Metamath Proof Explorer


Theorem trlnidatb

Description: A lattice translation is not the identity iff its trace is an atom. TODO: Can proofs be reorganized so this goes with trlnidat ? Why do both this and ltrnideq need trlnidat ? (Contributed by NM, 4-Jun-2013)

Ref Expression
Hypotheses trlnidatb.b B = Base K
trlnidatb.a A = Atoms K
trlnidatb.h H = LHyp K
trlnidatb.t T = LTrn K W
trlnidatb.r R = trL K W
Assertion trlnidatb K HL W H F T F I B R F A

Proof

Step Hyp Ref Expression
1 trlnidatb.b B = Base K
2 trlnidatb.a A = Atoms K
3 trlnidatb.h H = LHyp K
4 trlnidatb.t T = LTrn K W
5 trlnidatb.r R = trL K W
6 1 2 3 4 5 trlnidat K HL W H F T F I B R F A
7 6 3expia K HL W H F T F I B R F A
8 eqid K = K
9 8 2 3 lhpexnle K HL W H p A ¬ p K W
10 9 adantr K HL W H F T p A ¬ p K W
11 1 8 2 3 4 ltrnideq K HL W H F T p A ¬ p K W F = I B F p = p
12 11 3expa K HL W H F T p A ¬ p K W F = I B F p = p
13 simp1l K HL W H F T p A ¬ p K W F p = p K HL W H
14 simp2 K HL W H F T p A ¬ p K W F p = p p A ¬ p K W
15 simp1r K HL W H F T p A ¬ p K W F p = p F T
16 simp3 K HL W H F T p A ¬ p K W F p = p F p = p
17 eqid 0. K = 0. K
18 8 17 2 3 4 5 trl0 K HL W H p A ¬ p K W F T F p = p R F = 0. K
19 13 14 15 16 18 syl112anc K HL W H F T p A ¬ p K W F p = p R F = 0. K
20 19 3expia K HL W H F T p A ¬ p K W F p = p R F = 0. K
21 simplll K HL W H F T p A ¬ p K W K HL
22 hlatl K HL K AtLat
23 17 2 atn0 K AtLat R F A R F 0. K
24 23 ex K AtLat R F A R F 0. K
25 21 22 24 3syl K HL W H F T p A ¬ p K W R F A R F 0. K
26 25 necon2bd K HL W H F T p A ¬ p K W R F = 0. K ¬ R F A
27 20 26 syld K HL W H F T p A ¬ p K W F p = p ¬ R F A
28 12 27 sylbid K HL W H F T p A ¬ p K W F = I B ¬ R F A
29 10 28 rexlimddv K HL W H F T F = I B ¬ R F A
30 29 necon2ad K HL W H F T R F A F I B
31 7 30 impbid K HL W H F T F I B R F A