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 = ( 0g𝐷 )
lshpset2.h 𝐻 = ( LSHyp ‘ 𝑊 )
lshpset2.f 𝐹 = ( LFnl ‘ 𝑊 )
lshpset2.k 𝐾 = ( LKer ‘ 𝑊 )
Assertion islshpkrN ( 𝑊 ∈ LVec → ( 𝑈𝐻 ↔ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) ) )

Proof

Step Hyp Ref Expression
1 lshpset2.v 𝑉 = ( Base ‘ 𝑊 )
2 lshpset2.d 𝐷 = ( Scalar ‘ 𝑊 )
3 lshpset2.z 0 = ( 0g𝐷 )
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 ∣ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑠 = ( 𝐾𝑔 ) ) } )
8 7 eleq2d ( 𝑊 ∈ LVec → ( 𝑈𝐻𝑈 ∈ { 𝑠 ∣ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑠 = ( 𝐾𝑔 ) ) } ) )
9 elex ( 𝑈 ∈ { 𝑠 ∣ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑠 = ( 𝐾𝑔 ) ) } → 𝑈 ∈ V )
10 9 adantl ( ( 𝑊 ∈ LVec ∧ 𝑈 ∈ { 𝑠 ∣ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑠 = ( 𝐾𝑔 ) ) } ) → 𝑈 ∈ V )
11 fvex ( 𝐾𝑔 ) ∈ V
12 eleq1 ( 𝑈 = ( 𝐾𝑔 ) → ( 𝑈 ∈ V ↔ ( 𝐾𝑔 ) ∈ V ) )
13 11 12 mpbiri ( 𝑈 = ( 𝐾𝑔 ) → 𝑈 ∈ V )
14 13 adantl ( ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) → 𝑈 ∈ V )
15 14 rexlimivw ( ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) → 𝑈 ∈ V )
16 15 adantl ( ( 𝑊 ∈ LVec ∧ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) ) → 𝑈 ∈ V )
17 eqeq1 ( 𝑠 = 𝑈 → ( 𝑠 = ( 𝐾𝑔 ) ↔ 𝑈 = ( 𝐾𝑔 ) ) )
18 17 anbi2d ( 𝑠 = 𝑈 → ( ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑠 = ( 𝐾𝑔 ) ) ↔ ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) ) )
19 18 rexbidv ( 𝑠 = 𝑈 → ( ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑠 = ( 𝐾𝑔 ) ) ↔ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) ) )
20 19 elabg ( 𝑈 ∈ V → ( 𝑈 ∈ { 𝑠 ∣ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑠 = ( 𝐾𝑔 ) ) } ↔ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) ) )
21 10 16 20 pm5.21nd ( 𝑊 ∈ LVec → ( 𝑈 ∈ { 𝑠 ∣ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑠 = ( 𝐾𝑔 ) ) } ↔ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) ) )
22 8 21 bitrd ( 𝑊 ∈ LVec → ( 𝑈𝐻 ↔ ∃ 𝑔𝐹 ( 𝑔 ≠ ( 𝑉 × { 0 } ) ∧ 𝑈 = ( 𝐾𝑔 ) ) ) )