Metamath Proof Explorer


Theorem lspdisjb

Description: A nonzero vector is not in a subspace iff its span is disjoint with the subspace. (Contributed by NM, 23-Apr-2015)

Ref Expression
Hypotheses lspdisjb.v ⊢ V = Base W
lspdisjb.o ⊢ 0 ˙ = 0 W
lspdisjb.n ⊢ N = LSpan ⁡ W
lspdisjb.s ⊢ S = LSubSp ⁡ W
lspdisjb.w ⊢ φ → W ∈ LVec
lspdisjb.u ⊢ φ → U ∈ S
lspdisjb.x ⊢ φ → X ∈ V ∖ 0 ˙
Assertion lspdisjb ⊢ φ → ¬ X ∈ U ↔ N ⁡ X ∩ U = 0 ˙

Proof

Step Hyp Ref Expression
1 lspdisjb.v ⊢ V = Base W
2 lspdisjb.o ⊢ 0 ˙ = 0 W
3 lspdisjb.n ⊢ N = LSpan ⁡ W
4 lspdisjb.s ⊢ S = LSubSp ⁡ W
5 lspdisjb.w ⊢ φ → W ∈ LVec
6 lspdisjb.u ⊢ φ → U ∈ S
7 lspdisjb.x ⊢ φ → X ∈ V ∖ 0 ˙
8 5 adantr ⊢ φ ∧ ¬ X ∈ U → W ∈ LVec
9 6 adantr ⊢ φ ∧ ¬ X ∈ U → U ∈ S
10 7 eldifad ⊢ φ → X ∈ V
11 10 adantr ⊢ φ ∧ ¬ X ∈ U → X ∈ V
12 simpr ⊢ φ ∧ ¬ X ∈ U → ¬ X ∈ U
13 1 2 3 4 8 9 11 12 lspdisj ⊢ φ ∧ ¬ X ∈ U → N ⁡ X ∩ U = 0 ˙
14 eldifsni ⊢ X ∈ V ∖ 0 ˙ → X ≠ 0 ˙
15 7 14 syl ⊢ φ → X ≠ 0 ˙
16 15 adantr ⊢ φ ∧ N ⁡ X ∩ U = 0 ˙ → X ≠ 0 ˙
17 lveclmod ⊢ W ∈ LVec → W ∈ LMod
18 5 17 syl ⊢ φ → W ∈ LMod
19 1 3 lspsnid ⊢ W ∈ LMod ∧ X ∈ V → X ∈ N ⁡ X
20 18 10 19 syl2anc ⊢ φ → X ∈ N ⁡ X
21 elin ⊢ X ∈ N ⁡ X ∩ U ↔ X ∈ N ⁡ X ∧ X ∈ U
22 eleq2 ⊢ N ⁡ X ∩ U = 0 ˙ → X ∈ N ⁡ X ∩ U ↔ X ∈ 0 ˙
23 elsni ⊢ X ∈ 0 ˙ → X = 0 ˙
24 22 23 biimtrdi ⊢ N ⁡ X ∩ U = 0 ˙ → X ∈ N ⁡ X ∩ U → X = 0 ˙
25 21 24 biimtrrid ⊢ N ⁡ X ∩ U = 0 ˙ → X ∈ N ⁡ X ∧ X ∈ U → X = 0 ˙
26 25 expd ⊢ N ⁡ X ∩ U = 0 ˙ → X ∈ N ⁡ X → X ∈ U → X = 0 ˙
27 20 26 mpan9 ⊢ φ ∧ N ⁡ X ∩ U = 0 ˙ → X ∈ U → X = 0 ˙
28 27 necon3ad ⊢ φ ∧ N ⁡ X ∩ U = 0 ˙ → X ≠ 0 ˙ → ¬ X ∈ U
29 16 28 mpd ⊢ φ ∧ N ⁡ X ∩ U = 0 ˙ → ¬ X ∈ U
30 13 29 impbida ⊢ φ → ¬ X ∈ U ↔ N ⁡ X ∩ U = 0 ˙