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