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 = ( 0g ‘ 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 = ( 0g ‘ 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 } ) )