Metamath Proof Explorer


Theorem lspsnne1

Description: Two ways to express that vectors have different spans. (Contributed by NM, 28-May-2015)

Ref Expression
Hypotheses lspsnne1.v 𝑉 = ( Base ‘ 𝑊 )
lspsnne1.o 0 = ( 0g𝑊 )
lspsnne1.n 𝑁 = ( LSpan ‘ 𝑊 )
lspsnne1.w ( 𝜑𝑊 ∈ LVec )
lspsnne1.x ( 𝜑𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
lspsnne1.y ( 𝜑𝑌𝑉 )
lspsnne1.e ( 𝜑 → ( 𝑁 ‘ { 𝑋 } ) ≠ ( 𝑁 ‘ { 𝑌 } ) )
Assertion lspsnne1 ( 𝜑 → ¬ 𝑋 ∈ ( 𝑁 ‘ { 𝑌 } ) )

Proof

Step Hyp Ref Expression
1 lspsnne1.v 𝑉 = ( Base ‘ 𝑊 )
2 lspsnne1.o 0 = ( 0g𝑊 )
3 lspsnne1.n 𝑁 = ( LSpan ‘ 𝑊 )
4 lspsnne1.w ( 𝜑𝑊 ∈ LVec )
5 lspsnne1.x ( 𝜑𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
6 lspsnne1.y ( 𝜑𝑌𝑉 )
7 lspsnne1.e ( 𝜑 → ( 𝑁 ‘ { 𝑋 } ) ≠ ( 𝑁 ‘ { 𝑌 } ) )
8 eqid ( LSubSp ‘ 𝑊 ) = ( LSubSp ‘ 𝑊 )
9 lveclmod ( 𝑊 ∈ LVec → 𝑊 ∈ LMod )
10 4 9 syl ( 𝜑𝑊 ∈ LMod )
11 1 8 3 lspsncl ( ( 𝑊 ∈ LMod ∧ 𝑌𝑉 ) → ( 𝑁 ‘ { 𝑌 } ) ∈ ( LSubSp ‘ 𝑊 ) )
12 10 6 11 syl2anc ( 𝜑 → ( 𝑁 ‘ { 𝑌 } ) ∈ ( LSubSp ‘ 𝑊 ) )
13 5 eldifad ( 𝜑𝑋𝑉 )
14 1 8 3 10 12 13 lspsnel5 ( 𝜑 → ( 𝑋 ∈ ( 𝑁 ‘ { 𝑌 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑌 } ) ) )
15 14 notbid ( 𝜑 → ( ¬ 𝑋 ∈ ( 𝑁 ‘ { 𝑌 } ) ↔ ¬ ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑌 } ) ) )
16 1 2 3 4 5 6 lspsncmp ( 𝜑 → ( ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑌 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) = ( 𝑁 ‘ { 𝑌 } ) ) )
17 16 necon3bbid ( 𝜑 → ( ¬ ( 𝑁 ‘ { 𝑋 } ) ⊆ ( 𝑁 ‘ { 𝑌 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) ≠ ( 𝑁 ‘ { 𝑌 } ) ) )
18 15 17 bitrd ( 𝜑 → ( ¬ 𝑋 ∈ ( 𝑁 ‘ { 𝑌 } ) ↔ ( 𝑁 ‘ { 𝑋 } ) ≠ ( 𝑁 ‘ { 𝑌 } ) ) )
19 7 18 mpbird ( 𝜑 → ¬ 𝑋 ∈ ( 𝑁 ‘ { 𝑌 } ) )