Metamath Proof Explorer


Theorem lcfrlem12N

Description: Lemma for lcfr . (Contributed by NM, 23-Feb-2015) (New usage is discouraged.)

Ref Expression
Hypotheses lcf1o.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
lcf1o.o ⊢ ⊥ = ( ( ocH ‘ 𝐾 ) ‘ 𝑊 )
lcf1o.u ⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
lcf1o.v ⊢ 𝑉 = ( Base ‘ 𝑈 )
lcf1o.a ⊢ + = ( +g ‘ 𝑈 )
lcf1o.t ⊢ · = ( ·𝑠 ‘ 𝑈 )
lcf1o.s ⊢ 𝑆 = ( Scalar ‘ 𝑈 )
lcf1o.r ⊢ 𝑅 = ( Base ‘ 𝑆 )
lcf1o.z ⊢ 0 = ( 0g ‘ 𝑈 )
lcf1o.f ⊢ 𝐹 = ( LFnl ‘ 𝑈 )
lcf1o.l ⊢ 𝐿 = ( LKer ‘ 𝑈 )
lcf1o.d ⊢ 𝐷 = ( LDual ‘ 𝑈 )
lcf1o.q ⊢ 𝑄 = ( 0g ‘ 𝐷 )
lcf1o.c ⊢ 𝐶 = { 𝑓 ∈ 𝐹 ∣ ( ⊥ ‘ ( ⊥ ‘ ( 𝐿 ‘ 𝑓 ) ) ) = ( 𝐿 ‘ 𝑓 ) }
lcf1o.j ⊢ 𝐽 = ( 𝑥 ∈ ( 𝑉 ∖ { 0 } ) ↦ ( 𝑣 ∈ 𝑉 ↦ ( ℩ 𝑘 ∈ 𝑅 ∃ 𝑤 ∈ ( ⊥ ‘ { 𝑥 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑥 ) ) ) ) )
lcflo.k ⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
lcfrlem10.x ⊢ ( 𝜑 → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
lcfrlem12.b ⊢ 𝐵 = ( 0g ‘ 𝑆 )
lcfrlem12.y ⊢ ( 𝜑 → 𝑌 ∈ ( ⊥ ‘ { 𝑋 } ) )
Assertion lcfrlem12N ( 𝜑 → ( ( 𝐽 ‘ 𝑋 ) ‘ 𝑌 ) = 𝐵 )

Proof

Step Hyp Ref Expression
1 lcf1o.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
2 lcf1o.o ⊢ ⊥ = ( ( ocH ‘ 𝐾 ) ‘ 𝑊 )
3 lcf1o.u ⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
4 lcf1o.v ⊢ 𝑉 = ( Base ‘ 𝑈 )
5 lcf1o.a ⊢ + = ( +g ‘ 𝑈 )
6 lcf1o.t ⊢ · = ( ·𝑠 ‘ 𝑈 )
7 lcf1o.s ⊢ 𝑆 = ( Scalar ‘ 𝑈 )
8 lcf1o.r ⊢ 𝑅 = ( Base ‘ 𝑆 )
9 lcf1o.z ⊢ 0 = ( 0g ‘ 𝑈 )
10 lcf1o.f ⊢ 𝐹 = ( LFnl ‘ 𝑈 )
11 lcf1o.l ⊢ 𝐿 = ( LKer ‘ 𝑈 )
12 lcf1o.d ⊢ 𝐷 = ( LDual ‘ 𝑈 )
13 lcf1o.q ⊢ 𝑄 = ( 0g ‘ 𝐷 )
14 lcf1o.c ⊢ 𝐶 = { 𝑓 ∈ 𝐹 ∣ ( ⊥ ‘ ( ⊥ ‘ ( 𝐿 ‘ 𝑓 ) ) ) = ( 𝐿 ‘ 𝑓 ) }
15 lcf1o.j ⊢ 𝐽 = ( 𝑥 ∈ ( 𝑉 ∖ { 0 } ) ↦ ( 𝑣 ∈ 𝑉 ↦ ( ℩ 𝑘 ∈ 𝑅 ∃ 𝑤 ∈ ( ⊥ ‘ { 𝑥 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑥 ) ) ) ) )
16 lcflo.k ⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
17 lcfrlem10.x ⊢ ( 𝜑 → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
18 lcfrlem12.b ⊢ 𝐵 = ( 0g ‘ 𝑆 )
19 lcfrlem12.y ⊢ ( 𝜑 → 𝑌 ∈ ( ⊥ ‘ { 𝑋 } ) )
20 1 3 16 dvhlmod ⊢ ( 𝜑 → 𝑈 ∈ LMod )
21 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 lcfrlem10 ⊢ ( 𝜑 → ( 𝐽 ‘ 𝑋 ) ∈ 𝐹 )
22 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 lcfrlem11 ⊢ ( 𝜑 → ( 𝐿 ‘ ( 𝐽 ‘ 𝑋 ) ) = ( ⊥ ‘ { 𝑋 } ) )
23 19 22 eleqtrrd ⊢ ( 𝜑 → 𝑌 ∈ ( 𝐿 ‘ ( 𝐽 ‘ 𝑋 ) ) )
24 7 18 10 11 lkrf0 ⊢ ( ( 𝑈 ∈ LMod ∧ ( 𝐽 ‘ 𝑋 ) ∈ 𝐹 ∧ 𝑌 ∈ ( 𝐿 ‘ ( 𝐽 ‘ 𝑋 ) ) ) → ( ( 𝐽 ‘ 𝑋 ) ‘ 𝑌 ) = 𝐵 )
25 20 21 23 24 syl3anc ⊢ ( 𝜑 → ( ( 𝐽 ‘ 𝑋 ) ‘ 𝑌 ) = 𝐵 )