Metamath Proof Explorer


Theorem lspsnat

Description: There is no subspace strictly between the zero subspace and the span of a vector (i.e. a 1-dimensional subspace is an atom). ( h1datomi analog.) (Contributed by NM, 20-Apr-2014) (Proof shortened by Mario Carneiro, 22-Jun-2014)

Ref Expression
Hypotheses lspsnat.v V=BaseW
lspsnat.z 0˙=0W
lspsnat.s S=LSubSpW
lspsnat.n N=LSpanW
Assertion lspsnat WLVecUSXVUNXU=NXU=0˙

Proof

Step Hyp Ref Expression
1 lspsnat.v V=BaseW
2 lspsnat.z 0˙=0W
3 lspsnat.s S=LSubSpW
4 lspsnat.n N=LSpanW
5 n0 U0˙xxU0˙
6 simprl WLVecUSXVUNXxU0˙UNX
7 simpl1 WLVecUSXVUNXxU0˙WLVec
8 lveclmod WLVecWLMod
9 7 8 syl WLVecUSXVUNXxU0˙WLMod
10 simpl2 WLVecUSXVUNXxU0˙US
11 simprr WLVecUSXVUNXxU0˙xU0˙
12 11 eldifad WLVecUSXVUNXxU0˙xU
13 3 4 9 10 12 lspsnel5a WLVecUSXVUNXxU0˙NxU
14 0ss V
15 14 a1i WLVecUSXVUNXxU0˙V
16 simpl3 WLVecUSXVUNXxU0˙XV
17 ssdif UNXU0˙NX0˙
18 17 ad2antrl WLVecUSXVUNXxU0˙U0˙NX0˙
19 18 11 sseldd WLVecUSXVUNXxU0˙xNX0˙
20 uncom X=X
21 un0 X=X
22 20 21 eqtri X=X
23 22 fveq2i NX=NX
24 23 a1i WLVecUSXVUNXxU0˙NX=NX
25 2 4 lsp0 WLModN=0˙
26 9 25 syl WLVecUSXVUNXxU0˙N=0˙
27 24 26 difeq12d WLVecUSXVUNXxU0˙NXN=NX0˙
28 19 27 eleqtrrd WLVecUSXVUNXxU0˙xNXN
29 1 3 4 lspsolv WLVecVXVxNXNXNx
30 7 15 16 28 29 syl13anc WLVecUSXVUNXxU0˙XNx
31 uncom x=x
32 un0 x=x
33 31 32 eqtri x=x
34 33 fveq2i Nx=Nx
35 30 34 eleqtrdi WLVecUSXVUNXxU0˙XNx
36 13 35 sseldd WLVecUSXVUNXxU0˙XU
37 3 4 9 10 36 lspsnel5a WLVecUSXVUNXxU0˙NXU
38 6 37 eqssd WLVecUSXVUNXxU0˙U=NX
39 38 expr WLVecUSXVUNXxU0˙U=NX
40 39 exlimdv WLVecUSXVUNXxxU0˙U=NX
41 5 40 biimtrid WLVecUSXVUNXU0˙U=NX
42 41 necon1bd WLVecUSXVUNX¬U=NXU0˙=
43 ssdif0 U0˙U0˙=
44 42 43 imbitrrdi WLVecUSXVUNX¬U=NXU0˙
45 simpl1 WLVecUSXVUNXWLVec
46 45 8 syl WLVecUSXVUNXWLMod
47 simpl2 WLVecUSXVUNXUS
48 2 3 lssle0 WLModUSU0˙U=0˙
49 46 47 48 syl2anc WLVecUSXVUNXU0˙U=0˙
50 44 49 sylibd WLVecUSXVUNX¬U=NXU=0˙
51 50 orrd WLVecUSXVUNXU=NXU=0˙