Metamath Proof Explorer


Theorem islshpkrN

Description: The predicate "is a hyperplane" (of a left module or left vector space). TODO: should it be U = ( Kg ) or ( Kg ) = U as in lshpkrex ? Both standards seem to be used randomly throughout set.mm; we should decide on a preferred one. (Contributed by NM, 7-Oct-2014) (New usage is discouraged.)

Ref Expression
Hypotheses lshpset2.v V=BaseW
lshpset2.d D=ScalarW
lshpset2.z 0˙=0D
lshpset2.h H=LSHypW
lshpset2.f F=LFnlW
lshpset2.k K=LKerW
Assertion islshpkrN WLVecUHgFgV×0˙U=Kg

Proof

Step Hyp Ref Expression
1 lshpset2.v V=BaseW
2 lshpset2.d D=ScalarW
3 lshpset2.z 0˙=0D
4 lshpset2.h H=LSHypW
5 lshpset2.f F=LFnlW
6 lshpset2.k K=LKerW
7 1 2 3 4 5 6 lshpset2N WLVecH=s|gFgV×0˙s=Kg
8 7 eleq2d WLVecUHUs|gFgV×0˙s=Kg
9 elex Us|gFgV×0˙s=KgUV
10 9 adantl WLVecUs|gFgV×0˙s=KgUV
11 fvex KgV
12 eleq1 U=KgUVKgV
13 11 12 mpbiri U=KgUV
14 13 adantl gV×0˙U=KgUV
15 14 rexlimivw gFgV×0˙U=KgUV
16 15 adantl WLVecgFgV×0˙U=KgUV
17 eqeq1 s=Us=KgU=Kg
18 17 anbi2d s=UgV×0˙s=KggV×0˙U=Kg
19 18 rexbidv s=UgFgV×0˙s=KggFgV×0˙U=Kg
20 19 elabg UVUs|gFgV×0˙s=KggFgV×0˙U=Kg
21 10 16 20 pm5.21nd WLVecUs|gFgV×0˙s=KggFgV×0˙U=Kg
22 8 21 bitrd WLVecUHgFgV×0˙U=Kg