Metamath Proof Explorer


Theorem lbsind

Description: A basis is linearly independent; that is, every element has a span which trivially intersects the span of the remainder of the basis. (Contributed by Mario Carneiro, 12-Jan-2015)

Ref Expression
Hypotheses lbsss.v V=BaseW
lbsss.j J=LBasisW
lbssp.n N=LSpanW
lbsind.f F=ScalarW
lbsind.s ·˙=W
lbsind.k K=BaseF
lbsind.z 0˙=0F
Assertion lbsind BJEBAKA0˙¬A·˙ENBE

Proof

Step Hyp Ref Expression
1 lbsss.v V=BaseW
2 lbsss.j J=LBasisW
3 lbssp.n N=LSpanW
4 lbsind.f F=ScalarW
5 lbsind.s ·˙=W
6 lbsind.k K=BaseF
7 lbsind.z 0˙=0F
8 eldifsn AK0˙AKA0˙
9 elfvdm BLBasisWWdomLBasis
10 9 2 eleq2s BJWdomLBasis
11 1 4 5 6 2 3 7 islbs WdomLBasisBJBVNB=VxByK0˙¬y·˙xNBx
12 10 11 syl BJBJBVNB=VxByK0˙¬y·˙xNBx
13 12 ibi BJBVNB=VxByK0˙¬y·˙xNBx
14 13 simp3d BJxByK0˙¬y·˙xNBx
15 oveq2 x=Ey·˙x=y·˙E
16 sneq x=Ex=E
17 16 difeq2d x=EBx=BE
18 17 fveq2d x=ENBx=NBE
19 15 18 eleq12d x=Ey·˙xNBxy·˙ENBE
20 19 notbid x=E¬y·˙xNBx¬y·˙ENBE
21 oveq1 y=Ay·˙E=A·˙E
22 21 eleq1d y=Ay·˙ENBEA·˙ENBE
23 22 notbid y=A¬y·˙ENBE¬A·˙ENBE
24 20 23 rspc2v EBAK0˙xByK0˙¬y·˙xNBx¬A·˙ENBE
25 14 24 syl5com BJEBAK0˙¬A·˙ENBE
26 25 impl BJEBAK0˙¬A·˙ENBE
27 8 26 sylan2br BJEBAKA0˙¬A·˙ENBE