringLModR𝑠IG=F-ringLModR𝑠I𝑠BG
28 23 5 6 27 syl3anc φF-ringLModR𝑠IG=F-ringLModR𝑠I𝑠BG
29 lmodgrp ringLModRLModringLModRGrp
30 3 14 29 3syl φringLModRGrp
31 eqid BaseR=BaseR
32 1 31 2 frlmbasmap IWFBFBaseRI
33 4 5 32 syl2anc φFBaseRI
34 rlmbas BaseR=BaseringLModR
35 16 34 pwsbas ringLModRGrpIWBaseRI=BaseringLModR𝑠I
36 30 4 35 syl2anc φBaseRI=BaseringLModR𝑠I
37 33 36 eleqtrd φFBaseringLModR𝑠I
38 1 31 2 frlmbasmap IWGBGBaseRI
39 4 6 38 syl2anc φGBaseRI
40 39 36 eleqtrd φGBaseringLModR𝑠I
41 eqid BaseringLModR𝑠I=BaseringLModR𝑠I
42 rlmsub -R=-ringLModR
43 7 42 eqtri -˙=-ringLModR
44 16 41 43 24 pwssub ringLModRGrpIWFBaseringLModR𝑠IGBaseringLModR𝑠IF-ringLModR𝑠IG=F-˙fG
45 30 4 37 40 44 syl22anc φF-ringLModR𝑠IG=F-˙fG
46 13 28 45 3eqtr2d φFMG=F-˙fG