Metamath Proof Explorer


Theorem dih0cnv

Description: The isomorphism H converse value of the zero subspace is the lattice zero. (Contributed by NM, 19-Jun-2014)

Ref Expression
Hypotheses dih0cnv.h 𝐻 = ( LHyp ‘ 𝐾 )
dih0cnv.o 0 = ( 0. ‘ 𝐾 )
dih0cnv.i 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
dih0cnv.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
dih0cnv.z 𝑍 = ( 0g𝑈 )
Assertion dih0cnv ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝐼 ‘ { 𝑍 } ) = 0 )

Proof

Step Hyp Ref Expression
1 dih0cnv.h 𝐻 = ( LHyp ‘ 𝐾 )
2 dih0cnv.o 0 = ( 0. ‘ 𝐾 )
3 dih0cnv.i 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
4 dih0cnv.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
5 dih0cnv.z 𝑍 = ( 0g𝑈 )
6 2 1 3 4 5 dih0 ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝐼0 ) = { 𝑍 } )
7 6 fveq2d ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝐼 ‘ ( 𝐼0 ) ) = ( 𝐼 ‘ { 𝑍 } ) )
8 hlatl ( 𝐾 ∈ HL → 𝐾 ∈ AtLat )
9 8 adantr ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 𝐾 ∈ AtLat )
10 eqid ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
11 10 2 atl0cl ( 𝐾 ∈ AtLat → 0 ∈ ( Base ‘ 𝐾 ) )
12 9 11 syl ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 0 ∈ ( Base ‘ 𝐾 ) )
13 10 1 3 dihcnvid1 ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 0 ∈ ( Base ‘ 𝐾 ) ) → ( 𝐼 ‘ ( 𝐼0 ) ) = 0 )
14 12 13 mpdan ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝐼 ‘ ( 𝐼0 ) ) = 0 )
15 7 14 eqtr3d ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝐼 ‘ { 𝑍 } ) = 0 )