Metamath Proof Explorer


Theorem lsmss1

Description: Subgroup sum with a subset. (Contributed by NM, 27-Mar-2014) (Revised by Mario Carneiro, 19-Apr-2016)

Ref Expression
Hypothesis lsmub1.p ⊢ ⊕ ˙ = LSSum ⁡ G
Assertion lsmss1 ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ T ⊆ U → T ⊕ ˙ U = U

Proof

Step Hyp Ref Expression
1 lsmub1.p ⊢ ⊕ ˙ = LSSum ⁡ G
2 ssid ⊢ U ⊆ U
3 1 lsmlub ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → T ⊆ U ∧ U ⊆ U ↔ T ⊕ ˙ U ⊆ U
4 3 3anidm23 ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → T ⊆ U ∧ U ⊆ U ↔ T ⊕ ˙ U ⊆ U
5 4 biimpd ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → T ⊆ U ∧ U ⊆ U → T ⊕ ˙ U ⊆ U
6 2 5 mpan2i ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → T ⊆ U → T ⊕ ˙ U ⊆ U
7 6 3impia ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ T ⊆ U → T ⊕ ˙ U ⊆ U
8 1 lsmub2 ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → U ⊆ T ⊕ ˙ U
9 8 3adant3 ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ T ⊆ U → U ⊆ T ⊕ ˙ U
10 7 9 eqssd ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ T ⊆ U → T ⊕ ˙ U = U