Metamath Proof Explorer


Theorem trlnid

Description: Different translations with the same trace cannot be the identity. (Contributed by NM, 26-Jul-2013)

Ref Expression
Hypotheses trlnid.b ⊢ B = Base K
trlnid.h ⊢ H = LHyp ⁡ K
trlnid.t ⊢ T = LTrn ⁡ K ⁡ W
trlnid.r ⊢ R = trL ⁡ K ⁡ W
Assertion trlnid ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → F ≠ I ↾ B

Proof

Step Hyp Ref Expression
1 trlnid.b ⊢ B = Base K
2 trlnid.h ⊢ H = LHyp ⁡ K
3 trlnid.t ⊢ T = LTrn ⁡ K ⁡ W
4 trlnid.r ⊢ R = trL ⁡ K ⁡ W
5 simp3l ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → F ≠ G
6 simp1 ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → K ∈ HL ∧ W ∈ H
7 simp2l ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → F ∈ T
8 eqid ⊢ 0. ⁡ K = 0. ⁡ K
9 1 8 2 3 4 trlid0b ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T → F = I ↾ B ↔ R ⁡ F = 0. ⁡ K
10 6 7 9 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → F = I ↾ B ↔ R ⁡ F = 0. ⁡ K
11 10 biimpar ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G ∧ R ⁡ F = 0. ⁡ K → F = I ↾ B
12 simp3r ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → R ⁡ F = R ⁡ G
13 12 eqeq1d ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → R ⁡ F = 0. ⁡ K ↔ R ⁡ G = 0. ⁡ K
14 13 biimpa ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G ∧ R ⁡ F = 0. ⁡ K → R ⁡ G = 0. ⁡ K
15 simpl1 ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G ∧ R ⁡ F = 0. ⁡ K → K ∈ HL ∧ W ∈ H
16 simpl2r ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G ∧ R ⁡ F = 0. ⁡ K → G ∈ T
17 1 8 2 3 4 trlid0b ⊢ K ∈ HL ∧ W ∈ H ∧ G ∈ T → G = I ↾ B ↔ R ⁡ G = 0. ⁡ K
18 15 16 17 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G ∧ R ⁡ F = 0. ⁡ K → G = I ↾ B ↔ R ⁡ G = 0. ⁡ K
19 14 18 mpbird ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G ∧ R ⁡ F = 0. ⁡ K → G = I ↾ B
20 11 19 eqtr4d ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G ∧ R ⁡ F = 0. ⁡ K → F = G
21 20 ex ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → R ⁡ F = 0. ⁡ K → F = G
22 10 21 sylbid ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → F = I ↾ B → F = G
23 22 necon3d ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → F ≠ G → F ≠ I ↾ B
24 5 23 mpd ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ G ∈ T ∧ F ≠ G ∧ R ⁡ F = R ⁡ G → F ≠ I ↾ B