Metamath Proof Explorer


Theorem lcfrlem15

Description: Lemma for lcfr . (Contributed by NM, 9-Mar-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 + = ( +g𝑈 )
lcf1o.t · = ( ·𝑠𝑈 )
lcf1o.s 𝑆 = ( Scalar ‘ 𝑈 )
lcf1o.r 𝑅 = ( Base ‘ 𝑆 )
lcf1o.z 0 = ( 0g𝑈 )
lcf1o.f 𝐹 = ( LFnl ‘ 𝑈 )
lcf1o.l 𝐿 = ( LKer ‘ 𝑈 )
lcf1o.d 𝐷 = ( LDual ‘ 𝑈 )
lcf1o.q 𝑄 = ( 0g𝐷 )
lcf1o.c 𝐶 = { 𝑓𝐹 ∣ ( ‘ ( ‘ ( 𝐿𝑓 ) ) ) = ( 𝐿𝑓 ) }
lcf1o.j 𝐽 = ( 𝑥 ∈ ( 𝑉 ∖ { 0 } ) ↦ ( 𝑣𝑉 ↦ ( 𝑘𝑅𝑤 ∈ ( ‘ { 𝑥 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑥 ) ) ) ) )
lcflo.k ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
lcfrlem10.x ( 𝜑𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
Assertion lcfrlem15 ( 𝜑𝑋 ∈ ( ‘ ( 𝐿 ‘ ( 𝐽𝑋 ) ) ) )

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 + = ( +g𝑈 )
6 lcf1o.t · = ( ·𝑠𝑈 )
7 lcf1o.s 𝑆 = ( Scalar ‘ 𝑈 )
8 lcf1o.r 𝑅 = ( Base ‘ 𝑆 )
9 lcf1o.z 0 = ( 0g𝑈 )
10 lcf1o.f 𝐹 = ( LFnl ‘ 𝑈 )
11 lcf1o.l 𝐿 = ( LKer ‘ 𝑈 )
12 lcf1o.d 𝐷 = ( LDual ‘ 𝑈 )
13 lcf1o.q 𝑄 = ( 0g𝐷 )
14 lcf1o.c 𝐶 = { 𝑓𝐹 ∣ ( ‘ ( ‘ ( 𝐿𝑓 ) ) ) = ( 𝐿𝑓 ) }
15 lcf1o.j 𝐽 = ( 𝑥 ∈ ( 𝑉 ∖ { 0 } ) ↦ ( 𝑣𝑉 ↦ ( 𝑘𝑅𝑤 ∈ ( ‘ { 𝑥 } ) 𝑣 = ( 𝑤 + ( 𝑘 · 𝑥 ) ) ) ) )
16 lcflo.k ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
17 lcfrlem10.x ( 𝜑𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
18 1 3 16 dvhlmod ( 𝜑𝑈 ∈ LMod )
19 17 eldifad ( 𝜑𝑋𝑉 )
20 eqid ( LSpan ‘ 𝑈 ) = ( LSpan ‘ 𝑈 )
21 4 20 lspsnid ( ( 𝑈 ∈ LMod ∧ 𝑋𝑉 ) → 𝑋 ∈ ( ( LSpan ‘ 𝑈 ) ‘ { 𝑋 } ) )
22 18 19 21 syl2anc ( 𝜑𝑋 ∈ ( ( LSpan ‘ 𝑈 ) ‘ { 𝑋 } ) )
23 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 20 lcfrlem14 ( 𝜑 → ( ‘ ( 𝐿 ‘ ( 𝐽𝑋 ) ) ) = ( ( LSpan ‘ 𝑈 ) ‘ { 𝑋 } ) )
24 22 23 eleqtrrd ( 𝜑𝑋 ∈ ( ‘ ( 𝐿 ‘ ( 𝐽𝑋 ) ) ) )