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 = ( 0g ‘ 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 = ( 0g ‘ 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 } ) )