Metamath Proof Explorer


Theorem pwssub

Description: Subtraction in a group power. (Contributed by Mario Carneiro, 12-Jan-2015)

Ref Expression
Hypotheses pwsgrp.y Y=R𝑠I
pwsinvg.b B=BaseY
pwssub.m M=-R
pwssub.n -˙=-Y
Assertion pwssub RGrpIVFBGBF-˙G=FMfG

Proof

Step Hyp Ref Expression
1 pwsgrp.y Y=R𝑠I
2 pwsinvg.b B=BaseY
3 pwssub.m M=-R
4 pwssub.n -˙=-Y
5 simplr RGrpIVFBGBIV
6 eqid BaseR=BaseR
7 simpll RGrpIVFBGBRGrp
8 simprl RGrpIVFBGBFB
9 1 6 2 7 5 8 pwselbas RGrpIVFBGBF:IBaseR
10 9 ffvelrnda RGrpIVFBGBxIFxBaseR
11 fvexd RGrpIVFBGBxIinvgRGxV
12 9 feqmptd RGrpIVFBGBF=xIFx
13 simprr RGrpIVFBGBGB
14 eqid invgR=invgR
15 eqid invgY=invgY
16 1 2 14 15 pwsinvg RGrpIVGBinvgYG=invgRG
17 7 5 13 16 syl3anc RGrpIVFBGBinvgYG=invgRG
18 1 6 2 7 5 13 pwselbas RGrpIVFBGBG:IBaseR
19 18 ffvelrnda RGrpIVFBGBxIGxBaseR
20 18 feqmptd RGrpIVFBGBG=xIGx
21 6 14 grpinvf RGrpinvgR:BaseRBaseR
22 21 ad2antrr RGrpIVFBGBinvgR:BaseRBaseR
23 22 feqmptd RGrpIVFBGBinvgR=yBaseRinvgRy
24 fveq2 y=GxinvgRy=invgRGx
25 19 20 23 24 fmptco RGrpIVFBGBinvgRG=xIinvgRGx
26 17 25 eqtrd RGrpIVFBGBinvgYG=xIinvgRGx
27 5 10 11 12 26 offval2 RGrpIVFBGBF+RfinvgYG=xIFx+RinvgRGx
28 1 pwsgrp RGrpIVYGrp
29 2 15 grpinvcl YGrpGBinvgYGB
30 28 13 29 syl2an2r RGrpIVFBGBinvgYGB
31 eqid +R=+R
32 eqid +Y=+Y
33 1 2 7 5 8 30 31 32 pwsplusgval RGrpIVFBGBF+YinvgYG=F+RfinvgYG
34 6 31 14 3 grpsubval FxBaseRGxBaseRFxMGx=Fx+RinvgRGx
35 10 19 34 syl2anc RGrpIVFBGBxIFxMGx=Fx+RinvgRGx
36 35 mpteq2dva RGrpIVFBGBxIFxMGx=xIFx+RinvgRGx
37 27 33 36 3eqtr4d RGrpIVFBGBF+YinvgYG=xIFxMGx
38 2 32 15 4 grpsubval FBGBF-˙G=F+YinvgYG
39 38 adantl RGrpIVFBGBF-˙G=F+YinvgYG
40 5 10 19 12 20 offval2 RGrpIVFBGBFMfG=xIFxMGx
41 37 39 40 3eqtr4d RGrpIVFBGBF-˙G=FMfG