Metamath Proof Explorer


Theorem lsmless1

Description: Subset implies subgroup sum subset. (Contributed by NM, 6-Feb-2014) (Revised by Mario Carneiro, 19-Apr-2016)

Ref Expression
Hypothesis lsmub1.p ˙ = LSSum G
Assertion lsmless1 T SubGrp G U SubGrp G S T S ˙ U T ˙ U

Proof

Step Hyp Ref Expression
1 lsmub1.p ˙ = LSSum G
2 subgrcl T SubGrp G G Grp
3 2 3ad2ant1 T SubGrp G U SubGrp G S T G Grp
4 eqid Base G = Base G
5 4 subgss T SubGrp G T Base G
6 5 3ad2ant1 T SubGrp G U SubGrp G S T T Base G
7 4 subgss U SubGrp G U Base G
8 7 3ad2ant2 T SubGrp G U SubGrp G S T U Base G
9 simp3 T SubGrp G U SubGrp G S T S T
10 4 1 lsmless1x G Grp T Base G U Base G S T S ˙ U T ˙ U
11 3 6 8 9 10 syl31anc T SubGrp G U SubGrp G S T S ˙ U T ˙ U