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 ˙=joinK
cdleme0.m ˙=meetK
cdleme0.a A=AtomsK
cdleme0.h H=LHypK
cdleme0.u U=P˙Q˙W
Assertion cdleme0cp KHLWHPA¬P˙WQAP˙U=P˙Q

Proof

Step Hyp Ref Expression
1 cdleme0.l ˙=K
2 cdleme0.j ˙=joinK
3 cdleme0.m ˙=meetK
4 cdleme0.a A=AtomsK
5 cdleme0.h H=LHypK
6 cdleme0.u U=P˙Q˙W
7 6 oveq2i P˙U=P˙P˙Q˙W
8 simpll KHLWHPA¬P˙WQAKHL
9 simprll KHLWHPA¬P˙WQAPA
10 hllat KHLKLat
11 10 ad2antrr KHLWHPA¬P˙WQAKLat
12 eqid BaseK=BaseK
13 12 4 atbase PAPBaseK
14 9 13 syl KHLWHPA¬P˙WQAPBaseK
15 simprr KHLWHPA¬P˙WQAQA
16 12 4 atbase QAQBaseK
17 15 16 syl KHLWHPA¬P˙WQAQBaseK
18 12 2 latjcl KLatPBaseKQBaseKP˙QBaseK
19 11 14 17 18 syl3anc KHLWHPA¬P˙WQAP˙QBaseK
20 12 5 lhpbase WHWBaseK
21 20 ad2antlr KHLWHPA¬P˙WQAWBaseK
22 1 2 4 hlatlej1 KHLPAQAP˙P˙Q
23 8 9 15 22 syl3anc KHLWHPA¬P˙WQAP˙P˙Q
24 12 1 2 3 4 atmod3i1 KHLPAP˙QBaseKWBaseKP˙P˙QP˙P˙Q˙W=P˙Q˙P˙W
25 8 9 19 21 23 24 syl131anc KHLWHPA¬P˙WQAP˙P˙Q˙W=P˙Q˙P˙W
26 eqid 1.K=1.K
27 1 2 26 4 5 lhpjat2 KHLWHPA¬P˙WP˙W=1.K
28 27 adantrr KHLWHPA¬P˙WQAP˙W=1.K
29 28 oveq2d KHLWHPA¬P˙WQAP˙Q˙P˙W=P˙Q˙1.K
30 hlol KHLKOL
31 30 ad2antrr KHLWHPA¬P˙WQAKOL
32 12 3 26 olm11 KOLP˙QBaseKP˙Q˙1.K=P˙Q
33 31 19 32 syl2anc KHLWHPA¬P˙WQAP˙Q˙1.K=P˙Q
34 25 29 33 3eqtrd KHLWHPA¬P˙WQAP˙P˙Q˙W=P˙Q
35 7 34 eqtrid KHLWHPA¬P˙WQAP˙U=P˙Q