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