Metamath Proof Explorer


Theorem hgmaprnlem3N

Description: Lemma for hgmaprnN . Eliminate k . (Contributed by NM, 7-Jun-2015) (New usage is discouraged.)

Ref Expression
Hypotheses hgmaprnlem1.h ⊢ H = LHyp ⁡ K
hgmaprnlem1.u ⊢ U = DVecH ⁡ K ⁡ W
hgmaprnlem1.v ⊢ V = Base U
hgmaprnlem1.r ⊢ R = Scalar ⁡ U
hgmaprnlem1.b ⊢ B = Base R
hgmaprnlem1.t ⊢ · ˙ = ⋅ U
hgmaprnlem1.o ⊢ 0 ˙ = 0 U
hgmaprnlem1.c ⊢ C = LCDual ⁡ K ⁡ W
hgmaprnlem1.d ⊢ D = Base C
hgmaprnlem1.p ⊢ P = Scalar ⁡ C
hgmaprnlem1.a ⊢ A = Base P
hgmaprnlem1.e ⊢ ∙ ˙ = ⋅ C
hgmaprnlem1.q ⊢ Q = 0 C
hgmaprnlem1.s ⊢ S = HDMap ⁡ K ⁡ W
hgmaprnlem1.g ⊢ G = HGMap ⁡ K ⁡ W
hgmaprnlem1.k ⊢ φ → K ∈ HL ∧ W ∈ H
hgmaprnlem1.z ⊢ φ → z ∈ A
hgmaprnlem1.t2 ⊢ φ → t ∈ V ∖ 0 ˙
hgmaprnlem1.s2 ⊢ φ → s ∈ V
hgmaprnlem1.sz ⊢ φ → S ⁡ s = z ∙ ˙ S ⁡ t
hgmaprnlem1.m ⊢ M = mapd ⁡ K ⁡ W
hgmaprnlem1.n ⊢ N = LSpan ⁡ U
hgmaprnlem1.l ⊢ L = LSpan ⁡ C
Assertion hgmaprnlem3N ⊢ φ → z ∈ ran ⁡ G

Proof

Step Hyp Ref Expression
1 hgmaprnlem1.h ⊢ H = LHyp ⁡ K
2 hgmaprnlem1.u ⊢ U = DVecH ⁡ K ⁡ W
3 hgmaprnlem1.v ⊢ V = Base U
4 hgmaprnlem1.r ⊢ R = Scalar ⁡ U
5 hgmaprnlem1.b ⊢ B = Base R
6 hgmaprnlem1.t ⊢ · ˙ = ⋅ U
7 hgmaprnlem1.o ⊢ 0 ˙ = 0 U
8 hgmaprnlem1.c ⊢ C = LCDual ⁡ K ⁡ W
9 hgmaprnlem1.d ⊢ D = Base C
10 hgmaprnlem1.p ⊢ P = Scalar ⁡ C
11 hgmaprnlem1.a ⊢ A = Base P
12 hgmaprnlem1.e ⊢ ∙ ˙ = ⋅ C
13 hgmaprnlem1.q ⊢ Q = 0 C
14 hgmaprnlem1.s ⊢ S = HDMap ⁡ K ⁡ W
15 hgmaprnlem1.g ⊢ G = HGMap ⁡ K ⁡ W
16 hgmaprnlem1.k ⊢ φ → K ∈ HL ∧ W ∈ H
17 hgmaprnlem1.z ⊢ φ → z ∈ A
18 hgmaprnlem1.t2 ⊢ φ → t ∈ V ∖ 0 ˙
19 hgmaprnlem1.s2 ⊢ φ → s ∈ V
20 hgmaprnlem1.sz ⊢ φ → S ⁡ s = z ∙ ˙ S ⁡ t
21 hgmaprnlem1.m ⊢ M = mapd ⁡ K ⁡ W
22 hgmaprnlem1.n ⊢ N = LSpan ⁡ U
23 hgmaprnlem1.l ⊢ L = LSpan ⁡ C
24 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 hgmaprnlem2N ⊢ φ → N ⁡ s ⊆ N ⁡ t
25 1 2 16 dvhlmod ⊢ φ → U ∈ LMod
26 18 eldifad ⊢ φ → t ∈ V
27 3 4 5 6 22 25 19 26 lspsnss2 ⊢ φ → N ⁡ s ⊆ N ⁡ t ↔ ∃ k ∈ B s = k · ˙ t
28 24 27 mpbid ⊢ φ → ∃ k ∈ B s = k · ˙ t
29 16 3ad2ant1 ⊢ φ ∧ k ∈ B ∧ s = k · ˙ t → K ∈ HL ∧ W ∈ H
30 17 3ad2ant1 ⊢ φ ∧ k ∈ B ∧ s = k · ˙ t → z ∈ A
31 18 3ad2ant1 ⊢ φ ∧ k ∈ B ∧ s = k · ˙ t → t ∈ V ∖ 0 ˙
32 19 3ad2ant1 ⊢ φ ∧ k ∈ B ∧ s = k · ˙ t → s ∈ V
33 20 3ad2ant1 ⊢ φ ∧ k ∈ B ∧ s = k · ˙ t → S ⁡ s = z ∙ ˙ S ⁡ t
34 simp2 ⊢ φ ∧ k ∈ B ∧ s = k · ˙ t → k ∈ B
35 simp3 ⊢ φ ∧ k ∈ B ∧ s = k · ˙ t → s = k · ˙ t
36 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 29 30 31 32 33 34 35 hgmaprnlem1N ⊢ φ ∧ k ∈ B ∧ s = k · ˙ t → z ∈ ran ⁡ G
37 36 rexlimdv3a ⊢ φ → ∃ k ∈ B s = k · ˙ t → z ∈ ran ⁡ G
38 28 37 mpd ⊢ φ → z ∈ ran ⁡ G