Metamath Proof Explorer


Theorem dochsatshp

Description: The orthocomplement of a subspace atom is a hyperplane. (Contributed by NM, 27-Jul-2014) (Revised by Mario Carneiro, 1-Oct-2014)

Ref Expression
Hypotheses dochsatshp.h H = LHyp K
dochsatshp.u U = DVecH K W
dochsatshp.o ˙ = ocH K W
dochsatshp.a A = LSAtoms U
dochsatshp.y Y = LSHyp U
dochsatshp.k φ K HL W H
dochsatshp.q φ Q A
Assertion dochsatshp φ ˙ Q Y

Proof

Step Hyp Ref Expression
1 dochsatshp.h H = LHyp K
2 dochsatshp.u U = DVecH K W
3 dochsatshp.o ˙ = ocH K W
4 dochsatshp.a A = LSAtoms U
5 dochsatshp.y Y = LSHyp U
6 dochsatshp.k φ K HL W H
7 dochsatshp.q φ Q A
8 eqid Base U = Base U
9 1 2 6 dvhlmod φ U LMod
10 8 4 9 7 lsatssv φ Q Base U
11 eqid LSubSp U = LSubSp U
12 1 2 8 11 3 dochlss K HL W H Q Base U ˙ Q LSubSp U
13 6 10 12 syl2anc φ ˙ Q LSubSp U
14 eqid 0 U = 0 U
15 14 4 9 7 lsatn0 φ Q 0 U
16 1 2 3 8 14 doch0 K HL W H ˙ 0 U = Base U
17 6 16 syl φ ˙ 0 U = Base U
18 17 eqeq2d φ ˙ Q = ˙ 0 U ˙ Q = Base U
19 eqid DIsoH K W = DIsoH K W
20 1 2 19 4 dih1dimat K HL W H Q A Q ran DIsoH K W
21 6 7 20 syl2anc φ Q ran DIsoH K W
22 1 19 2 14 dih0rn K HL W H 0 U ran DIsoH K W
23 6 22 syl φ 0 U ran DIsoH K W
24 1 19 3 6 21 23 doch11 φ ˙ Q = ˙ 0 U Q = 0 U
25 18 24 bitr3d φ ˙ Q = Base U Q = 0 U
26 25 necon3bid φ ˙ Q Base U Q 0 U
27 15 26 mpbird φ ˙ Q Base U
28 eqid LSpan U = LSpan U
29 8 28 14 4 islsat U LMod Q A v Base U 0 U Q = LSpan U v
30 9 29 syl φ Q A v Base U 0 U Q = LSpan U v
31 7 30 mpbid φ v Base U 0 U Q = LSpan U v
32 eldifi v Base U 0 U v Base U
33 32 adantr v Base U 0 U Q = LSpan U v v Base U
34 33 a1i φ v Base U 0 U Q = LSpan U v v Base U
35 11 28 lspid U LMod ˙ Q LSubSp U LSpan U ˙ Q = ˙ Q
36 9 13 35 syl2anc φ LSpan U ˙ Q = ˙ Q
37 36 uneq1d φ LSpan U ˙ Q LSpan U v = ˙ Q LSpan U v
38 37 fveq2d φ LSpan U LSpan U ˙ Q LSpan U v = LSpan U ˙ Q LSpan U v
39 38 adantr φ v Base U 0 U Q = LSpan U v LSpan U LSpan U ˙ Q LSpan U v = LSpan U ˙ Q LSpan U v
40 9 adantr φ v Base U 0 U Q = LSpan U v U LMod
41 8 11 lssss ˙ Q LSubSp U ˙ Q Base U
42 13 41 syl φ ˙ Q Base U
43 42 adantr φ v Base U 0 U Q = LSpan U v ˙ Q Base U
44 32 snssd v Base U 0 U v Base U
45 44 adantr v Base U 0 U Q = LSpan U v v Base U
46 45 adantl φ v Base U 0 U Q = LSpan U v v Base U
47 8 28 lspun U LMod ˙ Q Base U v Base U LSpan U ˙ Q v = LSpan U LSpan U ˙ Q LSpan U v
48 40 43 46 47 syl3anc φ v Base U 0 U Q = LSpan U v LSpan U ˙ Q v = LSpan U LSpan U ˙ Q LSpan U v
49 uneq2 Q = LSpan U v ˙ Q Q = ˙ Q LSpan U v
50 49 fveq2d Q = LSpan U v LSpan U ˙ Q Q = LSpan U ˙ Q LSpan U v
51 50 adantl v Base U 0 U Q = LSpan U v LSpan U ˙ Q Q = LSpan U ˙ Q LSpan U v
52 51 adantl φ v Base U 0 U Q = LSpan U v LSpan U ˙ Q Q = LSpan U ˙ Q LSpan U v
53 39 48 52 3eqtr4d φ v Base U 0 U Q = LSpan U v LSpan U ˙ Q v = LSpan U ˙ Q Q
54 eqid joinH K W = joinH K W
55 eqid LSSum U = LSSum U
56 1 19 2 8 3 dochcl K HL W H Q Base U ˙ Q ran DIsoH K W
57 6 10 56 syl2anc φ ˙ Q ran DIsoH K W
58 1 19 54 2 55 4 6 57 7 dihjat2 φ ˙ Q joinH K W Q = ˙ Q LSSum U Q
59 1 2 8 54 6 42 10 djhcom φ ˙ Q joinH K W Q = Q joinH K W ˙ Q
60 11 4 9 7 lsatlssel φ Q LSubSp U
61 11 28 55 lsmsp U LMod ˙ Q LSubSp U Q LSubSp U ˙ Q LSSum U Q = LSpan U ˙ Q Q
62 9 13 60 61 syl3anc φ ˙ Q LSSum U Q = LSpan U ˙ Q Q
63 58 59 62 3eqtr3rd φ LSpan U ˙ Q Q = Q joinH K W ˙ Q
64 1 2 8 3 54 djhexmid K HL W H Q Base U Q joinH K W ˙ Q = Base U
65 6 10 64 syl2anc φ Q joinH K W ˙ Q = Base U
66 63 65 eqtrd φ LSpan U ˙ Q Q = Base U
67 66 adantr φ v Base U 0 U Q = LSpan U v LSpan U ˙ Q Q = Base U
68 53 67 eqtrd φ v Base U 0 U Q = LSpan U v LSpan U ˙ Q v = Base U
69 68 ex φ v Base U 0 U Q = LSpan U v LSpan U ˙ Q v = Base U
70 34 69 jcad φ v Base U 0 U Q = LSpan U v v Base U LSpan U ˙ Q v = Base U
71 70 reximdv2 φ v Base U 0 U Q = LSpan U v v Base U LSpan U ˙ Q v = Base U
72 31 71 mpd φ v Base U LSpan U ˙ Q v = Base U
73 1 2 6 dvhlvec φ U LVec
74 8 28 11 5 islshp U LVec ˙ Q Y ˙ Q LSubSp U ˙ Q Base U v Base U LSpan U ˙ Q v = Base U
75 73 74 syl φ ˙ Q Y ˙ Q LSubSp U ˙ Q Base U v Base U LSpan U ˙ Q v = Base U
76 13 27 72 75 mpbir3and φ ˙ Q Y