Metamath Proof Explorer


Theorem ldualvsubval

Description: The value of the value of vector subtraction in the dual of a vector space. TODO: shorten with ldualvsub ? (Requires D to oppR conversion.) (Contributed by NM, 26-Feb-2015)

Ref Expression
Hypotheses ldualvsubval.v V = Base W
ldualvsubval.r R = Scalar W
ldualvsubval.s S = - R
ldualvsubval.f F = LFnl W
ldualvsubval.d D = LDual W
ldualvsubval.m - ˙ = - D
ldualvsubval.w φ W LMod
ldualvsubval.g φ G F
ldualvsubval.h φ H F
ldualvsubval.x φ X V
Assertion ldualvsubval φ G - ˙ H X = G X S H X

Proof

Step Hyp Ref Expression
1 ldualvsubval.v V = Base W
2 ldualvsubval.r R = Scalar W
3 ldualvsubval.s S = - R
4 ldualvsubval.f F = LFnl W
5 ldualvsubval.d D = LDual W
6 ldualvsubval.m - ˙ = - D
7 ldualvsubval.w φ W LMod
8 ldualvsubval.g φ G F
9 ldualvsubval.h φ H F
10 ldualvsubval.x φ X V
11 5 7 lduallmod φ D LMod
12 eqid Base D = Base D
13 4 5 12 7 8 ldualelvbase φ G Base D
14 4 5 12 7 9 ldualelvbase φ H Base D
15 eqid + D = + D
16 eqid Scalar D = Scalar D
17 eqid D = D
18 eqid inv g Scalar D = inv g Scalar D
19 eqid 1 Scalar D = 1 Scalar D
20 12 15 6 16 17 18 19 lmodvsubval2 D LMod G Base D H Base D G - ˙ H = G + D inv g Scalar D 1 Scalar D D H
21 11 13 14 20 syl3anc φ G - ˙ H = G + D inv g Scalar D 1 Scalar D D H
22 21 fveq1d φ G - ˙ H X = G + D inv g Scalar D 1 Scalar D D H X
23 eqid + R = + R
24 eqid Base R = Base R
25 16 lmodfgrp D LMod Scalar D Grp
26 11 25 syl φ Scalar D Grp
27 16 lmodring D LMod Scalar D Ring
28 11 27 syl φ Scalar D Ring
29 eqid Base Scalar D = Base Scalar D
30 29 19 ringidcl Scalar D Ring 1 Scalar D Base Scalar D
31 28 30 syl φ 1 Scalar D Base Scalar D
32 29 18 grpinvcl Scalar D Grp 1 Scalar D Base Scalar D inv g Scalar D 1 Scalar D Base Scalar D
33 26 31 32 syl2anc φ inv g Scalar D 1 Scalar D Base Scalar D
34 2 24 5 16 29 7 ldualsbase φ Base Scalar D = Base R
35 33 34 eleqtrd φ inv g Scalar D 1 Scalar D Base R
36 4 2 24 5 17 7 35 9 ldualvscl φ inv g Scalar D 1 Scalar D D H F
37 1 2 23 4 5 15 7 8 36 10 ldualvaddval φ G + D inv g Scalar D 1 Scalar D D H X = G X + R inv g Scalar D 1 Scalar D D H X
38 eqid inv g R = inv g R
39 2 38 5 16 18 7 ldualneg φ inv g Scalar D = inv g R
40 eqid 1 R = 1 R
41 2 40 5 16 19 7 ldual1 φ 1 Scalar D = 1 R
42 39 41 fveq12d φ inv g Scalar D 1 Scalar D = inv g R 1 R
43 42 oveq1d φ inv g Scalar D 1 Scalar D D H = inv g R 1 R D H
44 43 fveq1d φ inv g Scalar D 1 Scalar D D H X = inv g R 1 R D H X
45 eqid R = R
46 2 lmodring W LMod R Ring
47 7 46 syl φ R Ring
48 ringgrp R Ring R Grp
49 47 48 syl φ R Grp
50 2 24 40 lmod1cl W LMod 1 R Base R
51 7 50 syl φ 1 R Base R
52 24 38 grpinvcl R Grp 1 R Base R inv g R 1 R Base R
53 49 51 52 syl2anc φ inv g R 1 R Base R
54 4 1 2 24 45 5 17 7 53 9 10 ldualvsval φ inv g R 1 R D H X = H X R inv g R 1 R
55 2 24 1 4 lflcl W LMod H F X V H X Base R
56 7 9 10 55 syl3anc φ H X Base R
57 24 45 40 38 47 56 rngnegr φ H X R inv g R 1 R = inv g R H X
58 44 54 57 3eqtrd φ inv g Scalar D 1 Scalar D D H X = inv g R H X
59 58 oveq2d φ G X + R inv g Scalar D 1 Scalar D D H X = G X + R inv g R H X
60 2 24 1 4 lflcl W LMod G F X V G X Base R
61 7 8 10 60 syl3anc φ G X Base R
62 24 23 38 3 grpsubval G X Base R H X Base R G X S H X = G X + R inv g R H X
63 61 56 62 syl2anc φ G X S H X = G X + R inv g R H X
64 59 63 eqtr4d φ G X + R inv g Scalar D 1 Scalar D D H X = G X S H X
65 22 37 64 3eqtrd φ G - ˙ H X = G X S H X