Metamath Proof Explorer


Theorem elspancl

Description: A member of a span is a vector. (Contributed by NM, 17-Dec-2004) (New usage is discouraged.)

Ref Expression
Assertion elspancl ( ( 𝐴 ⊆ ℋ ∧ 𝐵 ∈ ( span ‘ 𝐴 ) ) → 𝐵 ∈ ℋ )

Proof

Step Hyp Ref Expression
1 spancl ( 𝐴 ⊆ ℋ → ( span ‘ 𝐴 ) ∈ S )
2 shel ( ( ( span ‘ 𝐴 ) ∈ S𝐵 ∈ ( span ‘ 𝐴 ) ) → 𝐵 ∈ ℋ )
3 1 2 sylan ( ( 𝐴 ⊆ ℋ ∧ 𝐵 ∈ ( span ‘ 𝐴 ) ) → 𝐵 ∈ ℋ )