Metamath Proof Explorer


Theorem hdmapneg

Description: Part of proof of part 12 in Baer p. 49 line 4. The sigma map of a negative is the negative of the sigma map. (Contributed by NM, 24-May-2015)

Ref Expression
Hypotheses hdmap12b.h ⊢ H = LHyp ⁡ K
hdmap12b.u ⊢ U = DVecH ⁡ K ⁡ W
hdmap12b.v ⊢ V = Base U
hdmap12b.m ⊢ M = inv g ⁡ U
hdmap12b.c ⊢ C = LCDual ⁡ K ⁡ W
hdmap12b.i ⊢ I = inv g ⁡ C
hdmap12b.s ⊢ S = HDMap ⁡ K ⁡ W
hdmap12b.k ⊢ φ → K ∈ HL ∧ W ∈ H
hdmap12b.x ⊢ φ → T ∈ V
Assertion hdmapneg ⊢ φ → S ⁡ M ⁡ T = I ⁡ S ⁡ T

Proof

Step Hyp Ref Expression
1 hdmap12b.h ⊢ H = LHyp ⁡ K
2 hdmap12b.u ⊢ U = DVecH ⁡ K ⁡ W
3 hdmap12b.v ⊢ V = Base U
4 hdmap12b.m ⊢ M = inv g ⁡ U
5 hdmap12b.c ⊢ C = LCDual ⁡ K ⁡ W
6 hdmap12b.i ⊢ I = inv g ⁡ C
7 hdmap12b.s ⊢ S = HDMap ⁡ K ⁡ W
8 hdmap12b.k ⊢ φ → K ∈ HL ∧ W ∈ H
9 hdmap12b.x ⊢ φ → T ∈ V
10 1 5 8 lcdlmod ⊢ φ → C ∈ LMod
11 eqid ⊢ Base C = Base C
12 1 2 3 5 11 7 8 9 hdmapcl ⊢ φ → S ⁡ T ∈ Base C
13 eqid ⊢ + C = + C
14 eqid ⊢ 0 C = 0 C
15 11 13 14 6 lmodvnegid ⊢ C ∈ LMod ∧ S ⁡ T ∈ Base C → S ⁡ T + C I ⁡ S ⁡ T = 0 C
16 10 12 15 syl2anc ⊢ φ → S ⁡ T + C I ⁡ S ⁡ T = 0 C
17 1 2 8 dvhlmod ⊢ φ → U ∈ LMod
18 eqid ⊢ + U = + U
19 eqid ⊢ 0 U = 0 U
20 3 18 19 4 lmodvnegid ⊢ U ∈ LMod ∧ T ∈ V → T + U M ⁡ T = 0 U
21 17 9 20 syl2anc ⊢ φ → T + U M ⁡ T = 0 U
22 3 4 lmodvnegcl ⊢ U ∈ LMod ∧ T ∈ V → M ⁡ T ∈ V
23 17 9 22 syl2anc ⊢ φ → M ⁡ T ∈ V
24 3 18 lmodvacl ⊢ U ∈ LMod ∧ T ∈ V ∧ M ⁡ T ∈ V → T + U M ⁡ T ∈ V
25 17 9 23 24 syl3anc ⊢ φ → T + U M ⁡ T ∈ V
26 1 2 3 19 5 14 7 8 25 hdmapeq0 ⊢ φ → S ⁡ T + U M ⁡ T = 0 C ↔ T + U M ⁡ T = 0 U
27 21 26 mpbird ⊢ φ → S ⁡ T + U M ⁡ T = 0 C
28 1 2 3 18 5 13 7 8 9 23 hdmapadd ⊢ φ → S ⁡ T + U M ⁡ T = S ⁡ T + C S ⁡ M ⁡ T
29 16 27 28 3eqtr2rd ⊢ φ → S ⁡ T + C S ⁡ M ⁡ T = S ⁡ T + C I ⁡ S ⁡ T
30 1 2 3 5 11 7 8 23 hdmapcl ⊢ φ → S ⁡ M ⁡ T ∈ Base C
31 11 6 lmodvnegcl ⊢ C ∈ LMod ∧ S ⁡ T ∈ Base C → I ⁡ S ⁡ T ∈ Base C
32 10 12 31 syl2anc ⊢ φ → I ⁡ S ⁡ T ∈ Base C
33 11 13 lmodlcan ⊢ C ∈ LMod ∧ S ⁡ M ⁡ T ∈ Base C ∧ I ⁡ S ⁡ T ∈ Base C ∧ S ⁡ T ∈ Base C → S ⁡ T + C S ⁡ M ⁡ T = S ⁡ T + C I ⁡ S ⁡ T ↔ S ⁡ M ⁡ T = I ⁡ S ⁡ T
34 10 30 32 12 33 syl13anc ⊢ φ → S ⁡ T + C S ⁡ M ⁡ T = S ⁡ T + C I ⁡ S ⁡ T ↔ S ⁡ M ⁡ T = I ⁡ S ⁡ T
35 29 34 mpbid ⊢ φ → S ⁡ M ⁡ T = I ⁡ S ⁡ T