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