Metamath Proof Explorer


Theorem lkreqN

Description: Proportional functionals have equal kernels. (Contributed by NM, 28-Mar-2015) (New usage is discouraged.)

Ref Expression
Hypotheses lkreq.s ⊢ S = Scalar ⁡ W
lkreq.r ⊢ R = Base S
lkreq.o ⊢ 0 ˙ = 0 S
lkreq.f ⊢ F = LFnl ⁡ W
lkreq.k ⊢ K = LKer ⁡ W
lkreq.d ⊢ D = LDual ⁡ W
lkreq.t ⊢ · ˙ = ⋅ D
lkreq.w ⊢ φ → W ∈ LVec
lkreq.a ⊢ φ → A ∈ R ∖ 0 ˙
lkreq.h ⊢ φ → H ∈ F
lkreq.g ⊢ φ → G = A · ˙ H
Assertion lkreqN ⊢ φ → K ⁡ G = K ⁡ H

Proof

Step Hyp Ref Expression
1 lkreq.s ⊢ S = Scalar ⁡ W
2 lkreq.r ⊢ R = Base S
3 lkreq.o ⊢ 0 ˙ = 0 S
4 lkreq.f ⊢ F = LFnl ⁡ W
5 lkreq.k ⊢ K = LKer ⁡ W
6 lkreq.d ⊢ D = LDual ⁡ W
7 lkreq.t ⊢ · ˙ = ⋅ D
8 lkreq.w ⊢ φ → W ∈ LVec
9 lkreq.a ⊢ φ → A ∈ R ∖ 0 ˙
10 lkreq.h ⊢ φ → H ∈ F
11 lkreq.g ⊢ φ → G = A · ˙ H
12 11 eqeq1d ⊢ φ → G = 0 D ↔ A · ˙ H = 0 D
13 eqid ⊢ Base D = Base D
14 eqid ⊢ Scalar ⁡ D = Scalar ⁡ D
15 eqid ⊢ Base Scalar ⁡ D = Base Scalar ⁡ D
16 eqid ⊢ 0 Scalar ⁡ D = 0 Scalar ⁡ D
17 eqid ⊢ 0 D = 0 D
18 6 8 lduallvec ⊢ φ → D ∈ LVec
19 9 eldifad ⊢ φ → A ∈ R
20 1 2 6 14 15 8 ldualsbase ⊢ φ → Base Scalar ⁡ D = R
21 19 20 eleqtrrd ⊢ φ → A ∈ Base Scalar ⁡ D
22 4 6 13 8 10 ldualelvbase ⊢ φ → H ∈ Base D
23 13 7 14 15 16 17 18 21 22 lvecvs0or ⊢ φ → A · ˙ H = 0 D ↔ A = 0 Scalar ⁡ D ∨ H = 0 D
24 lveclmod ⊢ W ∈ LVec → W ∈ LMod
25 8 24 syl ⊢ φ → W ∈ LMod
26 1 3 6 14 16 25 ldual0 ⊢ φ → 0 Scalar ⁡ D = 0 ˙
27 26 eqeq2d ⊢ φ → A = 0 Scalar ⁡ D ↔ A = 0 ˙
28 eldifsni ⊢ A ∈ R ∖ 0 ˙ → A ≠ 0 ˙
29 9 28 syl ⊢ φ → A ≠ 0 ˙
30 29 a1d ⊢ φ → H ≠ 0 D → A ≠ 0 ˙
31 30 necon4d ⊢ φ → A = 0 ˙ → H = 0 D
32 27 31 sylbid ⊢ φ → A = 0 Scalar ⁡ D → H = 0 D
33 idd ⊢ φ → H = 0 D → H = 0 D
34 32 33 jaod ⊢ φ → A = 0 Scalar ⁡ D ∨ H = 0 D → H = 0 D
35 23 34 sylbid ⊢ φ → A · ˙ H = 0 D → H = 0 D
36 12 35 sylbid ⊢ φ → G = 0 D → H = 0 D
37 nne ⊢ ¬ H ≠ 0 D ↔ H = 0 D
38 36 37 imbitrrdi ⊢ φ → G = 0 D → ¬ H ≠ 0 D
39 38 con3d ⊢ φ → ¬ ¬ H ≠ 0 D → ¬ G = 0 D
40 39 orrd ⊢ φ → ¬ H ≠ 0 D ∨ ¬ G = 0 D
41 ianor ⊢ ¬ H ≠ 0 D ∧ G = 0 D ↔ ¬ H ≠ 0 D ∨ ¬ G = 0 D
42 40 41 sylibr ⊢ φ → ¬ H ≠ 0 D ∧ G = 0 D
43 4 1 2 6 7 25 19 10 ldualvscl ⊢ φ → A · ˙ H ∈ F
44 11 43 eqeltrd ⊢ φ → G ∈ F
45 4 5 6 17 8 10 44 lkrpssN ⊢ φ → K ⁡ H ⊂ K ⁡ G ↔ H ≠ 0 D ∧ G = 0 D
46 df-pss ⊢ K ⁡ H ⊂ K ⁡ G ↔ K ⁡ H ⊆ K ⁡ G ∧ K ⁡ H ≠ K ⁡ G
47 45 46 bitr3di ⊢ φ → H ≠ 0 D ∧ G = 0 D ↔ K ⁡ H ⊆ K ⁡ G ∧ K ⁡ H ≠ K ⁡ G
48 1 2 4 5 6 7 8 10 19 lkrss ⊢ φ → K ⁡ H ⊆ K ⁡ A · ˙ H
49 11 fveq2d ⊢ φ → K ⁡ G = K ⁡ A · ˙ H
50 48 49 sseqtrrd ⊢ φ → K ⁡ H ⊆ K ⁡ G
51 50 biantrurd ⊢ φ → K ⁡ H ≠ K ⁡ G ↔ K ⁡ H ⊆ K ⁡ G ∧ K ⁡ H ≠ K ⁡ G
52 47 51 bitr4d ⊢ φ → H ≠ 0 D ∧ G = 0 D ↔ K ⁡ H ≠ K ⁡ G
53 52 necon2bbid ⊢ φ → K ⁡ H = K ⁡ G ↔ ¬ H ≠ 0 D ∧ G = 0 D
54 42 53 mpbird ⊢ φ → K ⁡ H = K ⁡ G
55 54 eqcomd ⊢ φ → K ⁡ G = K ⁡ H