Metamath Proof Explorer


Theorem lspprid1

Description: A member of a pair of vectors belongs to their span. (Contributed by NM, 14-May-2015)

Ref Expression
Hypotheses lspprid.v 𝑉 = ( Base ‘ 𝑊 )
lspprid.n 𝑁 = ( LSpan ‘ 𝑊 )
lspprid.w ( 𝜑𝑊 ∈ LMod )
lspprid.x ( 𝜑𝑋𝑉 )
lspprid.y ( 𝜑𝑌𝑉 )
Assertion lspprid1 ( 𝜑𝑋 ∈ ( 𝑁 ‘ { 𝑋 , 𝑌 } ) )

Proof

Step Hyp Ref Expression
1 lspprid.v 𝑉 = ( Base ‘ 𝑊 )
2 lspprid.n 𝑁 = ( LSpan ‘ 𝑊 )
3 lspprid.w ( 𝜑𝑊 ∈ LMod )
4 lspprid.x ( 𝜑𝑋𝑉 )
5 lspprid.y ( 𝜑𝑌𝑉 )
6 4 5 prssd ( 𝜑 → { 𝑋 , 𝑌 } ⊆ 𝑉 )
7 snsspr1 { 𝑋 } ⊆ { 𝑋 , 𝑌 }
8 7 a1i ( 𝜑 → { 𝑋 } ⊆ { 𝑋 , 𝑌 } )
9 1 2 lspss ( ( 𝑊 ∈ LMod ∧ { 𝑋 , 𝑌 } ⊆ 𝑉 ∧ { 𝑋 } ⊆ { 𝑋 , 𝑌 } ) → ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑋 , 𝑌 } ) )
10 3 6 8 9 syl3anc ( 𝜑 → ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑋 , 𝑌 } ) )
11 eqid ( LSubSp ‘ 𝑊 ) = ( LSubSp ‘ 𝑊 )
12 1 11 2 3 4 5 lspprcl ( 𝜑 → ( 𝑁 ‘ { 𝑋 , 𝑌 } ) ∈ ( LSubSp ‘ 𝑊 ) )
13 1 11 2 3 12 4 lspsnel5 ( 𝜑 → ( 𝑋 ∈ ( 𝑁 ‘ { 𝑋 , 𝑌 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑋 , 𝑌 } ) ) )
14 10 13 mpbird ( 𝜑𝑋 ∈ ( 𝑁 ‘ { 𝑋 , 𝑌 } ) )