Metamath Proof Explorer
Description: A subspace member is a vector. (Contributed by NM, 11-Jan-2014)
(Revised by Mario Carneiro, 8-Jan-2015)
|
|
Ref |
Expression |
|
Hypotheses |
lssss.v |
⊢ 𝑉 = ( Base ‘ 𝑊 ) |
|
|
lssss.s |
⊢ 𝑆 = ( LSubSp ‘ 𝑊 ) |
|
Assertion |
lssel |
⊢ ( ( 𝑈 ∈ 𝑆 ∧ 𝑋 ∈ 𝑈 ) → 𝑋 ∈ 𝑉 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
lssss.v |
⊢ 𝑉 = ( Base ‘ 𝑊 ) |
2 |
|
lssss.s |
⊢ 𝑆 = ( LSubSp ‘ 𝑊 ) |
3 |
1 2
|
lssss |
⊢ ( 𝑈 ∈ 𝑆 → 𝑈 ⊆ 𝑉 ) |
4 |
3
|
sselda |
⊢ ( ( 𝑈 ∈ 𝑆 ∧ 𝑋 ∈ 𝑈 ) → 𝑋 ∈ 𝑉 ) |