Metamath Proof Explorer


Theorem lssn0

Description: A subspace is not empty. (Contributed by NM, 12-Jan-2014) (Revised by Mario Carneiro, 8-Jan-2015)

Ref Expression
Hypothesis lssn0.s 𝑆 = ( LSubSp ‘ 𝑊 )
Assertion lssn0 ( 𝑈𝑆𝑈 ≠ ∅ )

Proof

Step Hyp Ref Expression
1 lssn0.s 𝑆 = ( LSubSp ‘ 𝑊 )
2 eqid ( Scalar ‘ 𝑊 ) = ( Scalar ‘ 𝑊 )
3 eqid ( Base ‘ ( Scalar ‘ 𝑊 ) ) = ( Base ‘ ( Scalar ‘ 𝑊 ) )
4 eqid ( Base ‘ 𝑊 ) = ( Base ‘ 𝑊 )
5 eqid ( +g𝑊 ) = ( +g𝑊 )
6 eqid ( ·𝑠𝑊 ) = ( ·𝑠𝑊 )
7 2 3 4 5 6 1 islss ( 𝑈𝑆 ↔ ( 𝑈 ⊆ ( Base ‘ 𝑊 ) ∧ 𝑈 ≠ ∅ ∧ ∀ 𝑥 ∈ ( Base ‘ ( Scalar ‘ 𝑊 ) ) ∀ 𝑎𝑈𝑏𝑈 ( ( 𝑥 ( ·𝑠𝑊 ) 𝑎 ) ( +g𝑊 ) 𝑏 ) ∈ 𝑈 ) )
8 7 simp2bi ( 𝑈𝑆𝑈 ≠ ∅ )