Metamath Proof Explorer


Theorem lclkrlem2a

Description: Lemma for lclkr . Use lshpat to show that the intersection of a hyperplane with a noncomparable sum of atoms is an atom. (Contributed by NM, 16-Jan-2015)

Ref Expression
Hypotheses lclkrlem2a.h H = LHyp K
lclkrlem2a.o ˙ = ocH K W
lclkrlem2a.u U = DVecH K W
lclkrlem2a.v V = Base U
lclkrlem2a.z 0 ˙ = 0 U
lclkrlem2a.p ˙ = LSSum U
lclkrlem2a.n N = LSpan U
lclkrlem2a.a A = LSAtoms U
lclkrlem2a.k φ K HL W H
lclkrlem2a.b φ B V 0 ˙
lclkrlem2a.x φ X V 0 ˙
lclkrlem2a.y φ Y V 0 ˙
lclkrlem2a.e φ ˙ X ˙ Y
lclkrlem2a.d φ ¬ X ˙ B
Assertion lclkrlem2a φ N X ˙ N Y ˙ B A

Proof

Step Hyp Ref Expression
1 lclkrlem2a.h H = LHyp K
2 lclkrlem2a.o ˙ = ocH K W
3 lclkrlem2a.u U = DVecH K W
4 lclkrlem2a.v V = Base U
5 lclkrlem2a.z 0 ˙ = 0 U
6 lclkrlem2a.p ˙ = LSSum U
7 lclkrlem2a.n N = LSpan U
8 lclkrlem2a.a A = LSAtoms U
9 lclkrlem2a.k φ K HL W H
10 lclkrlem2a.b φ B V 0 ˙
11 lclkrlem2a.x φ X V 0 ˙
12 lclkrlem2a.y φ Y V 0 ˙
13 lclkrlem2a.e φ ˙ X ˙ Y
14 lclkrlem2a.d φ ¬ X ˙ B
15 eqid LSubSp U = LSubSp U
16 eqid LSHyp U = LSHyp U
17 1 3 9 dvhlvec φ U LVec
18 1 2 3 4 5 16 9 10 dochsnshp φ ˙ B LSHyp U
19 1 3 9 dvhlmod φ U LMod
20 4 7 5 8 19 11 lsatlspsn φ N X A
21 4 7 5 8 19 12 lsatlspsn φ N Y A
22 11 eldifad φ X V
23 22 snssd φ X V
24 1 3 2 4 7 9 23 dochocsp φ ˙ N X = ˙ X
25 12 eldifad φ Y V
26 25 snssd φ Y V
27 1 3 2 4 7 9 26 dochocsp φ ˙ N Y = ˙ Y
28 24 27 eqeq12d φ ˙ N X = ˙ N Y ˙ X = ˙ Y
29 eqid DIsoH K W = DIsoH K W
30 1 3 4 7 29 dihlsprn K HL W H X V N X ran DIsoH K W
31 9 22 30 syl2anc φ N X ran DIsoH K W
32 1 3 4 7 29 dihlsprn K HL W H Y V N Y ran DIsoH K W
33 9 25 32 syl2anc φ N Y ran DIsoH K W
34 1 29 2 9 31 33 doch11 φ ˙ N X = ˙ N Y N X = N Y
35 28 34 bitr3d φ ˙ X = ˙ Y N X = N Y
36 35 necon3bid φ ˙ X ˙ Y N X N Y
37 13 36 mpbid φ N X N Y
38 10 eldifad φ B V
39 38 snssd φ B V
40 1 3 4 15 2 dochlss K HL W H B V ˙ B LSubSp U
41 9 39 40 syl2anc φ ˙ B LSubSp U
42 4 15 7 19 41 22 lspsnel5 φ X ˙ B N X ˙ B
43 14 42 mtbid φ ¬ N X ˙ B
44 15 6 16 8 17 18 20 21 37 43 lshpat φ N X ˙ N Y ˙ B A