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Metamath Proof Explorer

Theorem dih0sb

Description: A subspace is zero iff the converse of its isomorphism is lattice zero. (Contributed by NM, 17-Aug-2014)

Ref Expression
Hypotheses dih0sb.h 
|- H = ( LHyp ` K )
dih0sb.o 
|- .0. = ( 0. ` K )
dih0sb.i 
|- I = ( ( DIsoH ` K ) ` W )
dih0sb.u 
|- U = ( ( DVecH ` K ) ` W )
dih0sb.v 
|- V = ( Base ` U )
dih0sb.z 
|- Z = ( 0g ` U )
dih0sb.n 
|- N = ( LSpan ` U )
dih0sb.k 
|- ( ph -> ( K e. HL /\ W e. H ) )
dih0sb.x 
|- ( ph -> X e. ran I )
Assertion dih0sb 
|- ( ph -> ( X = { Z } <-> ( `' I ` X ) = .0. ) )

Proof

Step Hyp Ref Expression
1 dih0sb.h 
 |-  H = ( LHyp ` K )
2 dih0sb.o 
 |-  .0. = ( 0. ` K )
3 dih0sb.i 
 |-  I = ( ( DIsoH ` K ) ` W )
4 dih0sb.u 
 |-  U = ( ( DVecH ` K ) ` W )
5 dih0sb.v 
 |-  V = ( Base ` U )
6 dih0sb.z 
 |-  Z = ( 0g ` U )
7 dih0sb.n 
 |-  N = ( LSpan ` U )
8 dih0sb.k 
 |-  ( ph -> ( K e. HL /\ W e. H ) )
9 dih0sb.x 
 |-  ( ph -> X e. ran I )
10 1 3 4 6 dih0rn 
 |-  ( ( K e. HL /\ W e. H ) -> { Z } e. ran I )
11 8 10 syl 
 |-  ( ph -> { Z } e. ran I )
12 1 3 8 9 11 dihcnv11 
 |-  ( ph -> ( ( `' I ` X ) = ( `' I ` { Z } ) <-> X = { Z } ) )
13 1 2 3 4 6 dih0cnv 
 |-  ( ( K e. HL /\ W e. H ) -> ( `' I ` { Z } ) = .0. )
14 8 13 syl 
 |-  ( ph -> ( `' I ` { Z } ) = .0. )
15 14 eqeq2d 
 |-  ( ph -> ( ( `' I ` X ) = ( `' I ` { Z } ) <-> ( `' I ` X ) = .0. ) )
16 12 15 bitr3d 
 |-  ( ph -> ( X = { Z } <-> ( `' I ` X ) = .0. ) )