Metamath Proof Explorer


Theorem dihlspsnat

Description: The inverse isomorphism H of the span of a singleton is a Hilbert lattice atom. (Contributed by NM, 27-Apr-2014)

Ref Expression
Hypotheses dihlspsnat.a A = Atoms K
dihlspsnat.h H = LHyp K
dihlspsnat.u U = DVecH K W
dihlspsnat.v V = Base U
dihlspsnat.o 0 ˙ = 0 U
dihlspsnat.n N = LSpan U
dihlspsnat.i I = DIsoH K W
Assertion dihlspsnat K HL W H X V X 0 ˙ I -1 N X A

Proof

Step Hyp Ref Expression
1 dihlspsnat.a A = Atoms K
2 dihlspsnat.h H = LHyp K
3 dihlspsnat.u U = DVecH K W
4 dihlspsnat.v V = Base U
5 dihlspsnat.o 0 ˙ = 0 U
6 dihlspsnat.n N = LSpan U
7 dihlspsnat.i I = DIsoH K W
8 eqid Base K = Base K
9 eqid LSubSp U = LSubSp U
10 8 2 7 3 9 dihf11 K HL W H I : Base K 1-1 LSubSp U
11 10 3ad2ant1 K HL W H X V X 0 ˙ I : Base K 1-1 LSubSp U
12 f1f1orn I : Base K 1-1 LSubSp U I : Base K 1-1 onto ran I
13 11 12 syl K HL W H X V X 0 ˙ I : Base K 1-1 onto ran I
14 2 3 4 6 7 dihlsprn K HL W H X V N X ran I
15 14 3adant3 K HL W H X V X 0 ˙ N X ran I
16 f1ocnvdm I : Base K 1-1 onto ran I N X ran I I -1 N X Base K
17 13 15 16 syl2anc K HL W H X V X 0 ˙ I -1 N X Base K
18 fveq2 I -1 N X = 0. K I I -1 N X = I 0. K
19 2 7 dihcnvid2 K HL W H N X ran I I I -1 N X = N X
20 14 19 syldan K HL W H X V I I -1 N X = N X
21 eqid 0. K = 0. K
22 21 2 7 3 5 dih0 K HL W H I 0. K = 0 ˙
23 22 adantr K HL W H X V I 0. K = 0 ˙
24 20 23 eqeq12d K HL W H X V I I -1 N X = I 0. K N X = 0 ˙
25 id K HL W H K HL W H
26 2 3 25 dvhlmod K HL W H U LMod
27 4 5 6 lspsneq0 U LMod X V N X = 0 ˙ X = 0 ˙
28 26 27 sylan K HL W H X V N X = 0 ˙ X = 0 ˙
29 24 28 bitrd K HL W H X V I I -1 N X = I 0. K X = 0 ˙
30 18 29 syl5ib K HL W H X V I -1 N X = 0. K X = 0 ˙
31 30 necon3d K HL W H X V X 0 ˙ I -1 N X 0. K
32 31 3impia K HL W H X V X 0 ˙ I -1 N X 0. K
33 simpll1 K HL W H X V X 0 ˙ x Base K I x N X K HL W H
34 2 3 33 dvhlvec K HL W H X V X 0 ˙ x Base K I x N X U LVec
35 simplr K HL W H X V X 0 ˙ x Base K I x N X x Base K
36 8 2 7 3 9 dihlss K HL W H x Base K I x LSubSp U
37 33 35 36 syl2anc K HL W H X V X 0 ˙ x Base K I x N X I x LSubSp U
38 simpll2 K HL W H X V X 0 ˙ x Base K I x N X X V
39 simpr K HL W H X V X 0 ˙ x Base K I x N X I x N X
40 4 5 9 6 lspsnat U LVec I x LSubSp U X V I x N X I x = N X I x = 0 ˙
41 34 37 38 39 40 syl31anc K HL W H X V X 0 ˙ x Base K I x N X I x = N X I x = 0 ˙
42 41 ex K HL W H X V X 0 ˙ x Base K I x N X I x = N X I x = 0 ˙
43 simp1 K HL W H X V X 0 ˙ K HL W H
44 43 15 19 syl2anc K HL W H X V X 0 ˙ I I -1 N X = N X
45 44 adantr K HL W H X V X 0 ˙ x Base K I I -1 N X = N X
46 45 sseq2d K HL W H X V X 0 ˙ x Base K I x I I -1 N X I x N X
47 simpl1 K HL W H X V X 0 ˙ x Base K K HL W H
48 simpr K HL W H X V X 0 ˙ x Base K x Base K
49 17 adantr K HL W H X V X 0 ˙ x Base K I -1 N X Base K
50 eqid K = K
51 8 50 2 7 dihord K HL W H x Base K I -1 N X Base K I x I I -1 N X x K I -1 N X
52 47 48 49 51 syl3anc K HL W H X V X 0 ˙ x Base K I x I I -1 N X x K I -1 N X
53 46 52 bitr3d K HL W H X V X 0 ˙ x Base K I x N X x K I -1 N X
54 45 eqeq2d K HL W H X V X 0 ˙ x Base K I x = I I -1 N X I x = N X
55 8 2 7 dih11 K HL W H x Base K I -1 N X Base K I x = I I -1 N X x = I -1 N X
56 47 48 49 55 syl3anc K HL W H X V X 0 ˙ x Base K I x = I I -1 N X x = I -1 N X
57 54 56 bitr3d K HL W H X V X 0 ˙ x Base K I x = N X x = I -1 N X
58 47 22 syl K HL W H X V X 0 ˙ x Base K I 0. K = 0 ˙
59 58 eqeq2d K HL W H X V X 0 ˙ x Base K I x = I 0. K I x = 0 ˙
60 simpl1l K HL W H X V X 0 ˙ x Base K K HL
61 hlop K HL K OP
62 8 21 op0cl K OP 0. K Base K
63 60 61 62 3syl K HL W H X V X 0 ˙ x Base K 0. K Base K
64 8 2 7 dih11 K HL W H x Base K 0. K Base K I x = I 0. K x = 0. K
65 47 48 63 64 syl3anc K HL W H X V X 0 ˙ x Base K I x = I 0. K x = 0. K
66 59 65 bitr3d K HL W H X V X 0 ˙ x Base K I x = 0 ˙ x = 0. K
67 57 66 orbi12d K HL W H X V X 0 ˙ x Base K I x = N X I x = 0 ˙ x = I -1 N X x = 0. K
68 42 53 67 3imtr3d K HL W H X V X 0 ˙ x Base K x K I -1 N X x = I -1 N X x = 0. K
69 68 ralrimiva K HL W H X V X 0 ˙ x Base K x K I -1 N X x = I -1 N X x = 0. K
70 simp1l K HL W H X V X 0 ˙ K HL
71 hlatl K HL K AtLat
72 8 50 21 1 isat3 K AtLat I -1 N X A I -1 N X Base K I -1 N X 0. K x Base K x K I -1 N X x = I -1 N X x = 0. K
73 70 71 72 3syl K HL W H X V X 0 ˙ I -1 N X A I -1 N X Base K I -1 N X 0. K x Base K x K I -1 N X x = I -1 N X x = 0. K
74 17 32 69 73 mpbir3and K HL W H X V X 0 ˙ I -1 N X A