Metamath Proof Explorer


Theorem ogrpinv0le

Description: In an ordered group, the ordering is compatible with group inverse. (Contributed by Thierry Arnoux, 3-Sep-2018)

Ref Expression
Hypotheses ogrpsub.0 B = Base G
ogrpsub.1 ˙ = G
ogrpinv.2 I = inv g G
ogrpinv.3 0 ˙ = 0 G
Assertion ogrpinv0le G oGrp X B 0 ˙ ˙ X I X ˙ 0 ˙

Proof

Step Hyp Ref Expression
1 ogrpsub.0 B = Base G
2 ogrpsub.1 ˙ = G
3 ogrpinv.2 I = inv g G
4 ogrpinv.3 0 ˙ = 0 G
5 isogrp G oGrp G Grp G oMnd
6 5 simprbi G oGrp G oMnd
7 6 ad2antrr G oGrp X B 0 ˙ ˙ X G oMnd
8 omndmnd G oMnd G Mnd
9 1 4 mndidcl G Mnd 0 ˙ B
10 7 8 9 3syl G oGrp X B 0 ˙ ˙ X 0 ˙ B
11 simplr G oGrp X B 0 ˙ ˙ X X B
12 ogrpgrp G oGrp G Grp
13 12 ad2antrr G oGrp X B 0 ˙ ˙ X G Grp
14 1 3 grpinvcl G Grp X B I X B
15 13 11 14 syl2anc G oGrp X B 0 ˙ ˙ X I X B
16 simpr G oGrp X B 0 ˙ ˙ X 0 ˙ ˙ X
17 eqid + G = + G
18 1 2 17 omndadd G oMnd 0 ˙ B X B I X B 0 ˙ ˙ X 0 ˙ + G I X ˙ X + G I X
19 7 10 11 15 16 18 syl131anc G oGrp X B 0 ˙ ˙ X 0 ˙ + G I X ˙ X + G I X
20 1 17 4 grplid G Grp I X B 0 ˙ + G I X = I X
21 13 15 20 syl2anc G oGrp X B 0 ˙ ˙ X 0 ˙ + G I X = I X
22 1 17 4 3 grprinv G Grp X B X + G I X = 0 ˙
23 13 11 22 syl2anc G oGrp X B 0 ˙ ˙ X X + G I X = 0 ˙
24 19 21 23 3brtr3d G oGrp X B 0 ˙ ˙ X I X ˙ 0 ˙
25 6 ad2antrr G oGrp X B I X ˙ 0 ˙ G oMnd
26 12 ad2antrr G oGrp X B I X ˙ 0 ˙ G Grp
27 simplr G oGrp X B I X ˙ 0 ˙ X B
28 26 27 14 syl2anc G oGrp X B I X ˙ 0 ˙ I X B
29 25 8 9 3syl G oGrp X B I X ˙ 0 ˙ 0 ˙ B
30 simpr G oGrp X B I X ˙ 0 ˙ I X ˙ 0 ˙
31 1 2 17 omndadd G oMnd I X B 0 ˙ B X B I X ˙ 0 ˙ I X + G X ˙ 0 ˙ + G X
32 25 28 29 27 30 31 syl131anc G oGrp X B I X ˙ 0 ˙ I X + G X ˙ 0 ˙ + G X
33 1 17 4 3 grplinv G Grp X B I X + G X = 0 ˙
34 26 27 33 syl2anc G oGrp X B I X ˙ 0 ˙ I X + G X = 0 ˙
35 1 17 4 grplid G Grp X B 0 ˙ + G X = X
36 26 27 35 syl2anc G oGrp X B I X ˙ 0 ˙ 0 ˙ + G X = X
37 32 34 36 3brtr3d G oGrp X B I X ˙ 0 ˙ 0 ˙ ˙ X
38 24 37 impbida G oGrp X B 0 ˙ ˙ X I X ˙ 0 ˙