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 = Base W
lbsss.j J = LBasis W
lbssp.n N = LSpan W
lbsind.f F = Scalar W
lbsind.s · ˙ = W
lbsind.k K = Base F
lbsind.z 0 ˙ = 0 F
Assertion lbsind B J E B A K A 0 ˙ ¬ A · ˙ E N B E

Proof

Step Hyp Ref Expression
1 lbsss.v V = Base W
2 lbsss.j J = LBasis W
3 lbssp.n N = LSpan W
4 lbsind.f F = Scalar W
5 lbsind.s · ˙ = W
6 lbsind.k K = Base F
7 lbsind.z 0 ˙ = 0 F
8 eldifsn A K 0 ˙ A K A 0 ˙
9 elfvdm B LBasis W W dom LBasis
10 9 2 eleq2s B J W dom LBasis
11 1 4 5 6 2 3 7 islbs W dom LBasis B J B V N B = V x B y K 0 ˙ ¬ y · ˙ x N B x
12 10 11 syl B J B J B V N B = V x B y K 0 ˙ ¬ y · ˙ x N B x
13 12 ibi B J B V N B = V x B y K 0 ˙ ¬ y · ˙ x N B x
14 13 simp3d B J x B y K 0 ˙ ¬ y · ˙ x N B x
15 oveq2 x = E y · ˙ x = y · ˙ E
16 sneq x = E x = E
17 16 difeq2d x = E B x = B E
18 17 fveq2d x = E N B x = N B E
19 15 18 eleq12d x = E y · ˙ x N B x y · ˙ E N B E
20 19 notbid x = E ¬ y · ˙ x N B x ¬ y · ˙ E N B E
21 oveq1 y = A y · ˙ E = A · ˙ E
22 21 eleq1d y = A y · ˙ E N B E A · ˙ E N B E
23 22 notbid y = A ¬ y · ˙ E N B E ¬ A · ˙ E N B E
24 20 23 rspc2v E B A K 0 ˙ x B y K 0 ˙ ¬ y · ˙ x N B x ¬ A · ˙ E N B E
25 14 24 syl5com B J E B A K 0 ˙ ¬ A · ˙ E N B E
26 25 impl B J E B A K 0 ˙ ¬ A · ˙ E N B E
27 8 26 sylan2br B J E B A K A 0 ˙ ¬ A · ˙ E N B E