Metamath Proof Explorer


Theorem lhpelim

Description: Eliminate an atom not under a lattice hyperplane. TODO: Look at proofs using lhpmat to see if this can be used to shorten them. (Contributed by NM, 27-Apr-2013)

Ref Expression
Hypotheses lhpelim.b B = Base K
lhpelim.l ˙ = K
lhpelim.j ˙ = join K
lhpelim.m ˙ = meet K
lhpelim.a A = Atoms K
lhpelim.h H = LHyp K
Assertion lhpelim K HL W H P A ¬ P ˙ W X B P ˙ X ˙ W ˙ W = X ˙ W

Proof

Step Hyp Ref Expression
1 lhpelim.b B = Base K
2 lhpelim.l ˙ = K
3 lhpelim.j ˙ = join K
4 lhpelim.m ˙ = meet K
5 lhpelim.a A = Atoms K
6 lhpelim.h H = LHyp K
7 eqid 0. K = 0. K
8 2 4 7 5 6 lhpmat K HL W H P A ¬ P ˙ W P ˙ W = 0. K
9 8 3adant3 K HL W H P A ¬ P ˙ W X B P ˙ W = 0. K
10 9 oveq1d K HL W H P A ¬ P ˙ W X B P ˙ W ˙ X ˙ W = 0. K ˙ X ˙ W
11 simp1l K HL W H P A ¬ P ˙ W X B K HL
12 simp2l K HL W H P A ¬ P ˙ W X B P A
13 11 hllatd K HL W H P A ¬ P ˙ W X B K Lat
14 simp3 K HL W H P A ¬ P ˙ W X B X B
15 simp1r K HL W H P A ¬ P ˙ W X B W H
16 1 6 lhpbase W H W B
17 15 16 syl K HL W H P A ¬ P ˙ W X B W B
18 1 4 latmcl K Lat X B W B X ˙ W B
19 13 14 17 18 syl3anc K HL W H P A ¬ P ˙ W X B X ˙ W B
20 1 2 4 latmle2 K Lat X B W B X ˙ W ˙ W
21 13 14 17 20 syl3anc K HL W H P A ¬ P ˙ W X B X ˙ W ˙ W
22 1 2 3 4 5 atmod4i2 K HL P A X ˙ W B W B X ˙ W ˙ W P ˙ W ˙ X ˙ W = P ˙ X ˙ W ˙ W
23 11 12 19 17 21 22 syl131anc K HL W H P A ¬ P ˙ W X B P ˙ W ˙ X ˙ W = P ˙ X ˙ W ˙ W
24 hlol K HL K OL
25 11 24 syl K HL W H P A ¬ P ˙ W X B K OL
26 1 3 7 olj02 K OL X ˙ W B 0. K ˙ X ˙ W = X ˙ W
27 25 19 26 syl2anc K HL W H P A ¬ P ˙ W X B 0. K ˙ X ˙ W = X ˙ W
28 10 23 27 3eqtr3d K HL W H P A ¬ P ˙ W X B P ˙ X ˙ W ˙ W = X ˙ W