Metamath Proof Explorer


Theorem dalem24

Description: Lemma for dath . Show that auxiliary atom G is outside of plane Y . (Contributed by NM, 2-Aug-2012)

Ref Expression
Hypotheses dalem.ph φKHLCBaseKPAQARASATAUAYOZO¬C˙P˙Q¬C˙Q˙R¬C˙R˙P¬C˙S˙T¬C˙T˙U¬C˙U˙SC˙P˙SC˙Q˙TC˙R˙U
dalem.l ˙=K
dalem.j ˙=joinK
dalem.a A=AtomsK
dalem.ps ψcAdA¬c˙Ydc¬d˙YC˙c˙d
dalem23.m ˙=meetK
dalem23.o O=LPlanesK
dalem23.y Y=P˙Q˙R
dalem23.z Z=S˙T˙U
dalem23.g G=c˙P˙d˙S
Assertion dalem24 φY=Zψ¬G˙Y

Proof

Step Hyp Ref Expression
1 dalem.ph φKHLCBaseKPAQARASATAUAYOZO¬C˙P˙Q¬C˙Q˙R¬C˙R˙P¬C˙S˙T¬C˙T˙U¬C˙U˙SC˙P˙SC˙Q˙TC˙R˙U
2 dalem.l ˙=K
3 dalem.j ˙=joinK
4 dalem.a A=AtomsK
5 dalem.ps ψcAdA¬c˙Ydc¬d˙YC˙c˙d
6 dalem23.m ˙=meetK
7 dalem23.o O=LPlanesK
8 dalem23.y Y=P˙Q˙R
9 dalem23.z Z=S˙T˙U
10 dalem23.g G=c˙P˙d˙S
11 10 oveq1i G˙Y=c˙P˙d˙S˙Y
12 1 dalemkehl φKHL
13 hlol KHLKOL
14 12 13 syl φKOL
15 14 3ad2ant1 φY=ZψKOL
16 12 3ad2ant1 φY=ZψKHL
17 5 dalemccea ψcA
18 17 3ad2ant3 φY=ZψcA
19 1 dalempea φPA
20 19 3ad2ant1 φY=ZψPA
21 eqid BaseK=BaseK
22 21 3 4 hlatjcl KHLcAPAc˙PBaseK
23 16 18 20 22 syl3anc φY=Zψc˙PBaseK
24 5 dalemddea ψdA
25 24 3ad2ant3 φY=ZψdA
26 1 dalemsea φSA
27 26 3ad2ant1 φY=ZψSA
28 21 3 4 hlatjcl KHLdASAd˙SBaseK
29 16 25 27 28 syl3anc φY=Zψd˙SBaseK
30 1 7 dalemyeb φYBaseK
31 30 3ad2ant1 φY=ZψYBaseK
32 21 6 latmmdir KOLc˙PBaseKd˙SBaseKYBaseKc˙P˙d˙S˙Y=c˙P˙Y˙d˙S˙Y
33 15 23 29 31 32 syl13anc φY=Zψc˙P˙d˙S˙Y=c˙P˙Y˙d˙S˙Y
34 11 33 eqtrid φY=ZψG˙Y=c˙P˙Y˙d˙S˙Y
35 3 4 hlatjcom KHLcAPAc˙P=P˙c
36 16 18 20 35 syl3anc φY=Zψc˙P=P˙c
37 36 oveq1d φY=Zψc˙P˙Y=P˙c˙Y
38 1 2 3 4 7 8 dalemply φP˙Y
39 38 3ad2ant1 φY=ZψP˙Y
40 5 dalem-ccly ψ¬c˙Y
41 40 3ad2ant3 φY=Zψ¬c˙Y
42 21 2 3 6 4 2atjm KHLPAcAYBaseKP˙Y¬c˙YP˙c˙Y=P
43 16 20 18 31 39 41 42 syl132anc φY=ZψP˙c˙Y=P
44 37 43 eqtrd φY=Zψc˙P˙Y=P
45 3 4 hlatjcom KHLdASAd˙S=S˙d
46 16 25 27 45 syl3anc φY=Zψd˙S=S˙d
47 46 oveq1d φY=Zψd˙S˙Y=S˙d˙Y
48 1 2 3 4 9 dalemsly φY=ZS˙Y
49 48 3adant3 φY=ZψS˙Y
50 5 dalem-ddly ψ¬d˙Y
51 50 3ad2ant3 φY=Zψ¬d˙Y
52 21 2 3 6 4 2atjm KHLSAdAYBaseKS˙Y¬d˙YS˙d˙Y=S
53 16 27 25 31 49 51 52 syl132anc φY=ZψS˙d˙Y=S
54 47 53 eqtrd φY=Zψd˙S˙Y=S
55 44 54 oveq12d φY=Zψc˙P˙Y˙d˙S˙Y=P˙S
56 1 2 3 4 7 8 dalempnes φPS
57 hlatl KHLKAtLat
58 12 57 syl φKAtLat
59 eqid 0.K=0.K
60 6 59 4 atnem0 KAtLatPASAPSP˙S=0.K
61 58 19 26 60 syl3anc φPSP˙S=0.K
62 56 61 mpbid φP˙S=0.K
63 62 3ad2ant1 φY=ZψP˙S=0.K
64 34 55 63 3eqtrd φY=ZψG˙Y=0.K
65 58 3ad2ant1 φY=ZψKAtLat
66 1 2 3 4 5 6 7 8 9 10 dalem23 φY=ZψGA
67 21 2 6 59 4 atnle KAtLatGAYBaseK¬G˙YG˙Y=0.K
68 65 66 31 67 syl3anc φY=Zψ¬G˙YG˙Y=0.K
69 64 68 mpbird φY=Zψ¬G˙Y