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