Metamath Proof Explorer


Theorem trl0

Description: If an atom not under the fiducial co-atom W equals its lattice translation, the trace of the translation is zero. (Contributed by NM, 24-May-2012)

Ref Expression
Hypotheses trl0.l ⊢ ≤ ˙ = ≤ K
trl0.z ⊢ 0 ˙ = 0. ⁡ K
trl0.a ⊢ A = Atoms ⁡ K
trl0.h ⊢ H = LHyp ⁡ K
trl0.t ⊢ T = LTrn ⁡ K ⁡ W
trl0.r ⊢ R = trL ⁡ K ⁡ W
Assertion trl0 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → R ⁡ F = 0 ˙

Proof

Step Hyp Ref Expression
1 trl0.l ⊢ ≤ ˙ = ≤ K
2 trl0.z ⊢ 0 ˙ = 0. ⁡ K
3 trl0.a ⊢ A = Atoms ⁡ K
4 trl0.h ⊢ H = LHyp ⁡ K
5 trl0.t ⊢ T = LTrn ⁡ K ⁡ W
6 trl0.r ⊢ R = trL ⁡ K ⁡ W
7 simp1 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → K ∈ HL ∧ W ∈ H
8 simp3l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → F ∈ T
9 simp2 ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → P ∈ A ∧ ¬ P ≤ ˙ W
10 eqid ⊢ join ⁡ K = join ⁡ K
11 eqid ⊢ meet ⁡ K = meet ⁡ K
12 1 10 11 3 4 5 6 trlval2 ⊢ K ∈ HL ∧ W ∈ H ∧ F ∈ T ∧ P ∈ A ∧ ¬ P ≤ ˙ W → R ⁡ F = P join ⁡ K F ⁡ P meet ⁡ K W
13 7 8 9 12 syl3anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → R ⁡ F = P join ⁡ K F ⁡ P meet ⁡ K W
14 simp3r ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → F ⁡ P = P
15 14 oveq2d ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → P join ⁡ K F ⁡ P = P join ⁡ K P
16 simp1l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → K ∈ HL
17 simp2l ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → P ∈ A
18 10 3 hlatjidm ⊢ K ∈ HL ∧ P ∈ A → P join ⁡ K P = P
19 16 17 18 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → P join ⁡ K P = P
20 15 19 eqtrd ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → P join ⁡ K F ⁡ P = P
21 20 oveq1d ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → P join ⁡ K F ⁡ P meet ⁡ K W = P meet ⁡ K W
22 1 11 2 3 4 lhpmat ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W → P meet ⁡ K W = 0 ˙
23 7 9 22 syl2anc ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → P meet ⁡ K W = 0 ˙
24 13 21 23 3eqtrd ⊢ K ∈ HL ∧ W ∈ H ∧ P ∈ A ∧ ¬ P ≤ ˙ W ∧ F ∈ T ∧ F ⁡ P = P → R ⁡ F = 0 ˙