Metamath Proof Explorer


Theorem lsmss2

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 lsmss2 ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ U ⊆ T → T ⊕ ˙ U = T

Proof

Step Hyp Ref Expression
1 lsmub1.p ⊢ ⊕ ˙ = LSSum ⁡ G
2 ssid ⊢ T ⊆ T
3 1 lsmlub ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ T ∈ SubGrp ⁡ G → T ⊆ T ∧ U ⊆ T ↔ T ⊕ ˙ U ⊆ T
4 3 3anidm13 ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → T ⊆ T ∧ U ⊆ T ↔ T ⊕ ˙ U ⊆ T
5 4 biimpd ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → T ⊆ T ∧ U ⊆ T → T ⊕ ˙ U ⊆ T
6 2 5 mpani ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → U ⊆ T → T ⊕ ˙ U ⊆ T
7 6 3impia ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ U ⊆ T → T ⊕ ˙ U ⊆ T
8 1 lsmub1 ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G → T ⊆ T ⊕ ˙ U
9 8 3adant3 ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ U ⊆ T → T ⊆ T ⊕ ˙ U
10 7 9 eqssd ⊢ T ∈ SubGrp ⁡ G ∧ U ∈ SubGrp ⁡ G ∧ U ⊆ T → T ⊕ ˙ U = T