Metamath Proof Explorer


Theorem lsmless12

Description: Subset implies subgroup sum subset. (Contributed by NM, 14-Jan-2015) (Revised by Mario Carneiro, 19-Apr-2016)

Ref Expression
Hypothesis lsmub1.p ˙ = LSSum G
Assertion lsmless12 S SubGrp G U SubGrp G R S T U R ˙ T S ˙ U

Proof

Step Hyp Ref Expression
1 lsmub1.p ˙ = LSSum G
2 subgrcl S SubGrp G G Grp
3 2 ad2antrr S SubGrp G U SubGrp G R S T U G Grp
4 eqid Base G = Base G
5 4 subgss S SubGrp G S Base G
6 5 ad2antrr S SubGrp G U SubGrp G R S T U S Base G
7 simprr S SubGrp G U SubGrp G R S T U T U
8 4 subgss U SubGrp G U Base G
9 8 ad2antlr S SubGrp G U SubGrp G R S T U U Base G
10 7 9 sstrd S SubGrp G U SubGrp G R S T U T Base G
11 simprl S SubGrp G U SubGrp G R S T U R S
12 4 1 lsmless1x G Grp S Base G T Base G R S R ˙ T S ˙ T
13 3 6 10 11 12 syl31anc S SubGrp G U SubGrp G R S T U R ˙ T S ˙ T
14 simpll S SubGrp G U SubGrp G R S T U S SubGrp G
15 simplr S SubGrp G U SubGrp G R S T U U SubGrp G
16 1 lsmless2 S SubGrp G U SubGrp G T U S ˙ T S ˙ U
17 14 15 7 16 syl3anc S SubGrp G U SubGrp G R S T U S ˙ T S ˙ U
18 13 17 sstrd S SubGrp G U SubGrp G R S T U R ˙ T S ˙ U