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 - ˙ = - D
ldualssvsubcl.s S = LSubSp D
ldualssvsubcl.w φ W LMod
ldualssvsubcl.u φ U S
ldualssvsubcl.x φ X U
ldualssvsubcl.y φ Y U
Assertion ldualssvsubcl φ X - ˙ Y U

Proof

Step Hyp Ref Expression
1 ldualssvsubcl.d D = LDual W
2 ldualssvsubcl.m - ˙ = - D
3 ldualssvsubcl.s S = LSubSp D
4 ldualssvsubcl.w φ W LMod
5 ldualssvsubcl.u φ U S
6 ldualssvsubcl.x φ X U
7 ldualssvsubcl.y φ Y U
8 1 4 lduallmod φ D LMod
9 2 3 lssvsubcl D LMod U S X U Y U X - ˙ Y U
10 8 5 6 7 9 syl22anc φ X - ˙ Y U