Metamath Proof Explorer


Theorem dochn0nv

Description: An orthocomplement is nonzero iff the double orthocomplement is not the whole vector space. (Contributed by NM, 1-Jan-2015)

Ref Expression
Hypotheses dochn0nv.h H = LHyp K
dochn0nv.o ˙ = ocH K W
dochn0nv.u U = DVecH K W
dochn0nv.v V = Base U
dochn0nv.z 0 ˙ = 0 U
dochn0nv.k φ K HL W H
dochn0nv.x φ X V
Assertion dochn0nv φ ˙ X 0 ˙ ˙ ˙ X V

Proof

Step Hyp Ref Expression
1 dochn0nv.h H = LHyp K
2 dochn0nv.o ˙ = ocH K W
3 dochn0nv.u U = DVecH K W
4 dochn0nv.v V = Base U
5 dochn0nv.z 0 ˙ = 0 U
6 dochn0nv.k φ K HL W H
7 dochn0nv.x φ X V
8 eqid DIsoH K W = DIsoH K W
9 1 8 3 4 2 dochcl K HL W H X V ˙ X ran DIsoH K W
10 6 7 9 syl2anc φ ˙ X ran DIsoH K W
11 1 8 2 dochoc K HL W H ˙ X ran DIsoH K W ˙ ˙ ˙ X = ˙ X
12 6 10 11 syl2anc φ ˙ ˙ ˙ X = ˙ X
13 1 3 2 4 5 doch1 K HL W H ˙ V = 0 ˙
14 6 13 syl φ ˙ V = 0 ˙
15 12 14 eqeq12d φ ˙ ˙ ˙ X = ˙ V ˙ X = 0 ˙
16 1 3 4 2 dochssv K HL W H X V ˙ X V
17 6 7 16 syl2anc φ ˙ X V
18 1 8 3 4 2 dochcl K HL W H ˙ X V ˙ ˙ X ran DIsoH K W
19 6 17 18 syl2anc φ ˙ ˙ X ran DIsoH K W
20 1 8 3 4 dih1rn K HL W H V ran DIsoH K W
21 6 20 syl φ V ran DIsoH K W
22 1 8 2 6 19 21 doch11 φ ˙ ˙ ˙ X = ˙ V ˙ ˙ X = V
23 15 22 bitr3d φ ˙ X = 0 ˙ ˙ ˙ X = V
24 23 necon3bid φ ˙ X 0 ˙ ˙ ˙ X V