Metamath Proof Explorer


Theorem lspdisjb

Description: A nonzero vector is not in a subspace iff its span is disjoint with the subspace. (Contributed by NM, 23-Apr-2015)

Ref Expression
Hypotheses lspdisjb.v V = Base W
lspdisjb.o 0 ˙ = 0 W
lspdisjb.n N = LSpan W
lspdisjb.s S = LSubSp W
lspdisjb.w φ W LVec
lspdisjb.u φ U S
lspdisjb.x φ X V 0 ˙
Assertion lspdisjb φ ¬ X U N X U = 0 ˙

Proof

Step Hyp Ref Expression
1 lspdisjb.v V = Base W
2 lspdisjb.o 0 ˙ = 0 W
3 lspdisjb.n N = LSpan W
4 lspdisjb.s S = LSubSp W
5 lspdisjb.w φ W LVec
6 lspdisjb.u φ U S
7 lspdisjb.x φ X V 0 ˙
8 5 adantr φ ¬ X U W LVec
9 6 adantr φ ¬ X U U S
10 7 eldifad φ X V
11 10 adantr φ ¬ X U X V
12 simpr φ ¬ X U ¬ X U
13 1 2 3 4 8 9 11 12 lspdisj φ ¬ X U N X U = 0 ˙
14 eldifsni X V 0 ˙ X 0 ˙
15 7 14 syl φ X 0 ˙
16 15 adantr φ N X U = 0 ˙ X 0 ˙
17 lveclmod W LVec W LMod
18 5 17 syl φ W LMod
19 1 3 lspsnid W LMod X V X N X
20 18 10 19 syl2anc φ X N X
21 elin X N X U X N X X U
22 eleq2 N X U = 0 ˙ X N X U X 0 ˙
23 elsni X 0 ˙ X = 0 ˙
24 22 23 syl6bi N X U = 0 ˙ X N X U X = 0 ˙
25 21 24 syl5bir N X U = 0 ˙ X N X X U X = 0 ˙
26 25 expd N X U = 0 ˙ X N X X U X = 0 ˙
27 20 26 mpan9 φ N X U = 0 ˙ X U X = 0 ˙
28 27 necon3ad φ N X U = 0 ˙ X 0 ˙ ¬ X U
29 16 28 mpd φ N X U = 0 ˙ ¬ X U
30 13 29 impbida φ ¬ X U N X U = 0 ˙