Theorem lclkrlem2h

Description: Lemma for lclkr . Eliminate the ( L( E .+ G ) ) e. J hypothesis. (Contributed by NM, 16-Jan-2015)

	Ref	Expression
Hypotheses	lclkrlem2f.h	⊢ 𝐻 = ( LHyp ‘ 𝐾 )
	lclkrlem2f.o	⊢ ⊥ = ( ( ocH ‘ 𝐾 ) ‘ 𝑊 )
	lclkrlem2f.u	⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
	lclkrlem2f.v	⊢ 𝑉 = ( Base ‘ 𝑈 )
	lclkrlem2f.s	⊢ 𝑆 = ( Scalar ‘ 𝑈 )
	lclkrlem2f.q	⊢ 𝑄 = ( 0g ‘ 𝑆 )
	lclkrlem2f.z	⊢ 0 = ( 0g ‘ 𝑈 )
	lclkrlem2f.a	⊢ ⊕ = ( LSSum ‘ 𝑈 )
	lclkrlem2f.n	⊢ 𝑁 = ( LSpan ‘ 𝑈 )
	lclkrlem2f.f	⊢ 𝐹 = ( LFnl ‘ 𝑈 )
	lclkrlem2f.j	⊢ 𝐽 = ( LSHyp ‘ 𝑈 )
	lclkrlem2f.l	⊢ 𝐿 = ( LKer ‘ 𝑈 )
	lclkrlem2f.d	⊢ 𝐷 = ( LDual ‘ 𝑈 )
	lclkrlem2f.p	⊢ + = ( +g ‘ 𝐷 )
	lclkrlem2f.k	⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
	lclkrlem2f.b	⊢ ( 𝜑 → 𝐵 ∈ ( 𝑉 ∖ { 0 } ) )
	lclkrlem2f.e	⊢ ( 𝜑 → 𝐸 ∈ 𝐹 )
	lclkrlem2f.g	⊢ ( 𝜑 → 𝐺 ∈ 𝐹 )
	lclkrlem2f.le	⊢ ( 𝜑 → ( 𝐿 ‘ 𝐸 ) = ( ⊥ ‘ { 𝑋 } ) )
	lclkrlem2f.lg	⊢ ( 𝜑 → ( 𝐿 ‘ 𝐺 ) = ( ⊥ ‘ { 𝑌 } ) )
	lclkrlem2f.kb	⊢ ( 𝜑 → ( ( 𝐸 + 𝐺 ) ‘ 𝐵 ) = 𝑄 )
	lclkrlem2f.nx	⊢ ( 𝜑 → ( ¬ 𝑋 ∈ ( ⊥ ‘ { 𝐵 } ) ∨ ¬ 𝑌 ∈ ( ⊥ ‘ { 𝐵 } ) ) )
	lclkrlem2h.x	⊢ ( 𝜑 → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
	lclkrlem2h.y	⊢ ( 𝜑 → 𝑌 ∈ ( 𝑉 ∖ { 0 } ) )
	lclkrlem2h.ne	⊢ ( 𝜑 → ( 𝐿 ‘ 𝐸 ) ≠ ( 𝐿 ‘ 𝐺 ) )
Assertion	lclkrlem2h	⊢ ( 𝜑 → ( ⊥ ‘ ( ⊥ ‘ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ) ) = ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) )

Proof

Step	Hyp	Ref	Expression
1		lclkrlem2f.h	⊢ 𝐻 = ( LHyp ‘ 𝐾 )
2		lclkrlem2f.o	⊢ ⊥ = ( ( ocH ‘ 𝐾 ) ‘ 𝑊 )
3		lclkrlem2f.u	⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
4		lclkrlem2f.v	⊢ 𝑉 = ( Base ‘ 𝑈 )
5		lclkrlem2f.s	⊢ 𝑆 = ( Scalar ‘ 𝑈 )
6		lclkrlem2f.q	⊢ 𝑄 = ( 0g ‘ 𝑆 )
7		lclkrlem2f.z	⊢ 0 = ( 0g ‘ 𝑈 )
8		lclkrlem2f.a	⊢ ⊕ = ( LSSum ‘ 𝑈 )
9		lclkrlem2f.n	⊢ 𝑁 = ( LSpan ‘ 𝑈 )
10		lclkrlem2f.f	⊢ 𝐹 = ( LFnl ‘ 𝑈 )
11		lclkrlem2f.j	⊢ 𝐽 = ( LSHyp ‘ 𝑈 )
12		lclkrlem2f.l	⊢ 𝐿 = ( LKer ‘ 𝑈 )
13		lclkrlem2f.d	⊢ 𝐷 = ( LDual ‘ 𝑈 )
14		lclkrlem2f.p	⊢ + = ( +g ‘ 𝐷 )
15		lclkrlem2f.k	⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
16		lclkrlem2f.b	⊢ ( 𝜑 → 𝐵 ∈ ( 𝑉 ∖ { 0 } ) )
17		lclkrlem2f.e	⊢ ( 𝜑 → 𝐸 ∈ 𝐹 )
18		lclkrlem2f.g	⊢ ( 𝜑 → 𝐺 ∈ 𝐹 )
19		lclkrlem2f.le	⊢ ( 𝜑 → ( 𝐿 ‘ 𝐸 ) = ( ⊥ ‘ { 𝑋 } ) )
20		lclkrlem2f.lg	⊢ ( 𝜑 → ( 𝐿 ‘ 𝐺 ) = ( ⊥ ‘ { 𝑌 } ) )
21		lclkrlem2f.kb	⊢ ( 𝜑 → ( ( 𝐸 + 𝐺 ) ‘ 𝐵 ) = 𝑄 )
22		lclkrlem2f.nx	⊢ ( 𝜑 → ( ¬ 𝑋 ∈ ( ⊥ ‘ { 𝐵 } ) ∨ ¬ 𝑌 ∈ ( ⊥ ‘ { 𝐵 } ) ) )
23		lclkrlem2h.x	⊢ ( 𝜑 → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
24		lclkrlem2h.y	⊢ ( 𝜑 → 𝑌 ∈ ( 𝑉 ∖ { 0 } ) )
25		lclkrlem2h.ne	⊢ ( 𝜑 → ( 𝐿 ‘ 𝐸 ) ≠ ( 𝐿 ‘ 𝐺 ) )
26	15	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
27	16	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → 𝐵 ∈ ( 𝑉 ∖ { 0 } ) )
28	17	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → 𝐸 ∈ 𝐹 )
29	18	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → 𝐺 ∈ 𝐹 )
30	19	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( 𝐿 ‘ 𝐸 ) = ( ⊥ ‘ { 𝑋 } ) )
31	20	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( 𝐿 ‘ 𝐺 ) = ( ⊥ ‘ { 𝑌 } ) )
32	21	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( ( 𝐸 + 𝐺 ) ‘ 𝐵 ) = 𝑄 )
33	22	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( ¬ 𝑋 ∈ ( ⊥ ‘ { 𝐵 } ) ∨ ¬ 𝑌 ∈ ( ⊥ ‘ { 𝐵 } ) ) )
34	23	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
35	24	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → 𝑌 ∈ ( 𝑉 ∖ { 0 } ) )
36	25	adantr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( 𝐿 ‘ 𝐸 ) ≠ ( 𝐿 ‘ 𝐺 ) )
37		simpr	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 )
38	1 2 3 4 5 6 7 8 9 10 11 12 13 14 26 27 28 29 30 31 32 33 34 35 36 37	lclkrlem2g	⊢ ( ( 𝜑 ∧ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( ⊥ ‘ ( ⊥ ‘ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ) ) = ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) )
39	1 3 2 4 15	dochoc1	⊢ ( 𝜑 → ( ⊥ ‘ ( ⊥ ‘ 𝑉 ) ) = 𝑉 )
40	39	adantr	⊢ ( ( 𝜑 ∧ ¬ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( ⊥ ‘ ( ⊥ ‘ 𝑉 ) ) = 𝑉 )
41	1 3 15	dvhlvec	⊢ ( 𝜑 → 𝑈 ∈ LVec )
42	1 3 15	dvhlmod	⊢ ( 𝜑 → 𝑈 ∈ LMod )
43	10 13 14 42 17 18	ldualvaddcl	⊢ ( 𝜑 → ( 𝐸 + 𝐺 ) ∈ 𝐹 )
44	4 11 10 12 41 43	lkrshpor	⊢ ( 𝜑 → ( ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ∨ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) = 𝑉 ) )
45	44	orcanai	⊢ ( ( 𝜑 ∧ ¬ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) = 𝑉 )
46	45	fveq2d	⊢ ( ( 𝜑 ∧ ¬ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( ⊥ ‘ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ) = ( ⊥ ‘ 𝑉 ) )
47	46	fveq2d	⊢ ( ( 𝜑 ∧ ¬ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( ⊥ ‘ ( ⊥ ‘ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ) ) = ( ⊥ ‘ ( ⊥ ‘ 𝑉 ) ) )
48	40 47 45	3eqtr4d	⊢ ( ( 𝜑 ∧ ¬ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ∈ 𝐽 ) → ( ⊥ ‘ ( ⊥ ‘ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ) ) = ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) )
49	38 48	pm2.61dan	⊢ ( 𝜑 → ( ⊥ ‘ ( ⊥ ‘ ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) ) ) = ( 𝐿 ‘ ( 𝐸 + 𝐺 ) ) )