Metamath Proof Explorer


Theorem lssvneln0

Description: A vector X which doesn't belong to a subspace U is nonzero. (Contributed by NM, 14-May-2015) (Revised by AV, 19-Jul-2022)

Ref Expression
Hypotheses lssvneln0.o ⊢ 0 ˙ = 0 W
lssvneln0.s ⊢ S = LSubSp ⁡ W
lssvneln0.w ⊢ φ → W ∈ LMod
lssvneln0.u ⊢ φ → U ∈ S
lssvneln0.n ⊢ φ → ¬ X ∈ U
Assertion lssvneln0 ⊢ φ → X ≠ 0 ˙

Proof

Step Hyp Ref Expression
1 lssvneln0.o ⊢ 0 ˙ = 0 W
2 lssvneln0.s ⊢ S = LSubSp ⁡ W
3 lssvneln0.w ⊢ φ → W ∈ LMod
4 lssvneln0.u ⊢ φ → U ∈ S
5 lssvneln0.n ⊢ φ → ¬ X ∈ U
6 1 2 lss0cl ⊢ W ∈ LMod ∧ U ∈ S → 0 ˙ ∈ U
7 3 4 6 syl2anc ⊢ φ → 0 ˙ ∈ U
8 eleq1a ⊢ 0 ˙ ∈ U → X = 0 ˙ → X ∈ U
9 7 8 syl ⊢ φ → X = 0 ˙ → X ∈ U
10 9 necon3bd ⊢ φ → ¬ X ∈ U → X ≠ 0 ˙
11 5 10 mpd ⊢ φ → X ≠ 0 ˙