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