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 syl6ibr 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