Metamath Proof Explorer


Theorem cdleme0cp

Description: Part of proof of Lemma E in Crawley p. 113. TODO: Reformat as in cdlemg3a - swap consequent equality; make antecedent use df-3an . (Contributed by NM, 13-Jun-2012)

Ref Expression
Hypotheses cdleme0.l ˙ = K
cdleme0.j ˙ = join K
cdleme0.m ˙ = meet K
cdleme0.a A = Atoms K
cdleme0.h H = LHyp K
cdleme0.u U = P ˙ Q ˙ W
Assertion cdleme0cp K HL W H P A ¬ P ˙ W Q A P ˙ U = P ˙ Q

Proof

Step Hyp Ref Expression
1 cdleme0.l ˙ = K
2 cdleme0.j ˙ = join K
3 cdleme0.m ˙ = meet K
4 cdleme0.a A = Atoms K
5 cdleme0.h H = LHyp K
6 cdleme0.u U = P ˙ Q ˙ W
7 6 oveq2i P ˙ U = P ˙ P ˙ Q ˙ W
8 simpll K HL W H P A ¬ P ˙ W Q A K HL
9 simprll K HL W H P A ¬ P ˙ W Q A P A
10 hllat K HL K Lat
11 10 ad2antrr K HL W H P A ¬ P ˙ W Q A K Lat
12 eqid Base K = Base K
13 12 4 atbase P A P Base K
14 9 13 syl K HL W H P A ¬ P ˙ W Q A P Base K
15 simprr K HL W H P A ¬ P ˙ W Q A Q A
16 12 4 atbase Q A Q Base K
17 15 16 syl K HL W H P A ¬ P ˙ W Q A Q Base K
18 12 2 latjcl K Lat P Base K Q Base K P ˙ Q Base K
19 11 14 17 18 syl3anc K HL W H P A ¬ P ˙ W Q A P ˙ Q Base K
20 12 5 lhpbase W H W Base K
21 20 ad2antlr K HL W H P A ¬ P ˙ W Q A W Base K
22 1 2 4 hlatlej1 K HL P A Q A P ˙ P ˙ Q
23 8 9 15 22 syl3anc K HL W H P A ¬ P ˙ W Q A P ˙ P ˙ Q
24 12 1 2 3 4 atmod3i1 K HL P A P ˙ Q Base K W Base K P ˙ P ˙ Q P ˙ P ˙ Q ˙ W = P ˙ Q ˙ P ˙ W
25 8 9 19 21 23 24 syl131anc K HL W H P A ¬ P ˙ W Q A P ˙ P ˙ Q ˙ W = P ˙ Q ˙ P ˙ W
26 eqid 1. K = 1. K
27 1 2 26 4 5 lhpjat2 K HL W H P A ¬ P ˙ W P ˙ W = 1. K
28 27 adantrr K HL W H P A ¬ P ˙ W Q A P ˙ W = 1. K
29 28 oveq2d K HL W H P A ¬ P ˙ W Q A P ˙ Q ˙ P ˙ W = P ˙ Q ˙ 1. K
30 hlol K HL K OL
31 30 ad2antrr K HL W H P A ¬ P ˙ W Q A K OL
32 12 3 26 olm11 K OL P ˙ Q Base K P ˙ Q ˙ 1. K = P ˙ Q
33 31 19 32 syl2anc K HL W H P A ¬ P ˙ W Q A P ˙ Q ˙ 1. K = P ˙ Q
34 25 29 33 3eqtrd K HL W H P A ¬ P ˙ W Q A P ˙ P ˙ Q ˙ W = P ˙ Q
35 7 34 syl5eq K HL W H P A ¬ P ˙ W Q A P ˙ U = P ˙ Q