Metamath Proof Explorer


Theorem cdlemg15a

Description: Eliminate the ( FP ) =/= P condition from cdlemg13 . TODO: FIX COMMENT. (Contributed by NM, 6-May-2013)

Ref Expression
Hypotheses cdlemg12.l ˙=K
cdlemg12.j ˙=joinK
cdlemg12.m ˙=meetK
cdlemg12.a A=AtomsK
cdlemg12.h H=LHypK
cdlemg12.t T=LTrnKW
cdlemg12b.r R=trLKW
Assertion cdlemg15a KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QP˙FGP˙W=Q˙FGQ˙W

Proof

Step Hyp Ref Expression
1 cdlemg12.l ˙=K
2 cdlemg12.j ˙=joinK
3 cdlemg12.m ˙=meetK
4 cdlemg12.a A=AtomsK
5 cdlemg12.h H=LHypK
6 cdlemg12.t T=LTrnKW
7 cdlemg12b.r R=trLKW
8 simpl11 KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFP=PKHLWH
9 simpl12 KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFP=PPA¬P˙W
10 simpl13 KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFP=PQA¬Q˙W
11 simpl2l KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFP=PFT
12 simpl2r KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFP=PGT
13 simpr KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFP=PFP=P
14 1 2 3 4 5 6 7 cdlemg14f KHLWHPA¬P˙WQA¬Q˙WFTGTFP=PP˙FGP˙W=Q˙FGQ˙W
15 8 9 10 11 12 13 14 syl123anc KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFP=PP˙FGP˙W=Q˙FGQ˙W
16 simpl1 KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFPPKHLWHPA¬P˙WQA¬Q˙W
17 simpl2 KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFPPFTGT
18 simpr KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFPPFPP
19 simpl3l KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFPPRF=RG
20 simpl3r KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFPPFGP˙FGQP˙Q
21 1 2 3 4 5 6 7 cdlemg13 KHLWHPA¬P˙WQA¬Q˙WFTGTFPPRF=RGFGP˙FGQP˙QP˙FGP˙W=Q˙FGQ˙W
22 16 17 18 19 20 21 syl113anc KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QFPPP˙FGP˙W=Q˙FGQ˙W
23 15 22 pm2.61dane KHLWHPA¬P˙WQA¬Q˙WFTGTRF=RGFGP˙FGQP˙QP˙FGP˙W=Q˙FGQ˙W