Metamath Proof Explorer


Theorem ldualssvsubcl

Description: Closure of vector subtraction in a dual subspace.) (Contributed by NM, 9-Mar-2015)

Ref Expression
Hypotheses ldualssvsubcl.d
|- D = ( LDual ` W )
ldualssvsubcl.m
|- .- = ( -g ` D )
ldualssvsubcl.s
|- S = ( LSubSp ` D )
ldualssvsubcl.w
|- ( ph -> W e. LMod )
ldualssvsubcl.u
|- ( ph -> U e. S )
ldualssvsubcl.x
|- ( ph -> X e. U )
ldualssvsubcl.y
|- ( ph -> Y e. U )
Assertion ldualssvsubcl
|- ( ph -> ( X .- Y ) e. U )

Proof

Step Hyp Ref Expression
1 ldualssvsubcl.d
 |-  D = ( LDual ` W )
2 ldualssvsubcl.m
 |-  .- = ( -g ` D )
3 ldualssvsubcl.s
 |-  S = ( LSubSp ` D )
4 ldualssvsubcl.w
 |-  ( ph -> W e. LMod )
5 ldualssvsubcl.u
 |-  ( ph -> U e. S )
6 ldualssvsubcl.x
 |-  ( ph -> X e. U )
7 ldualssvsubcl.y
 |-  ( ph -> Y e. U )
8 1 4 lduallmod
 |-  ( ph -> D e. LMod )
9 2 3 lssvsubcl
 |-  ( ( ( D e. LMod /\ U e. S ) /\ ( X e. U /\ Y e. U ) ) -> ( X .- Y ) e. U )
10 8 5 6 7 9 syl22anc
 |-  ( ph -> ( X .- Y ) e. U )