Theorem hdmap14lem7

Description: Combine cases of F . TODO: Can this be done at once in hdmap14lem3 , in order to get rid of hdmap14lem6 ? Perhaps modify lspsneu to become E! k e. K instead of E! k e. ( K \ { .0. } ) ? (Contributed by NM, 1-Jun-2015)

	Ref	Expression
Hypotheses	hdmap14lem7.h	⊢ 𝐻 = ( LHyp ‘ 𝐾 )
	hdmap14lem7.u	⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
	hdmap14lem7.v	⊢ 𝑉 = ( Base ‘ 𝑈 )
	hdmap14lem7.t	⊢ · = ( ·𝑠 ‘ 𝑈 )
	hdmap14lem7.o	⊢ 0 = ( 0g ‘ 𝑈 )
	hdmap14lem7.r	⊢ 𝑅 = ( Scalar ‘ 𝑈 )
	hdmap14lem7.b	⊢ 𝐵 = ( Base ‘ 𝑅 )
	hdmap14lem7.c	⊢ 𝐶 = ( ( LCDual ‘ 𝐾 ) ‘ 𝑊 )
	hdmap14lem7.e	⊢ ∙ = ( ·𝑠 ‘ 𝐶 )
	hdmap14lem7.p	⊢ 𝑃 = ( Scalar ‘ 𝐶 )
	hdmap14lem7.a	⊢ 𝐴 = ( Base ‘ 𝑃 )
	hdmap14lem7.s	⊢ 𝑆 = ( ( HDMap ‘ 𝐾 ) ‘ 𝑊 )
	hdmap14lem7.k	⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
	hdmap14lem7.x	⊢ ( 𝜑 → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
	hdmap14lem7.f	⊢ ( 𝜑 → 𝐹 ∈ 𝐵 )
Assertion	hdmap14lem7	⊢ ( 𝜑 → ∃! 𝑔 ∈ 𝐴 ( 𝑆 ‘ ( 𝐹 · 𝑋 ) ) = ( 𝑔 ∙ ( 𝑆 ‘ 𝑋 ) ) )

Proof

Step	Hyp	Ref	Expression
1		hdmap14lem7.h	⊢ 𝐻 = ( LHyp ‘ 𝐾 )
2		hdmap14lem7.u	⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
3		hdmap14lem7.v	⊢ 𝑉 = ( Base ‘ 𝑈 )
4		hdmap14lem7.t	⊢ · = ( ·𝑠 ‘ 𝑈 )
5		hdmap14lem7.o	⊢ 0 = ( 0g ‘ 𝑈 )
6		hdmap14lem7.r	⊢ 𝑅 = ( Scalar ‘ 𝑈 )
7		hdmap14lem7.b	⊢ 𝐵 = ( Base ‘ 𝑅 )
8		hdmap14lem7.c	⊢ 𝐶 = ( ( LCDual ‘ 𝐾 ) ‘ 𝑊 )
9		hdmap14lem7.e	⊢ ∙ = ( ·𝑠 ‘ 𝐶 )
10		hdmap14lem7.p	⊢ 𝑃 = ( Scalar ‘ 𝐶 )
11		hdmap14lem7.a	⊢ 𝐴 = ( Base ‘ 𝑃 )
12		hdmap14lem7.s	⊢ 𝑆 = ( ( HDMap ‘ 𝐾 ) ‘ 𝑊 )
13		hdmap14lem7.k	⊢ ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
14		hdmap14lem7.x	⊢ ( 𝜑 → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
15		hdmap14lem7.f	⊢ ( 𝜑 → 𝐹 ∈ 𝐵 )
16		eqid	⊢ ( 0g ‘ 𝑅 ) = ( 0g ‘ 𝑅 )
17		eqid	⊢ ( LSpan ‘ 𝐶 ) = ( LSpan ‘ 𝐶 )
18		eqid	⊢ ( 0g ‘ 𝑃 ) = ( 0g ‘ 𝑃 )
19	13	adantr	⊢ ( ( 𝜑 ∧ 𝐹 = ( 0g ‘ 𝑅 ) ) → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
20	14	adantr	⊢ ( ( 𝜑 ∧ 𝐹 = ( 0g ‘ 𝑅 ) ) → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
21		simpr	⊢ ( ( 𝜑 ∧ 𝐹 = ( 0g ‘ 𝑅 ) ) → 𝐹 = ( 0g ‘ 𝑅 ) )
22	1 2 3 4 5 6 7 16 8 9 17 10 11 18 12 19 20 21	hdmap14lem6	⊢ ( ( 𝜑 ∧ 𝐹 = ( 0g ‘ 𝑅 ) ) → ∃! 𝑔 ∈ 𝐴 ( 𝑆 ‘ ( 𝐹 · 𝑋 ) ) = ( 𝑔 ∙ ( 𝑆 ‘ 𝑋 ) ) )
23	13	adantr	⊢ ( ( 𝜑 ∧ 𝐹 ≠ ( 0g ‘ 𝑅 ) ) → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
24	14	adantr	⊢ ( ( 𝜑 ∧ 𝐹 ≠ ( 0g ‘ 𝑅 ) ) → 𝑋 ∈ ( 𝑉 ∖ { 0 } ) )
25	15	adantr	⊢ ( ( 𝜑 ∧ 𝐹 ≠ ( 0g ‘ 𝑅 ) ) → 𝐹 ∈ 𝐵 )
26		simpr	⊢ ( ( 𝜑 ∧ 𝐹 ≠ ( 0g ‘ 𝑅 ) ) → 𝐹 ≠ ( 0g ‘ 𝑅 ) )
27		eldifsn	⊢ ( 𝐹 ∈ ( 𝐵 ∖ { ( 0g ‘ 𝑅 ) } ) ↔ ( 𝐹 ∈ 𝐵 ∧ 𝐹 ≠ ( 0g ‘ 𝑅 ) ) )
28	25 26 27	sylanbrc	⊢ ( ( 𝜑 ∧ 𝐹 ≠ ( 0g ‘ 𝑅 ) ) → 𝐹 ∈ ( 𝐵 ∖ { ( 0g ‘ 𝑅 ) } ) )
29	1 2 3 4 5 6 7 16 8 9 17 10 11 18 12 23 24 28	hdmap14lem4	⊢ ( ( 𝜑 ∧ 𝐹 ≠ ( 0g ‘ 𝑅 ) ) → ∃! 𝑔 ∈ 𝐴 ( 𝑆 ‘ ( 𝐹 · 𝑋 ) ) = ( 𝑔 ∙ ( 𝑆 ‘ 𝑋 ) ) )
30	22 29	pm2.61dane	⊢ ( 𝜑 → ∃! 𝑔 ∈ 𝐴 ( 𝑆 ‘ ( 𝐹 · 𝑋 ) ) = ( 𝑔 ∙ ( 𝑆 ‘ 𝑋 ) ) )