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 = Base W
lshpset2.d D = Scalar W
lshpset2.z 0 ˙ = 0 D
lshpset2.h H = LSHyp W
lshpset2.f F = LFnl W
lshpset2.k K = LKer W
Assertion islshpkrN W LVec U H g F g V × 0 ˙ U = K g

Proof

Step Hyp Ref Expression
1 lshpset2.v V = Base W
2 lshpset2.d D = Scalar W
3 lshpset2.z 0 ˙ = 0 D
4 lshpset2.h H = LSHyp W
5 lshpset2.f F = LFnl W
6 lshpset2.k K = LKer W
7 1 2 3 4 5 6 lshpset2N W LVec H = s | g F g V × 0 ˙ s = K g
8 7 eleq2d W LVec U H U s | g F g V × 0 ˙ s = K g
9 elex U s | g F g V × 0 ˙ s = K g U V
10 9 adantl W LVec U s | g F g V × 0 ˙ s = K g U V
11 fvex K g V
12 eleq1 U = K g U V K g V
13 11 12 mpbiri U = K g U V
14 13 adantl g V × 0 ˙ U = K g U V
15 14 rexlimivw g F g V × 0 ˙ U = K g U V
16 15 adantl W LVec g F g V × 0 ˙ U = K g U V
17 eqeq1 s = U s = K g U = K g
18 17 anbi2d s = U g V × 0 ˙ s = K g g V × 0 ˙ U = K g
19 18 rexbidv s = U g F g V × 0 ˙ s = K g g F g V × 0 ˙ U = K g
20 19 elabg U V U s | g F g V × 0 ˙ s = K g g F g V × 0 ˙ U = K g
21 10 16 20 pm5.21nd W LVec U s | g F g V × 0 ˙ s = K g g F g V × 0 ˙ U = K g
22 8 21 bitrd W LVec U H g F g V × 0 ˙ U = K g