Metamath Proof Explorer


Theorem lclkrlem2h

Description: Lemma for lclkr . Eliminate the ( L( E .+ G ) ) e. J hypothesis. (Contributed by NM, 16-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
lclkrlem2h.x ⊢ φ → X ∈ V ∖ 0 ˙
lclkrlem2h.y ⊢ φ → Y ∈ V ∖ 0 ˙
lclkrlem2h.ne ⊢ φ → L ⁡ E ≠ L ⁡ G
Assertion lclkrlem2h ⊢ 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 lclkrlem2h.x ⊢ φ → X ∈ V ∖ 0 ˙
24 lclkrlem2h.y ⊢ φ → Y ∈ V ∖ 0 ˙
25 lclkrlem2h.ne ⊢ φ → L ⁡ E ≠ L ⁡ G
26 15 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → K ∈ HL ∧ W ∈ H
27 16 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → B ∈ V ∖ 0 ˙
28 17 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → E ∈ F
29 18 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → G ∈ F
30 19 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → L ⁡ E = ⊥ ˙ ⁡ X
31 20 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → L ⁡ G = ⊥ ˙ ⁡ Y
32 21 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → E + ˙ G ⁡ B = Q
33 22 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → ¬ X ∈ ⊥ ˙ ⁡ B ∨ ¬ Y ∈ ⊥ ˙ ⁡ B
34 23 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → X ∈ V ∖ 0 ˙
35 24 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → Y ∈ V ∖ 0 ˙
36 25 adantr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → L ⁡ E ≠ L ⁡ G
37 simpr ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → L ⁡ E + ˙ G ∈ J
38 1 2 3 4 5 6 7 8 9 10 11 12 13 14 26 27 28 29 30 31 32 33 34 35 36 37 lclkrlem2g ⊢ φ ∧ L ⁡ E + ˙ G ∈ J → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ E + ˙ G = L ⁡ E + ˙ G
39 1 3 2 4 15 dochoc1 ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ V = V
40 39 adantr ⊢ φ ∧ ¬ L ⁡ E + ˙ G ∈ J → ⊥ ˙ ⁡ ⊥ ˙ ⁡ V = V
41 1 3 15 dvhlvec ⊢ φ → U ∈ LVec
42 1 3 15 dvhlmod ⊢ φ → U ∈ LMod
43 10 13 14 42 17 18 ldualvaddcl ⊢ φ → E + ˙ G ∈ F
44 4 11 10 12 41 43 lkrshpor ⊢ φ → L ⁡ E + ˙ G ∈ J ∨ L ⁡ E + ˙ G = V
45 44 orcanai ⊢ φ ∧ ¬ L ⁡ E + ˙ G ∈ J → L ⁡ E + ˙ G = V
46 45 fveq2d ⊢ φ ∧ ¬ L ⁡ E + ˙ G ∈ J → ⊥ ˙ ⁡ L ⁡ E + ˙ G = ⊥ ˙ ⁡ V
47 46 fveq2d ⊢ φ ∧ ¬ L ⁡ E + ˙ G ∈ J → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ E + ˙ G = ⊥ ˙ ⁡ ⊥ ˙ ⁡ V
48 40 47 45 3eqtr4d ⊢ φ ∧ ¬ L ⁡ E + ˙ G ∈ J → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ E + ˙ G = L ⁡ E + ˙ G
49 38 48 pm2.61dan ⊢ φ → ⊥ ˙ ⁡ ⊥ ˙ ⁡ L ⁡ E + ˙ G = L ⁡ E + ˙ G