Metamath Proof Explorer


Theorem dihlatat

Description: The reverse isomorphism H of a 1-dim subspace is an atom. (Contributed by NM, 28-Apr-2014)

Ref Expression
Hypotheses dihlatat.a 𝐴 = ( Atoms ‘ 𝐾 )
dihlatat.h 𝐻 = ( LHyp ‘ 𝐾 )
dihlatat.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
dihlatat.i 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
dihlatat.l 𝐿 = ( LSAtoms ‘ 𝑈 )
Assertion dihlatat ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑄𝐿 ) → ( 𝐼𝑄 ) ∈ 𝐴 )

Proof

Step Hyp Ref Expression
1 dihlatat.a 𝐴 = ( Atoms ‘ 𝐾 )
2 dihlatat.h 𝐻 = ( LHyp ‘ 𝐾 )
3 dihlatat.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
4 dihlatat.i 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
5 dihlatat.l 𝐿 = ( LSAtoms ‘ 𝑈 )
6 id ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
7 2 3 6 dvhlvec ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 𝑈 ∈ LVec )
8 eqid ( Base ‘ 𝑈 ) = ( Base ‘ 𝑈 )
9 eqid ( LSpan ‘ 𝑈 ) = ( LSpan ‘ 𝑈 )
10 eqid ( 0g𝑈 ) = ( 0g𝑈 )
11 8 9 10 5 islsat ( 𝑈 ∈ LVec → ( 𝑄𝐿 ↔ ∃ 𝑣 ∈ ( ( Base ‘ 𝑈 ) ∖ { ( 0g𝑈 ) } ) 𝑄 = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) ) )
12 7 11 syl ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝑄𝐿 ↔ ∃ 𝑣 ∈ ( ( Base ‘ 𝑈 ) ∖ { ( 0g𝑈 ) } ) 𝑄 = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) ) )
13 12 biimpa ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑄𝐿 ) → ∃ 𝑣 ∈ ( ( Base ‘ 𝑈 ) ∖ { ( 0g𝑈 ) } ) 𝑄 = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) )
14 eldifsn ( 𝑣 ∈ ( ( Base ‘ 𝑈 ) ∖ { ( 0g𝑈 ) } ) ↔ ( 𝑣 ∈ ( Base ‘ 𝑈 ) ∧ 𝑣 ≠ ( 0g𝑈 ) ) )
15 1 2 3 8 10 9 4 dihlspsnat ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑣 ∈ ( Base ‘ 𝑈 ) ∧ 𝑣 ≠ ( 0g𝑈 ) ) → ( 𝐼 ‘ ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) ) ∈ 𝐴 )
16 15 3expb ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝑣 ∈ ( Base ‘ 𝑈 ) ∧ 𝑣 ≠ ( 0g𝑈 ) ) ) → ( 𝐼 ‘ ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) ) ∈ 𝐴 )
17 14 16 sylan2b ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑣 ∈ ( ( Base ‘ 𝑈 ) ∖ { ( 0g𝑈 ) } ) ) → ( 𝐼 ‘ ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) ) ∈ 𝐴 )
18 fveq2 ( 𝑄 = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) → ( 𝐼𝑄 ) = ( 𝐼 ‘ ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) ) )
19 18 eleq1d ( 𝑄 = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) → ( ( 𝐼𝑄 ) ∈ 𝐴 ↔ ( 𝐼 ‘ ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) ) ∈ 𝐴 ) )
20 17 19 syl5ibrcom ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑣 ∈ ( ( Base ‘ 𝑈 ) ∖ { ( 0g𝑈 ) } ) ) → ( 𝑄 = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) → ( 𝐼𝑄 ) ∈ 𝐴 ) )
21 20 rexlimdva ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( ∃ 𝑣 ∈ ( ( Base ‘ 𝑈 ) ∖ { ( 0g𝑈 ) } ) 𝑄 = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) → ( 𝐼𝑄 ) ∈ 𝐴 ) )
22 21 adantr ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑄𝐿 ) → ( ∃ 𝑣 ∈ ( ( Base ‘ 𝑈 ) ∖ { ( 0g𝑈 ) } ) 𝑄 = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑣 } ) → ( 𝐼𝑄 ) ∈ 𝐴 ) )
23 13 22 mpd ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑄𝐿 ) → ( 𝐼𝑄 ) ∈ 𝐴 )