Metamath Proof Explorer


Theorem lspprcl

Description: The span of a pair is a subspace (frequently used special case of lspcl ). (Contributed by NM, 11-Apr-2015)

Ref Expression
Hypotheses lspval.v 𝑉 = ( Base ‘ 𝑊 )
lspval.s 𝑆 = ( LSubSp ‘ 𝑊 )
lspval.n 𝑁 = ( LSpan ‘ 𝑊 )
lspprcl.w ( 𝜑𝑊 ∈ LMod )
lspprcl.x ( 𝜑𝑋𝑉 )
lspprcl.y ( 𝜑𝑌𝑉 )
Assertion lspprcl ( 𝜑 → ( 𝑁 ‘ { 𝑋 , 𝑌 } ) ∈ 𝑆 )

Proof

Step Hyp Ref Expression
1 lspval.v 𝑉 = ( Base ‘ 𝑊 )
2 lspval.s 𝑆 = ( LSubSp ‘ 𝑊 )
3 lspval.n 𝑁 = ( LSpan ‘ 𝑊 )
4 lspprcl.w ( 𝜑𝑊 ∈ LMod )
5 lspprcl.x ( 𝜑𝑋𝑉 )
6 lspprcl.y ( 𝜑𝑌𝑉 )
7 5 6 prssd ( 𝜑 → { 𝑋 , 𝑌 } ⊆ 𝑉 )
8 1 2 3 lspcl ( ( 𝑊 ∈ LMod ∧ { 𝑋 , 𝑌 } ⊆ 𝑉 ) → ( 𝑁 ‘ { 𝑋 , 𝑌 } ) ∈ 𝑆 )
9 4 7 8 syl2anc ( 𝜑 → ( 𝑁 ‘ { 𝑋 , 𝑌 } ) ∈ 𝑆 )