Metamath Proof Explorer


Theorem lspsnne1

Description: Two ways to express that vectors have different spans. (Contributed by NM, 28-May-2015)

Ref Expression
Hypotheses lspsnne1.v ⊢ V = Base W
lspsnne1.o ⊢ 0 ˙ = 0 W
lspsnne1.n ⊢ N = LSpan ⁡ W
lspsnne1.w ⊢ φ → W ∈ LVec
lspsnne1.x ⊢ φ → X ∈ V ∖ 0 ˙
lspsnne1.y ⊢ φ → Y ∈ V
lspsnne1.e ⊢ φ → N ⁡ X ≠ N ⁡ Y
Assertion lspsnne1 ⊢ φ → ¬ X ∈ N ⁡ Y

Proof

Step Hyp Ref Expression
1 lspsnne1.v ⊢ V = Base W
2 lspsnne1.o ⊢ 0 ˙ = 0 W
3 lspsnne1.n ⊢ N = LSpan ⁡ W
4 lspsnne1.w ⊢ φ → W ∈ LVec
5 lspsnne1.x ⊢ φ → X ∈ V ∖ 0 ˙
6 lspsnne1.y ⊢ φ → Y ∈ V
7 lspsnne1.e ⊢ φ → N ⁡ X ≠ N ⁡ Y
8 eqid ⊢ LSubSp ⁡ W = LSubSp ⁡ W
9 lveclmod ⊢ W ∈ LVec → W ∈ LMod
10 4 9 syl ⊢ φ → W ∈ LMod
11 1 8 3 lspsncl ⊢ W ∈ LMod ∧ Y ∈ V → N ⁡ Y ∈ LSubSp ⁡ W
12 10 6 11 syl2anc ⊢ φ → N ⁡ Y ∈ LSubSp ⁡ W
13 5 eldifad ⊢ φ → X ∈ V
14 1 8 3 10 12 13 ellspsn5b ⊢ φ → X ∈ N ⁡ Y ↔ N ⁡ X ⊆ N ⁡ Y
15 14 notbid ⊢ φ → ¬ X ∈ N ⁡ Y ↔ ¬ N ⁡ X ⊆ N ⁡ Y
16 1 2 3 4 5 6 lspsncmp ⊢ φ → N ⁡ X ⊆ N ⁡ Y ↔ N ⁡ X = N ⁡ Y
17 16 necon3bbid ⊢ φ → ¬ N ⁡ X ⊆ N ⁡ Y ↔ N ⁡ X ≠ N ⁡ Y
18 15 17 bitrd ⊢ φ → ¬ X ∈ N ⁡ Y ↔ N ⁡ X ≠ N ⁡ Y
19 7 18 mpbird ⊢ φ → ¬ X ∈ N ⁡ Y