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