Metamath Proof Explorer


Theorem lshpne0

Description: The member of the span in the hyperplane definition does not belong to the hyperplane. (Contributed by NM, 14-Jul-2014) (Proof shortened by AV, 19-Jul-2022)

Ref Expression
Hypotheses lshpne0.v
|- V = ( Base ` W )
lshpne0.n
|- N = ( LSpan ` W )
lshpne0.p
|- .(+) = ( LSSum ` W )
lshpne0.h
|- H = ( LSHyp ` W )
lshpne0.o
|- .0. = ( 0g ` W )
lshpne0.w
|- ( ph -> W e. LMod )
lshpne0.u
|- ( ph -> U e. H )
lshpne0.x
|- ( ph -> X e. V )
lshpne0.e
|- ( ph -> ( U .(+) ( N ` { X } ) ) = V )
Assertion lshpne0
|- ( ph -> X =/= .0. )

Proof

Step Hyp Ref Expression
1 lshpne0.v
 |-  V = ( Base ` W )
2 lshpne0.n
 |-  N = ( LSpan ` W )
3 lshpne0.p
 |-  .(+) = ( LSSum ` W )
4 lshpne0.h
 |-  H = ( LSHyp ` W )
5 lshpne0.o
 |-  .0. = ( 0g ` W )
6 lshpne0.w
 |-  ( ph -> W e. LMod )
7 lshpne0.u
 |-  ( ph -> U e. H )
8 lshpne0.x
 |-  ( ph -> X e. V )
9 lshpne0.e
 |-  ( ph -> ( U .(+) ( N ` { X } ) ) = V )
10 eqid
 |-  ( LSubSp ` W ) = ( LSubSp ` W )
11 10 4 6 7 lshplss
 |-  ( ph -> U e. ( LSubSp ` W ) )
12 1 2 3 4 6 7 8 9 lshpnel
 |-  ( ph -> -. X e. U )
13 5 10 6 11 12 lssvneln0
 |-  ( ph -> X =/= .0. )