Metamath Proof Explorer


Theorem lsmless2

Description: Subset implies subgroup sum subset. (Contributed by NM, 25-Feb-2014) (Revised by Mario Carneiro, 19-Apr-2016)

Ref Expression
Hypothesis lsmub1.p ˙ = LSSum G
Assertion lsmless2 S SubGrp G U SubGrp G T U S ˙ T S ˙ U

Proof

Step Hyp Ref Expression
1 lsmub1.p ˙ = LSSum G
2 subgrcl S SubGrp G G Grp
3 2 3ad2ant1 S SubGrp G U SubGrp G T U G Grp
4 eqid Base G = Base G
5 4 subgss S SubGrp G S Base G
6 5 3ad2ant1 S SubGrp G U SubGrp G T U S Base G
7 4 subgss U SubGrp G U Base G
8 7 3ad2ant2 S SubGrp G U SubGrp G T U U Base G
9 simp3 S SubGrp G U SubGrp G T U T U
10 4 1 lsmless2x G Grp S Base G U Base G T U S ˙ T S ˙ U
11 3 6 8 9 10 syl31anc S SubGrp G U SubGrp G T U S ˙ T S ˙ U