Metamath Proof Explorer


Theorem lsmelpr

Description: Two ways to say that a vector belongs to the span of a pair of vectors. (Contributed by NM, 14-Jan-2015)

Ref Expression
Hypotheses lsmelpr.v 𝑉 = ( Base ‘ 𝑊 )
lsmelpr.n 𝑁 = ( LSpan ‘ 𝑊 )
lsmelpr.p = ( LSSum ‘ 𝑊 )
lsmelpr.w ( 𝜑𝑊 ∈ LMod )
lsmelpr.x ( 𝜑𝑋𝑉 )
lsmelpr.y ( 𝜑𝑌𝑉 )
lsmelpr.z ( 𝜑𝑍𝑉 )
Assertion lsmelpr ( 𝜑 → ( 𝑋 ∈ ( 𝑁 ‘ { 𝑌 , 𝑍 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) ⊆ ( ( 𝑁 ‘ { 𝑌 } ) ( 𝑁 ‘ { 𝑍 } ) ) ) )

Proof

Step Hyp Ref Expression
1 lsmelpr.v 𝑉 = ( Base ‘ 𝑊 )
2 lsmelpr.n 𝑁 = ( LSpan ‘ 𝑊 )
3 lsmelpr.p = ( LSSum ‘ 𝑊 )
4 lsmelpr.w ( 𝜑𝑊 ∈ LMod )
5 lsmelpr.x ( 𝜑𝑋𝑉 )
6 lsmelpr.y ( 𝜑𝑌𝑉 )
7 lsmelpr.z ( 𝜑𝑍𝑉 )
8 eqid ( LSubSp ‘ 𝑊 ) = ( LSubSp ‘ 𝑊 )
9 1 8 2 4 6 7 lspprcl ( 𝜑 → ( 𝑁 ‘ { 𝑌 , 𝑍 } ) ∈ ( LSubSp ‘ 𝑊 ) )
10 1 8 2 4 9 5 lspsnel5 ( 𝜑 → ( 𝑋 ∈ ( 𝑁 ‘ { 𝑌 , 𝑍 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑌 , 𝑍 } ) ) )
11 1 2 3 4 6 7 lsmpr ( 𝜑 → ( 𝑁 ‘ { 𝑌 , 𝑍 } ) = ( ( 𝑁 ‘ { 𝑌 } ) ( 𝑁 ‘ { 𝑍 } ) ) )
12 11 sseq2d ( 𝜑 → ( ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑌 , 𝑍 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) ⊆ ( ( 𝑁 ‘ { 𝑌 } ) ( 𝑁 ‘ { 𝑍 } ) ) ) )
13 10 12 bitrd ( 𝜑 → ( 𝑋 ∈ ( 𝑁 ‘ { 𝑌 , 𝑍 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) ⊆ ( ( 𝑁 ‘ { 𝑌 } ) ( 𝑁 ‘ { 𝑍 } ) ) ) )