Metamath Proof Explorer


Theorem lssneln0

Description: A vector X which doesn't belong to a subspace U is nonzero. (Contributed by NM, 14-May-2015) (Revised by AV, 17-Jul-2022) (Proof shortened by AV, 19-Jul-2022)

Ref Expression
Hypotheses lssneln0.o ⊢ 0 ˙ = 0 W
lssneln0.s ⊢ S = LSubSp ⁡ W
lssneln0.w ⊢ φ → W ∈ LMod
lssneln0.u ⊢ φ → U ∈ S
lssneln0.x ⊢ φ → X ∈ V
lssneln0.n ⊢ φ → ¬ X ∈ U
Assertion lssneln0 ⊢ φ → X ∈ V ∖ 0 ˙

Proof

Step Hyp Ref Expression
1 lssneln0.o ⊢ 0 ˙ = 0 W
2 lssneln0.s ⊢ S = LSubSp ⁡ W
3 lssneln0.w ⊢ φ → W ∈ LMod
4 lssneln0.u ⊢ φ → U ∈ S
5 lssneln0.x ⊢ φ → X ∈ V
6 lssneln0.n ⊢ φ → ¬ X ∈ U
7 1 2 3 4 6 lssvneln0 ⊢ φ → X ≠ 0 ˙
8 eldifsn ⊢ X ∈ V ∖ 0 ˙ ↔ X ∈ V ∧ X ≠ 0 ˙
9 5 7 8 sylanbrc ⊢ φ → X ∈ V ∖ 0 ˙