Metamath Proof Explorer


Theorem trlcoat

Description: The trace of a composition of two translations is an atom if their traces are different. (Contributed by NM, 15-Jun-2013)

Ref Expression
Hypotheses trlcoat.a ⊢ A = Atoms ⁡ K
trlcoat.h ⊢ H = LHyp ⁡ K
trlcoat.t ⊢ T = LTrn ⁡ K ⁡ W
trlcoat.r ⊢ R = trL ⁡ K ⁡ W
Assertion trlcoat ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ R ⁡ F ≠ R ⁡ G → R ⁡ F ∘ G ∈ A

Proof

Step Hyp Ref Expression
1 trlcoat.a ⊢ A = Atoms ⁡ K
2 trlcoat.h ⊢ H = LHyp ⁡ K
3 trlcoat.t ⊢ T = LTrn ⁡ K ⁡ W
4 trlcoat.r ⊢ R = trL ⁡ K ⁡ W
5 2 3 ltrnco ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T → F ∘ G ∈ T
6 5 3expb ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T → F ∘ G ∈ T
7 eqid ⊢ Base K = Base K
8 eqid ⊢ 0. ⁡ K = 0. ⁡ K
9 7 8 2 3 4 trlid0b ⊢ K ∈ HL ∧ W ∈ H ∧ F ∘ G ∈ T → F ∘ G = I ↾ Base K ↔ R ⁡ F ∘ G = 0. ⁡ K
10 6 9 syldan ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T → F ∘ G = I ↾ Base K ↔ R ⁡ F ∘ G = 0. ⁡ K
11 coass ⊢ F -1 ∘ F ∘ G = F -1 ∘ F ∘ G
12 simpll ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → K ∈ HL ∧ W ∈ H
13 simplrl ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → F ∈ T
14 7 2 3 ltrn1o ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T → F : Base K ⟶ 1-1 onto Base K
15 12 13 14 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → F : Base K ⟶ 1-1 onto Base K
16 f1ococnv1 ⊢ F : Base K ⟶ 1-1 onto Base K → F -1 ∘ F = I ↾ Base K
17 15 16 syl ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → F -1 ∘ F = I ↾ Base K
18 17 coeq1d ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → F -1 ∘ F ∘ G = I ↾ Base K ∘ G
19 coeq2 ⊢ F ∘ G = I ↾ Base K → F -1 ∘ F ∘ G = F -1 ∘ I ↾ Base K
20 19 adantl ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → F -1 ∘ F ∘ G = F -1 ∘ I ↾ Base K
21 11 18 20 3eqtr3a ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → I ↾ Base K ∘ G = F -1 ∘ I ↾ Base K
22 simplrr ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → G ∈ T
23 7 2 3 ltrn1o ⊢ K ∈ HL ∧ W ∈ H ∧ G ∈ T → G : Base K ⟶ 1-1 onto Base K
24 12 22 23 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → G : Base K ⟶ 1-1 onto Base K
25 f1of ⊢ G : Base K ⟶ 1-1 onto Base K → G : Base K ⟶ Base K
26 fcoi2 ⊢ G : Base K ⟶ Base K → I ↾ Base K ∘ G = G
27 24 25 26 3syl ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → I ↾ Base K ∘ G = G
28 2 3 ltrncnv ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T → F -1 ∈ T
29 12 13 28 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → F -1 ∈ T
30 7 2 3 ltrn1o ⊢ K ∈ HL ∧ W ∈ H ∧ F -1 ∈ T → F -1 : Base K ⟶ 1-1 onto Base K
31 12 29 30 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → F -1 : Base K ⟶ 1-1 onto Base K
32 f1of ⊢ F -1 : Base K ⟶ 1-1 onto Base K → F -1 : Base K ⟶ Base K
33 fcoi1 ⊢ F -1 : Base K ⟶ Base K → F -1 ∘ I ↾ Base K = F -1
34 31 32 33 3syl ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → F -1 ∘ I ↾ Base K = F -1
35 21 27 34 3eqtr3d ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → G = F -1
36 35 fveq2d ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → R ⁡ G = R ⁡ F -1
37 2 3 4 trlcnv ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T → R ⁡ F -1 = R ⁡ F
38 12 13 37 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → R ⁡ F -1 = R ⁡ F
39 36 38 eqtr2d ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ∘ G = I ↾ Base K → R ⁡ F = R ⁡ G
40 39 ex ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T → F ∘ G = I ↾ Base K → R ⁡ F = R ⁡ G
41 10 40 sylbird ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T → R ⁡ F ∘ G = 0. ⁡ K → R ⁡ F = R ⁡ G
42 41 necon3d ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T → R ⁡ F ≠ R ⁡ G → R ⁡ F ∘ G ≠ 0. ⁡ K
43 8 1 2 3 4 trlatn0 ⊢ K ∈ HL ∧ W ∈ H ∧ F ∘ G ∈ T → R ⁡ F ∘ G ∈ A ↔ R ⁡ F ∘ G ≠ 0. ⁡ K
44 6 43 syldan ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T → R ⁡ F ∘ G ∈ A ↔ R ⁡ F ∘ G ≠ 0. ⁡ K
45 42 44 sylibrd ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T → R ⁡ F ≠ R ⁡ G → R ⁡ F ∘ G ∈ A
46 45 3impia ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ R ⁡ F ≠ R ⁡ G → R ⁡ F ∘ G ∈ A