Theorem lcfrlem11

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

	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 } ) )
Assertion	lcfrlem11	⊢ ( 𝜑 → ( 𝐿 ‘ ( 𝐽 ‘ 𝑋 ) ) = ( ⊥ ‘ { 𝑋 } ) )

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	1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17	lcfrlem8	⊢ ( 𝜑 → ( 𝐽 ‘ 𝑋 ) = ( 𝑣 ∈ 𝑉 ↦ ( ℩ 𝑘 ∈ 𝑅 ∃ 𝑤 ∈ ( ⊥ ‘ { 𝑋 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑋 ) ) ) ) )
19	18	fveq2d	⊢ ( 𝜑 → ( 𝐿 ‘ ( 𝐽 ‘ 𝑋 ) ) = ( 𝐿 ‘ ( 𝑣 ∈ 𝑉 ↦ ( ℩ 𝑘 ∈ 𝑅 ∃ 𝑤 ∈ ( ⊥ ‘ { 𝑋 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑋 ) ) ) ) ) )
20		eqid	⊢ ( 𝑣 ∈ 𝑉 ↦ ( ℩ 𝑘 ∈ 𝑅 ∃ 𝑤 ∈ ( ⊥ ‘ { 𝑋 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑋 ) ) ) ) = ( 𝑣 ∈ 𝑉 ↦ ( ℩ 𝑘 ∈ 𝑅 ∃ 𝑤 ∈ ( ⊥ ‘ { 𝑋 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑋 ) ) ) )
21	1 2 3 4 9 5 6 11 7 8 20 16 17	dochsnkr2	⊢ ( 𝜑 → ( 𝐿 ‘ ( 𝑣 ∈ 𝑉 ↦ ( ℩ 𝑘 ∈ 𝑅 ∃ 𝑤 ∈ ( ⊥ ‘ { 𝑋 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑋 ) ) ) ) ) = ( ⊥ ‘ { 𝑋 } ) )
22	19 21	eqtrd	⊢ ( 𝜑 → ( 𝐿 ‘ ( 𝐽 ‘ 𝑋 ) ) = ( ⊥ ‘ { 𝑋 } ) )