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 ˙