Metamath Proof Explorer


Theorem lsmdisj

Description: Disjointness from a subgroup sum. (Contributed by Mario Carneiro, 21-Apr-2016)

Ref Expression
Hypotheses lsmcntz.p ⊢ ⊕ ˙ = LSSum ⁡ G
lsmcntz.s ⊢ φ → S ∈ SubGrp ⁡ G
lsmcntz.t ⊢ φ → T ∈ SubGrp ⁡ G
lsmcntz.u ⊢ φ → U ∈ SubGrp ⁡ G
lsmdisj.o ⊢ 0 ˙ = 0 G
lsmdisj.i ⊢ φ → S ⊕ ˙ T ∩ U = 0 ˙
Assertion lsmdisj ⊢ φ → S ∩ U = 0 ˙ ∧ T ∩ U = 0 ˙

Proof

Step Hyp Ref Expression
1 lsmcntz.p ⊢ ⊕ ˙ = LSSum ⁡ G
2 lsmcntz.s ⊢ φ → S ∈ SubGrp ⁡ G
3 lsmcntz.t ⊢ φ → T ∈ SubGrp ⁡ G
4 lsmcntz.u ⊢ φ → U ∈ SubGrp ⁡ G
5 lsmdisj.o ⊢ 0 ˙ = 0 G
6 lsmdisj.i ⊢ φ → S ⊕ ˙ T ∩ U = 0 ˙
7 1 lsmub1 ⊢ S ∈ SubGrp ⁡ G ∧ T ∈ SubGrp ⁡ G → S ⊆ S ⊕ ˙ T
8 2 3 7 syl2anc ⊢ φ → S ⊆ S ⊕ ˙ T
9 8 ssrind ⊢ φ → S ∩ U ⊆ S ⊕ ˙ T ∩ U
10 9 6 sseqtrd ⊢ φ → S ∩ U ⊆ 0 ˙
11 5 subg0cl ⊢ S ∈ SubGrp ⁡ G → 0 ˙ ∈ S
12 2 11 syl ⊢ φ → 0 ˙ ∈ S
13 5 subg0cl ⊢ U ∈ SubGrp ⁡ G → 0 ˙ ∈ U
14 4 13 syl ⊢ φ → 0 ˙ ∈ U
15 12 14 elind ⊢ φ → 0 ˙ ∈ S ∩ U
16 15 snssd ⊢ φ → 0 ˙ ⊆ S ∩ U
17 10 16 eqssd ⊢ φ → S ∩ U = 0 ˙
18 1 lsmub2 ⊢ S ∈ SubGrp ⁡ G ∧ T ∈ SubGrp ⁡ G → T ⊆ S ⊕ ˙ T
19 2 3 18 syl2anc ⊢ φ → T ⊆ S ⊕ ˙ T
20 19 ssrind ⊢ φ → T ∩ U ⊆ S ⊕ ˙ T ∩ U
21 20 6 sseqtrd ⊢ φ → T ∩ U ⊆ 0 ˙
22 5 subg0cl ⊢ T ∈ SubGrp ⁡ G → 0 ˙ ∈ T
23 3 22 syl ⊢ φ → 0 ˙ ∈ T
24 23 14 elind ⊢ φ → 0 ˙ ∈ T ∩ U
25 24 snssd ⊢ φ → 0 ˙ ⊆ T ∩ U
26 21 25 eqssd ⊢ φ → T ∩ U = 0 ˙
27 17 26 jca ⊢ φ → S ∩ U = 0 ˙ ∧ T ∩ U = 0 ˙