Metamath Proof Explorer


Theorem dvh2dim

Description: There is a vector that is outside the span of another. (Contributed by NM, 25-Apr-2015)

Ref Expression
Hypotheses dvh3dim.h 𝐻 = ( LHyp ‘ 𝐾 )
dvh3dim.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
dvh3dim.v 𝑉 = ( Base ‘ 𝑈 )
dvh3dim.n 𝑁 = ( LSpan ‘ 𝑈 )
dvh3dim.k ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
dvh3dim.x ( 𝜑𝑋𝑉 )
Assertion dvh2dim ( 𝜑 → ∃ 𝑧𝑉 ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) )

Proof

Step Hyp Ref Expression
1 dvh3dim.h 𝐻 = ( LHyp ‘ 𝐾 )
2 dvh3dim.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
3 dvh3dim.v 𝑉 = ( Base ‘ 𝑈 )
4 dvh3dim.n 𝑁 = ( LSpan ‘ 𝑈 )
5 dvh3dim.k ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
6 dvh3dim.x ( 𝜑𝑋𝑉 )
7 eqid ( 0g𝑈 ) = ( 0g𝑈 )
8 1 2 3 7 5 dvh1dim ( 𝜑 → ∃ 𝑧𝑉 𝑧 ≠ ( 0g𝑈 ) )
9 8 adantr ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ∃ 𝑧𝑉 𝑧 ≠ ( 0g𝑈 ) )
10 simpr ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → 𝑋 = ( 0g𝑈 ) )
11 10 sneqd ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → { 𝑋 } = { ( 0g𝑈 ) } )
12 11 fveq2d ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ( 𝑁 ‘ { 𝑋 } ) = ( 𝑁 ‘ { ( 0g𝑈 ) } ) )
13 1 2 5 dvhlmod ( 𝜑𝑈 ∈ LMod )
14 7 4 lspsn0 ( 𝑈 ∈ LMod → ( 𝑁 ‘ { ( 0g𝑈 ) } ) = { ( 0g𝑈 ) } )
15 13 14 syl ( 𝜑 → ( 𝑁 ‘ { ( 0g𝑈 ) } ) = { ( 0g𝑈 ) } )
16 15 adantr ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ( 𝑁 ‘ { ( 0g𝑈 ) } ) = { ( 0g𝑈 ) } )
17 12 16 eqtrd ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ( 𝑁 ‘ { 𝑋 } ) = { ( 0g𝑈 ) } )
18 17 eleq2d ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ( 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) ↔ 𝑧 ∈ { ( 0g𝑈 ) } ) )
19 velsn ( 𝑧 ∈ { ( 0g𝑈 ) } ↔ 𝑧 = ( 0g𝑈 ) )
20 18 19 bitrdi ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ( 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) ↔ 𝑧 = ( 0g𝑈 ) ) )
21 20 necon3bbid ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ( ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) ↔ 𝑧 ≠ ( 0g𝑈 ) ) )
22 21 rexbidv ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ( ∃ 𝑧𝑉 ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) ↔ ∃ 𝑧𝑉 𝑧 ≠ ( 0g𝑈 ) ) )
23 9 22 mpbird ( ( 𝜑𝑋 = ( 0g𝑈 ) ) → ∃ 𝑧𝑉 ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) )
24 5 adantr ( ( 𝜑𝑋 ≠ ( 0g𝑈 ) ) → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
25 6 adantr ( ( 𝜑𝑋 ≠ ( 0g𝑈 ) ) → 𝑋𝑉 )
26 simpr ( ( 𝜑𝑋 ≠ ( 0g𝑈 ) ) → 𝑋 ≠ ( 0g𝑈 ) )
27 1 2 3 4 24 25 25 7 26 26 dvhdimlem ( ( 𝜑𝑋 ≠ ( 0g𝑈 ) ) → ∃ 𝑧𝑉 ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 , 𝑋 } ) )
28 dfsn2 { 𝑋 } = { 𝑋 , 𝑋 }
29 28 fveq2i ( 𝑁 ‘ { 𝑋 } ) = ( 𝑁 ‘ { 𝑋 , 𝑋 } )
30 29 eleq2i ( 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) ↔ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 , 𝑋 } ) )
31 30 notbii ( ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) ↔ ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 , 𝑋 } ) )
32 31 rexbii ( ∃ 𝑧𝑉 ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) ↔ ∃ 𝑧𝑉 ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 , 𝑋 } ) )
33 27 32 sylibr ( ( 𝜑𝑋 ≠ ( 0g𝑈 ) ) → ∃ 𝑧𝑉 ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) )
34 23 33 pm2.61dane ( 𝜑 → ∃ 𝑧𝑉 ¬ 𝑧 ∈ ( 𝑁 ‘ { 𝑋 } ) )