Metamath Proof Explorer


Theorem lss0cl

Description: The zero vector belongs to every subspace. (Contributed by NM, 12-Jan-2014) (Proof shortened by Mario Carneiro, 19-Jun-2014)

Ref Expression
Hypotheses lss0cl.z 0 ˙ = 0 W
lss0cl.s S = LSubSp W
Assertion lss0cl W LMod U S 0 ˙ U

Proof

Step Hyp Ref Expression
1 lss0cl.z 0 ˙ = 0 W
2 lss0cl.s S = LSubSp W
3 2 lssn0 U S U
4 n0 U x x U
5 3 4 sylib U S x x U
6 5 adantl W LMod U S x x U
7 simp1 W LMod U S x U W LMod
8 eqid Base W = Base W
9 8 2 lssel U S x U x Base W
10 9 3adant1 W LMod U S x U x Base W
11 eqid - W = - W
12 8 1 11 lmodsubid W LMod x Base W x - W x = 0 ˙
13 7 10 12 syl2anc W LMod U S x U x - W x = 0 ˙
14 11 2 lssvsubcl W LMod U S x U x U x - W x U
15 14 anabsan2 W LMod U S x U x - W x U
16 15 3impa W LMod U S x U x - W x U
17 13 16 eqeltrrd W LMod U S x U 0 ˙ U
18 17 3expia W LMod U S x U 0 ˙ U
19 18 exlimdv W LMod U S x x U 0 ˙ U
20 6 19 mpd W LMod U S 0 ˙ U