Metamath Proof Explorer


Theorem lclkrlem2j

Description: Lemma for lclkr . Kernel closure when Y is zero. (Contributed by NM, 18-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 = 0 S
lclkrlem2f.z ⊢ 0 ˙ = 0 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 ⊢ + ˙ = + 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
lclkrlem2j.x ⊢ φ → X ∈ V
lclkrlem2j.y ⊢ φ → Y = 0 ˙
Assertion lclkrlem2j ⊢ 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 = 0 S
7 lclkrlem2f.z ⊢ 0 ˙ = 0 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 ⊢ + ˙ = + 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 lclkrlem2j.x ⊢ φ → X ∈ V
24 lclkrlem2j.y ⊢ φ → Y = 0 ˙
25 23 snssd ⊢ φ → X ⊆ V
26 eqid ⊢ DIsoH ⁡ K ⁡ W = DIsoH ⁡ K ⁡ W
27 1 26 3 4 2 dochcl ⊢ K ∈ HL ∧ W ∈ H ∧ X ⊆ V → ⊥ ˙ ⁡ X ∈ ran ⁡ DIsoH ⁡ K ⁡ W
28 15 25 27 syl2anc ⊢ φ → ⊥ ˙ ⁡ X ∈ ran ⁡ DIsoH ⁡ K ⁡ W
29 1 26 2 dochoc ⊢ K ∈ HL ∧ W ∈ H ∧ ⊥ ˙ ⁡ X ∈ ran ⁡ DIsoH ⁡ K ⁡ W → ⊥ ˙ ⁡ ⊥ ˙ ⁡ ⊥ ˙ ⁡ X = ⊥ ˙ ⁡ X
30 15 28 29 syl2anc ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ ⊥ ˙ ⁡ X = ⊥ ˙ ⁡ X
31 24 sneqd ⊢ φ → Y = 0 ˙
32 31 fveq2d ⊢ φ → ⊥ ˙ ⁡ Y = ⊥ ˙ ⁡ 0 ˙
33 eqid ⊢ Base U = Base U
34 1 3 2 33 7 doch0 ⊢ K ∈ HL ∧ W ∈ H → ⊥ ˙ ⁡ 0 ˙ = Base U
35 15 34 syl ⊢ φ → ⊥ ˙ ⁡ 0 ˙ = Base U
36 20 32 35 3eqtrd ⊢ φ → L ⁡ G = Base U
37 1 3 15 dvhlmod ⊢ φ → U ∈ LMod
38 5 6 33 10 12 lkr0f ⊢ U ∈ LMod ∧ G ∈ F → L ⁡ G = Base U ↔ G = Base U × Q
39 37 18 38 syl2anc ⊢ φ → L ⁡ G = Base U ↔ G = Base U × Q
40 36 39 mpbid ⊢ φ → G = Base U × Q
41 eqid ⊢ 0 D = 0 D
42 33 5 6 13 41 37 ldual0v ⊢ φ → 0 D = Base U × Q
43 40 42 eqtr4d ⊢ φ → G = 0 D
44 43 oveq2d ⊢ φ → E + ˙ G = E + ˙ 0 D
45 13 37 lduallmod ⊢ φ → D ∈ LMod
46 eqid ⊢ Base D = Base D
47 10 13 46 37 17 ldualelvbase ⊢ φ → E ∈ Base D
48 46 14 41 lmod0vrid ⊢ D ∈ LMod ∧ E ∈ Base D → E + ˙ 0 D = E
49 45 47 48 syl2anc ⊢ φ → E + ˙ 0 D = E
50 44 49 eqtrd ⊢ φ → E + ˙ G = E
51 50 fveq2d ⊢ φ → L ⁡ E + ˙ G = L ⁡ E
52 51 19 eqtr2d ⊢ φ → ⊥ ˙ ⁡ X = L ⁡ E + ˙ G
53 52 fveq2d ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ X = ⊥ ˙ ⁡ L ⁡ E + ˙ G
54 53 fveq2d ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ ⊥ ˙ ⁡ X = ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ E + ˙ G
55 30 54 52 3eqtr3d ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ E + ˙ G = L ⁡ E + ˙ G