Metamath Proof Explorer


Theorem dvh2dim

Description: There is a vector that is outside the span of another. (Contributed by NM, 25-Apr-2015)

Ref Expression
Hypotheses dvh3dim.h H = LHyp K
dvh3dim.u U = DVecH K W
dvh3dim.v V = Base U
dvh3dim.n N = LSpan U
dvh3dim.k φ K HL W H
dvh3dim.x φ X V
Assertion dvh2dim φ z V ¬ z N X

Proof

Step Hyp Ref Expression
1 dvh3dim.h H = LHyp K
2 dvh3dim.u U = DVecH K W
3 dvh3dim.v V = Base U
4 dvh3dim.n N = LSpan U
5 dvh3dim.k φ K HL W H
6 dvh3dim.x φ X V
7 eqid 0 U = 0 U
8 1 2 3 7 5 dvh1dim φ z V z 0 U
9 8 adantr φ X = 0 U z V z 0 U
10 simpr φ X = 0 U X = 0 U
11 10 sneqd φ X = 0 U X = 0 U
12 11 fveq2d φ X = 0 U N X = N 0 U
13 1 2 5 dvhlmod φ U LMod
14 7 4 lspsn0 U LMod N 0 U = 0 U
15 13 14 syl φ N 0 U = 0 U
16 15 adantr φ X = 0 U N 0 U = 0 U
17 12 16 eqtrd φ X = 0 U N X = 0 U
18 17 eleq2d φ X = 0 U z N X z 0 U
19 velsn z 0 U z = 0 U
20 18 19 syl6bb φ X = 0 U z N X z = 0 U
21 20 necon3bbid φ X = 0 U ¬ z N X z 0 U
22 21 rexbidv φ X = 0 U z V ¬ z N X z V z 0 U
23 9 22 mpbird φ X = 0 U z V ¬ z N X
24 5 adantr φ X 0 U K HL W H
25 6 adantr φ X 0 U X V
26 simpr φ X 0 U X 0 U
27 1 2 3 4 24 25 25 7 26 26 dvhdimlem φ X 0 U z V ¬ z N X X
28 dfsn2 X = X X
29 28 fveq2i N X = N X X
30 29 eleq2i z N X z N X X
31 30 notbii ¬ z N X ¬ z N X X
32 31 rexbii z V ¬ z N X z V ¬ z N X X
33 27 32 sylibr φ X 0 U z V ¬ z N X
34 23 33 pm2.61dane φ z V ¬ z N X