Metamath Proof Explorer


Theorem dvh2dim

Description: There is a vector that is outside the span of another. (Contributed by NM, 25-Apr-2015)

Ref Expression
Hypotheses dvh3dim.h ⊢ H = LHyp ⁡ K
dvh3dim.u ⊢ U = DVecH ⁡ K ⁡ W
dvh3dim.v ⊢ V = Base U
dvh3dim.n ⊢ N = LSpan ⁡ U
dvh3dim.k ⊢ φ → K ∈ HL ∧ W ∈ H
dvh3dim.x ⊢ φ → X ∈ V
Assertion dvh2dim ⊢ φ → ∃ z ∈ V ¬ z ∈ N ⁡ X

Proof

Step Hyp Ref Expression
1 dvh3dim.h ⊢ H = LHyp ⁡ K
2 dvh3dim.u ⊢ U = DVecH ⁡ K ⁡ W
3 dvh3dim.v ⊢ V = Base U
4 dvh3dim.n ⊢ N = LSpan ⁡ U
5 dvh3dim.k ⊢ φ → K ∈ HL ∧ W ∈ H
6 dvh3dim.x ⊢ φ → X ∈ V
7 eqid ⊢ 0 U = 0 U
8 1 2 3 7 5 dvh1dim ⊢ φ → ∃ z ∈ V z ≠ 0 U
9 8 adantr ⊢ φ ∧ X = 0 U → ∃ z ∈ V z ≠ 0 U
10 simpr ⊢ φ ∧ X = 0 U → X = 0 U
11 10 sneqd ⊢ φ ∧ X = 0 U → X = 0 U
12 11 fveq2d ⊢ φ ∧ X = 0 U → N ⁡ X = N ⁡ 0 U
13 1 2 5 dvhlmod ⊢ φ → U ∈ LMod
14 7 4 lspsn0 ⊢ U ∈ LMod → N ⁡ 0 U = 0 U
15 13 14 syl ⊢ φ → N ⁡ 0 U = 0 U
16 15 adantr ⊢ φ ∧ X = 0 U → N ⁡ 0 U = 0 U
17 12 16 eqtrd ⊢ φ ∧ X = 0 U → N ⁡ X = 0 U
18 17 eleq2d ⊢ φ ∧ X = 0 U → z ∈ N ⁡ X ↔ z ∈ 0 U
19 velsn ⊢ z ∈ 0 U ↔ z = 0 U
20 18 19 bitrdi ⊢ φ ∧ X = 0 U → z ∈ N ⁡ X ↔ z = 0 U
21 20 necon3bbid ⊢ φ ∧ X = 0 U → ¬ z ∈ N ⁡ X ↔ z ≠ 0 U
22 21 rexbidv ⊢ φ ∧ X = 0 U → ∃ z ∈ V ¬ z ∈ N ⁡ X ↔ ∃ z ∈ V z ≠ 0 U
23 9 22 mpbird ⊢ φ ∧ X = 0 U → ∃ z ∈ V ¬ z ∈ N ⁡ X
24 5 adantr ⊢ φ ∧ X ≠ 0 U → K ∈ HL ∧ W ∈ H
25 6 adantr ⊢ φ ∧ X ≠ 0 U → X ∈ V
26 simpr ⊢ φ ∧ X ≠ 0 U → X ≠ 0 U
27 1 2 3 4 24 25 25 7 26 26 dvhdimlem ⊢ φ ∧ X ≠ 0 U → ∃ z ∈ V ¬ z ∈ N ⁡ X X
28 dfsn2 ⊢ X = X X
29 28 fveq2i ⊢ N ⁡ X = N ⁡ X X
30 29 eleq2i ⊢ z ∈ N ⁡ X ↔ z ∈ N ⁡ X X
31 30 notbii ⊢ ¬ z ∈ N ⁡ X ↔ ¬ z ∈ N ⁡ X X
32 31 rexbii ⊢ ∃ z ∈ V ¬ z ∈ N ⁡ X ↔ ∃ z ∈ V ¬ z ∈ N ⁡ X X
33 27 32 sylibr ⊢ φ ∧ X ≠ 0 U → ∃ z ∈ V ¬ z ∈ N ⁡ X
34 23 33 pm2.61dane ⊢ φ → ∃ z ∈ V ¬ z ∈ N ⁡ X