Metamath Proof Explorer


Theorem ldualvsub

Description: The value of vector subtraction in the dual of a vector space. (Contributed by NM, 27-Feb-2015)

Ref Expression
Hypotheses ldualvsub.r ⊢ R = Scalar ⁡ W
ldualvsub.n ⊢ N = inv g ⁡ R
ldualvsub.u ⊢ 1 ˙ = 1 R
ldualvsub.f ⊢ F = LFnl ⁡ W
ldualvsub.d ⊢ D = LDual ⁡ W
ldualvsub.p ⊢ + ˙ = + D
ldualvsub.t ⊢ · ˙ = ⋅ D
ldualvsub.m ⊢ - ˙ = - D
ldualvsub.w ⊢ φ → W ∈ LMod
ldualvsub.g ⊢ φ → G ∈ F
ldualvsub.h ⊢ φ → H ∈ F
Assertion ldualvsub ⊢ φ → G - ˙ H = G + ˙ N ⁡ 1 ˙ · ˙ H

Proof

Step Hyp Ref Expression
1 ldualvsub.r ⊢ R = Scalar ⁡ W
2 ldualvsub.n ⊢ N = inv g ⁡ R
3 ldualvsub.u ⊢ 1 ˙ = 1 R
4 ldualvsub.f ⊢ F = LFnl ⁡ W
5 ldualvsub.d ⊢ D = LDual ⁡ W
6 ldualvsub.p ⊢ + ˙ = + D
7 ldualvsub.t ⊢ · ˙ = ⋅ D
8 ldualvsub.m ⊢ - ˙ = - D
9 ldualvsub.w ⊢ φ → W ∈ LMod
10 ldualvsub.g ⊢ φ → G ∈ F
11 ldualvsub.h ⊢ φ → H ∈ F
12 5 9 lduallmod ⊢ φ → D ∈ LMod
13 eqid ⊢ Base D = Base D
14 4 5 13 9 10 ldualelvbase ⊢ φ → G ∈ Base D
15 4 5 13 9 11 ldualelvbase ⊢ φ → H ∈ Base D
16 eqid ⊢ Scalar ⁡ D = Scalar ⁡ D
17 eqid ⊢ inv g ⁡ Scalar ⁡ D = inv g ⁡ Scalar ⁡ D
18 eqid ⊢ 1 Scalar ⁡ D = 1 Scalar ⁡ D
19 13 6 8 16 7 17 18 lmodvsubval2 ⊢ D ∈ LMod ∧ G ∈ Base D ∧ H ∈ Base D → G - ˙ H = G + ˙ inv g ⁡ Scalar ⁡ D ⁡ 1 Scalar ⁡ D · ˙ H
20 12 14 15 19 syl3anc ⊢ φ → G - ˙ H = G + ˙ inv g ⁡ Scalar ⁡ D ⁡ 1 Scalar ⁡ D · ˙ H
21 eqid ⊢ opp r ⁡ R = opp r ⁡ R
22 21 2 opprneg ⊢ N = inv g ⁡ opp r ⁡ R
23 1 21 5 16 9 ldualsca ⊢ φ → Scalar ⁡ D = opp r ⁡ R
24 23 fveq2d ⊢ φ → inv g ⁡ Scalar ⁡ D = inv g ⁡ opp r ⁡ R
25 22 24 eqtr4id ⊢ φ → N = inv g ⁡ Scalar ⁡ D
26 21 3 oppr1 ⊢ 1 ˙ = 1 opp r ⁡ R
27 23 fveq2d ⊢ φ → 1 Scalar ⁡ D = 1 opp r ⁡ R
28 26 27 eqtr4id ⊢ φ → 1 ˙ = 1 Scalar ⁡ D
29 25 28 fveq12d ⊢ φ → N ⁡ 1 ˙ = inv g ⁡ Scalar ⁡ D ⁡ 1 Scalar ⁡ D
30 29 oveq1d ⊢ φ → N ⁡ 1 ˙ · ˙ H = inv g ⁡ Scalar ⁡ D ⁡ 1 Scalar ⁡ D · ˙ H
31 30 oveq2d ⊢ φ → G + ˙ N ⁡ 1 ˙ · ˙ H = G + ˙ inv g ⁡ Scalar ⁡ D ⁡ 1 Scalar ⁡ D · ˙ H
32 20 31 eqtr4d ⊢ φ → G - ˙ H = G + ˙ N ⁡ 1 ˙ · ˙ H