Metamath Proof Explorer


Theorem lsmssv

Description: Subgroup sum is a subset of the base. (Contributed by Mario Carneiro, 19-Apr-2016)

Ref Expression
Hypotheses lsmless2.v ⊢ B = Base G
lsmless2.s ⊢ ⊕ ˙ = LSSum ⁡ G
Assertion lsmssv ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B → T ⊕ ˙ U ⊆ B

Proof

Step Hyp Ref Expression
1 lsmless2.v ⊢ B = Base G
2 lsmless2.s ⊢ ⊕ ˙ = LSSum ⁡ G
3 eqid ⊢ + G = + G
4 1 3 2 lsmvalx ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B → T ⊕ ˙ U = ran ⁡ x ∈ T , y ∈ U ⟼ x + G y
5 simpl1 ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B ∧ x ∈ T ∧ y ∈ U → G ∈ Mnd
6 simp2 ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B → T ⊆ B
7 6 sselda ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B ∧ x ∈ T → x ∈ B
8 7 adantrr ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B ∧ x ∈ T ∧ y ∈ U → x ∈ B
9 simp3 ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B → U ⊆ B
10 9 sselda ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B ∧ y ∈ U → y ∈ B
11 10 adantrl ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B ∧ x ∈ T ∧ y ∈ U → y ∈ B
12 1 3 mndcl ⊢ G ∈ Mnd ∧ x ∈ B ∧ y ∈ B → x + G y ∈ B
13 5 8 11 12 syl3anc ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B ∧ x ∈ T ∧ y ∈ U → x + G y ∈ B
14 13 ralrimivva ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B → ∀ x ∈ T ∀ y ∈ U x + G y ∈ B
15 eqid ⊢ x ∈ T , y ∈ U ⟼ x + G y = x ∈ T , y ∈ U ⟼ x + G y
16 15 fmpo ⊢ ∀ x ∈ T ∀ y ∈ U x + G y ∈ B ↔ x ∈ T , y ∈ U ⟼ x + G y : T × U ⟶ B
17 14 16 sylib ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B → x ∈ T , y ∈ U ⟼ x + G y : T × U ⟶ B
18 17 frnd ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B → ran ⁡ x ∈ T , y ∈ U ⟼ x + G y ⊆ B
19 4 18 eqsstrd ⊢ G ∈ Mnd ∧ T ⊆ B ∧ U ⊆ B → T ⊕ ˙ U ⊆ B