Metamath Proof Explorer


Theorem ldual0v

Description: The zero vector of the dual of a vector space. (Contributed by NM, 24-Oct-2014)

Ref Expression
Hypotheses ldualv0.v ⊢ V = Base W
ldualv0.r ⊢ R = Scalar ⁡ W
ldualv0.z ⊢ 0 ˙ = 0 R
ldualv0.d ⊢ D = LDual ⁡ W
ldualv0.o ⊢ O = 0 D
ldualv0.w ⊢ φ → W ∈ LMod
Assertion ldual0v ⊢ φ → O = V × 0 ˙

Proof

Step Hyp Ref Expression
1 ldualv0.v ⊢ V = Base W
2 ldualv0.r ⊢ R = Scalar ⁡ W
3 ldualv0.z ⊢ 0 ˙ = 0 R
4 ldualv0.d ⊢ D = LDual ⁡ W
5 ldualv0.o ⊢ O = 0 D
6 ldualv0.w ⊢ φ → W ∈ LMod
7 eqid ⊢ LFnl ⁡ W = LFnl ⁡ W
8 eqid ⊢ + R = + R
9 eqid ⊢ + D = + D
10 2 3 1 7 lfl0f ⊢ W ∈ LMod → V × 0 ˙ ∈ LFnl ⁡ W
11 6 10 syl ⊢ φ → V × 0 ˙ ∈ LFnl ⁡ W
12 7 2 8 4 9 6 11 11 ldualvadd ⊢ φ → V × 0 ˙ + D V × 0 ˙ = V × 0 ˙ + R f V × 0 ˙
13 1 2 8 3 7 6 11 lfladd0l ⊢ φ → V × 0 ˙ + R f V × 0 ˙ = V × 0 ˙
14 12 13 eqtrd ⊢ φ → V × 0 ˙ + D V × 0 ˙ = V × 0 ˙
15 4 6 ldualgrp ⊢ φ → D ∈ Grp
16 eqid ⊢ Base D = Base D
17 7 4 16 6 11 ldualelvbase ⊢ φ → V × 0 ˙ ∈ Base D
18 16 9 5 grpid ⊢ D ∈ Grp ∧ V × 0 ˙ ∈ Base D → V × 0 ˙ + D V × 0 ˙ = V × 0 ˙ ↔ O = V × 0 ˙
19 15 17 18 syl2anc ⊢ φ → V × 0 ˙ + D V × 0 ˙ = V × 0 ˙ ↔ O = V × 0 ˙
20 14 19 mpbid ⊢ φ → O = V × 0 ˙