Metamath Proof Explorer


Theorem lspexchn2

Description: Exchange property for span of a pair with negated membership. TODO: look at uses of lspexch to see if this will shorten proofs. (Contributed by NM, 24-May-2015)

Ref Expression
Hypotheses lspexchn2.v 𝑉 = ( Base ‘ 𝑊 )
lspexchn2.n 𝑁 = ( LSpan ‘ 𝑊 )
lspexchn2.w ( 𝜑𝑊 ∈ LVec )
lspexchn2.x ( 𝜑𝑋𝑉 )
lspexchn2.y ( 𝜑𝑌𝑉 )
lspexchn2.z ( 𝜑𝑍𝑉 )
lspexchn2.q ( 𝜑 → ¬ 𝑌 ∈ ( 𝑁 ‘ { 𝑍 } ) )
lspexchn2.e ( 𝜑 → ¬ 𝑋 ∈ ( 𝑁 ‘ { 𝑍 , 𝑌 } ) )
Assertion lspexchn2 ( 𝜑 → ¬ 𝑌 ∈ ( 𝑁 ‘ { 𝑍 , 𝑋 } ) )

Proof

Step Hyp Ref Expression
1 lspexchn2.v 𝑉 = ( Base ‘ 𝑊 )
2 lspexchn2.n 𝑁 = ( LSpan ‘ 𝑊 )
3 lspexchn2.w ( 𝜑𝑊 ∈ LVec )
4 lspexchn2.x ( 𝜑𝑋𝑉 )
5 lspexchn2.y ( 𝜑𝑌𝑉 )
6 lspexchn2.z ( 𝜑𝑍𝑉 )
7 lspexchn2.q ( 𝜑 → ¬ 𝑌 ∈ ( 𝑁 ‘ { 𝑍 } ) )
8 lspexchn2.e ( 𝜑 → ¬ 𝑋 ∈ ( 𝑁 ‘ { 𝑍 , 𝑌 } ) )
9 prcom { 𝑍 , 𝑌 } = { 𝑌 , 𝑍 }
10 9 fveq2i ( 𝑁 ‘ { 𝑍 , 𝑌 } ) = ( 𝑁 ‘ { 𝑌 , 𝑍 } )
11 10 eleq2i ( 𝑋 ∈ ( 𝑁 ‘ { 𝑍 , 𝑌 } ) ↔ 𝑋 ∈ ( 𝑁 ‘ { 𝑌 , 𝑍 } ) )
12 8 11 sylnib ( 𝜑 → ¬ 𝑋 ∈ ( 𝑁 ‘ { 𝑌 , 𝑍 } ) )
13 1 2 3 4 5 6 7 12 lspexchn1 ( 𝜑 → ¬ 𝑌 ∈ ( 𝑁 ‘ { 𝑋 , 𝑍 } ) )
14 prcom { 𝑋 , 𝑍 } = { 𝑍 , 𝑋 }
15 14 fveq2i ( 𝑁 ‘ { 𝑋 , 𝑍 } ) = ( 𝑁 ‘ { 𝑍 , 𝑋 } )
16 15 eleq2i ( 𝑌 ∈ ( 𝑁 ‘ { 𝑋 , 𝑍 } ) ↔ 𝑌 ∈ ( 𝑁 ‘ { 𝑍 , 𝑋 } ) )
17 13 16 sylnib ( 𝜑 → ¬ 𝑌 ∈ ( 𝑁 ‘ { 𝑍 , 𝑋 } ) )