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 ( 𝑈 ∈ 𝑆 → 𝑈 ≠ ∅ )

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 ⊢ ( +g ‘ 𝑊 ) = ( +g ‘ 𝑊 )
6 eqid ⊢ ( ·𝑠 ‘ 𝑊 ) = ( ·𝑠 ‘ 𝑊 )
7 2 3 4 5 6 1 islss ⊢ ( 𝑈 ∈ 𝑆 ↔ ( 𝑈 ⊆ ( Base ‘ 𝑊 ) ∧ 𝑈 ≠ ∅ ∧ ∀ 𝑥 ∈ ( Base ‘ ( Scalar ‘ 𝑊 ) ) ∀ 𝑎 ∈ 𝑈 ∀ 𝑏 ∈ 𝑈 ( ( 𝑥 ( ·𝑠 ‘ 𝑊 ) 𝑎 ) ( +g ‘ 𝑊 ) 𝑏 ) ∈ 𝑈 ) )
8 7 simp2bi ⊢ ( 𝑈 ∈ 𝑆 → 𝑈 ≠ ∅ )