Metamath Proof Explorer


Theorem lsmdisjr

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
lsmdisjr.i ⊢ φ → S ∩ T ⊕ ˙ U = 0 ˙
Assertion lsmdisjr ⊢ φ → S ∩ T = 0 ˙ ∧ S ∩ 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 lsmdisjr.i ⊢ φ → S ∩ T ⊕ ˙ U = 0 ˙
7 incom ⊢ S ∩ T ⊕ ˙ U = T ⊕ ˙ U ∩ S
8 7 6 eqtr3id ⊢ φ → T ⊕ ˙ U ∩ S = 0 ˙
9 1 3 4 2 5 8 lsmdisj ⊢ φ → T ∩ S = 0 ˙ ∧ U ∩ S = 0 ˙
10 incom ⊢ T ∩ S = S ∩ T
11 10 eqeq1i ⊢ T ∩ S = 0 ˙ ↔ S ∩ T = 0 ˙
12 incom ⊢ U ∩ S = S ∩ U
13 12 eqeq1i ⊢ U ∩ S = 0 ˙ ↔ S ∩ U = 0 ˙
14 11 13 anbi12i ⊢ T ∩ S = 0 ˙ ∧ U ∩ S = 0 ˙ ↔ S ∩ T = 0 ˙ ∧ S ∩ U = 0 ˙
15 9 14 sylib ⊢ φ → S ∩ T = 0 ˙ ∧ S ∩ U = 0 ˙