Metamath Proof Explorer


Theorem ltrnatb

Description: The lattice translation of an atom is an atom. (Contributed by NM, 20-May-2012)

Ref Expression
Hypotheses ltrnatb.b B = Base K
ltrnatb.a A = Atoms K
ltrnatb.h H = LHyp K
ltrnatb.t T = LTrn K W
Assertion ltrnatb K HL W H F T P B P A F P A

Proof

Step Hyp Ref Expression
1 ltrnatb.b B = Base K
2 ltrnatb.a A = Atoms K
3 ltrnatb.h H = LHyp K
4 ltrnatb.t T = LTrn K W
5 simp3 K HL W H F T P B P B
6 1 3 4 ltrncl K HL W H F T P B F P B
7 5 6 2thd K HL W H F T P B P B F P B
8 simp1 K HL W H F T P B K HL W H
9 simp2 K HL W H F T P B F T
10 simp1l K HL W H F T P B K HL
11 hlop K HL K OP
12 eqid 0. K = 0. K
13 1 12 op0cl K OP 0. K B
14 10 11 13 3syl K HL W H F T P B 0. K B
15 eqid K = K
16 1 15 3 4 ltrncvr K HL W H F T 0. K B P B 0. K K P F 0. K K F P
17 8 9 14 5 16 syl112anc K HL W H F T P B 0. K K P F 0. K K F P
18 10 11 syl K HL W H F T P B K OP
19 simp1r K HL W H F T P B W H
20 1 3 lhpbase W H W B
21 19 20 syl K HL W H F T P B W B
22 eqid K = K
23 1 22 12 op0le K OP W B 0. K K W
24 18 21 23 syl2anc K HL W H F T P B 0. K K W
25 1 22 3 4 ltrnval1 K HL W H F T 0. K B 0. K K W F 0. K = 0. K
26 8 9 14 24 25 syl112anc K HL W H F T P B F 0. K = 0. K
27 26 breq1d K HL W H F T P B F 0. K K F P 0. K K F P
28 17 27 bitrd K HL W H F T P B 0. K K P 0. K K F P
29 7 28 anbi12d K HL W H F T P B P B 0. K K P F P B 0. K K F P
30 1 12 15 2 isat K HL P A P B 0. K K P
31 10 30 syl K HL W H F T P B P A P B 0. K K P
32 1 12 15 2 isat K HL F P A F P B 0. K K F P
33 10 32 syl K HL W H F T P B F P A F P B 0. K K F P
34 29 31 33 3bitr4d K HL W H F T P B P A F P A