Metamath Proof Explorer


Theorem lshplss

Description: A hyperplane is a subspace. (Contributed by NM, 3-Jul-2014)

Ref Expression
Hypotheses lshplss.s
|- S = ( LSubSp ` W )
lshplss.h
|- H = ( LSHyp ` W )
lshplss.w
|- ( ph -> W e. LMod )
lshplss.u
|- ( ph -> U e. H )
Assertion lshplss
|- ( ph -> U e. S )

Proof

Step Hyp Ref Expression
1 lshplss.s
 |-  S = ( LSubSp ` W )
2 lshplss.h
 |-  H = ( LSHyp ` W )
3 lshplss.w
 |-  ( ph -> W e. LMod )
4 lshplss.u
 |-  ( ph -> U e. H )
5 eqid
 |-  ( Base ` W ) = ( Base ` W )
6 eqid
 |-  ( LSpan ` W ) = ( LSpan ` W )
7 5 6 1 2 islshp
 |-  ( W e. LMod -> ( U e. H <-> ( U e. S /\ U =/= ( Base ` W ) /\ E. v e. ( Base ` W ) ( ( LSpan ` W ) ` ( U u. { v } ) ) = ( Base ` W ) ) ) )
8 3 7 syl
 |-  ( ph -> ( U e. H <-> ( U e. S /\ U =/= ( Base ` W ) /\ E. v e. ( Base ` W ) ( ( LSpan ` W ) ` ( U u. { v } ) ) = ( Base ` W ) ) ) )
9 4 8 mpbid
 |-  ( ph -> ( U e. S /\ U =/= ( Base ` W ) /\ E. v e. ( Base ` W ) ( ( LSpan ` W ) ` ( U u. { v } ) ) = ( Base ` W ) ) )
10 9 simp1d
 |-  ( ph -> U e. S )