Metamath Proof Explorer


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 ⊢ · ˙ = ⋅ U
hdmap14lem7.o ⊢ 0 ˙ = 0 U
hdmap14lem7.r ⊢ R = Scalar ⁡ U
hdmap14lem7.b ⊢ B = Base R
hdmap14lem7.c ⊢ C = LCDual ⁡ K ⁡ W
hdmap14lem7.e ⊢ ∙ ˙ = ⋅ 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 ⊢ · ˙ = ⋅ U
5 hdmap14lem7.o ⊢ 0 ˙ = 0 U
6 hdmap14lem7.r ⊢ R = Scalar ⁡ U
7 hdmap14lem7.b ⊢ B = Base R
8 hdmap14lem7.c ⊢ C = LCDual ⁡ K ⁡ W
9 hdmap14lem7.e ⊢ ∙ ˙ = ⋅ 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 ⊢ 0 R = 0 R
17 eqid ⊢ LSpan ⁡ C = LSpan ⁡ C
18 eqid ⊢ 0 P = 0 P
19 13 adantr ⊢ φ ∧ F = 0 R → K ∈ HL ∧ W ∈ H
20 14 adantr ⊢ φ ∧ F = 0 R → X ∈ V ∖ 0 ˙
21 simpr ⊢ φ ∧ F = 0 R → F = 0 R
22 1 2 3 4 5 6 7 16 8 9 17 10 11 18 12 19 20 21 hdmap14lem6 ⊢ φ ∧ F = 0 R → ∃! g ∈ A S ⁡ F · ˙ X = g ∙ ˙ S ⁡ X
23 13 adantr ⊢ φ ∧ F ≠ 0 R → K ∈ HL ∧ W ∈ H
24 14 adantr ⊢ φ ∧ F ≠ 0 R → X ∈ V ∖ 0 ˙
25 15 adantr ⊢ φ ∧ F ≠ 0 R → F ∈ B
26 simpr ⊢ φ ∧ F ≠ 0 R → F ≠ 0 R
27 eldifsn ⊢ F ∈ B ∖ 0 R ↔ F ∈ B ∧ F ≠ 0 R
28 25 26 27 sylanbrc ⊢ φ ∧ F ≠ 0 R → F ∈ B ∖ 0 R
29 1 2 3 4 5 6 7 16 8 9 17 10 11 18 12 23 24 28 hdmap14lem4 ⊢ φ ∧ F ≠ 0 R → ∃! g ∈ A S ⁡ F · ˙ X = g ∙ ˙ S ⁡ X
30 22 29 pm2.61dane ⊢ φ → ∃! g ∈ A S ⁡ F · ˙ X = g ∙ ˙ S ⁡ X