Metamath Proof Explorer


Theorem 3pthdlem1

Description: Lemma 1 for 3pthd . (Contributed by AV, 9-Feb-2021)

Ref Expression
Hypotheses 3wlkd.p P=⟨“ABCD”⟩
3wlkd.f F=⟨“JKL”⟩
3wlkd.s φAVBVCVDV
3wlkd.n φABACBCBDCD
Assertion 3pthdlem1 φk0..^Pj1..^FkjPkPj

Proof

Step Hyp Ref Expression
1 3wlkd.p P=⟨“ABCD”⟩
2 3wlkd.f F=⟨“JKL”⟩
3 3wlkd.s φAVBVCVDV
4 3wlkd.n φABACBCBDCD
5 1 2 3 3wlkdlem3 φP0=AP1=BP2=CP3=D
6 simpr1l P0=AP1=BP2=CP3=DABACBCBDCDAB
7 simpl P0=AP1=BP0=A
8 7 adantr P0=AP1=BP2=CP3=DP0=A
9 simpr P0=AP1=BP1=B
10 9 adantr P0=AP1=BP2=CP3=DP1=B
11 8 10 neeq12d P0=AP1=BP2=CP3=DP0P1AB
12 11 adantr P0=AP1=BP2=CP3=DABACBCBDCDP0P1AB
13 6 12 mpbird P0=AP1=BP2=CP3=DABACBCBDCDP0P1
14 13 a1d P0=AP1=BP2=CP3=DABACBCBDCD01P0P1
15 simpr1r P0=AP1=BP2=CP3=DABACBCBDCDAC
16 simpl P2=CP3=DP2=C
17 16 adantl P0=AP1=BP2=CP3=DP2=C
18 8 17 neeq12d P0=AP1=BP2=CP3=DP0P2AC
19 18 adantr P0=AP1=BP2=CP3=DABACBCBDCDP0P2AC
20 15 19 mpbird P0=AP1=BP2=CP3=DABACBCBDCDP0P2
21 20 a1d P0=AP1=BP2=CP3=DABACBCBDCD02P0P2
22 14 21 jca P0=AP1=BP2=CP3=DABACBCBDCD01P0P102P0P2
23 eqid 1=1
24 23 2a1i P0=AP1=BP2=CP3=DABACBCBDCDP1=P11=1
25 24 necon3d P0=AP1=BP2=CP3=DABACBCBDCD11P1P1
26 simpr2l P0=AP1=BP2=CP3=DABACBCBDCDBC
27 10 17 neeq12d P0=AP1=BP2=CP3=DP1P2BC
28 27 adantr P0=AP1=BP2=CP3=DABACBCBDCDP1P2BC
29 26 28 mpbird P0=AP1=BP2=CP3=DABACBCBDCDP1P2
30 29 a1d P0=AP1=BP2=CP3=DABACBCBDCD12P1P2
31 25 30 jca P0=AP1=BP2=CP3=DABACBCBDCD11P1P112P1P2
32 29 necomd P0=AP1=BP2=CP3=DABACBCBDCDP2P1
33 32 a1d P0=AP1=BP2=CP3=DABACBCBDCD21P2P1
34 eqid 2=2
35 34 2a1i P0=AP1=BP2=CP3=DABACBCBDCDP2=P22=2
36 35 necon3d P0=AP1=BP2=CP3=DABACBCBDCD22P2P2
37 simpr2r P0=AP1=BP2=CP3=DABACBCBDCDBD
38 simpr P2=CP3=DP3=D
39 38 adantl P0=AP1=BP2=CP3=DP3=D
40 10 39 neeq12d P0=AP1=BP2=CP3=DP1P3BD
41 40 adantr P0=AP1=BP2=CP3=DABACBCBDCDP1P3BD
42 37 41 mpbird P0=AP1=BP2=CP3=DABACBCBDCDP1P3
43 42 necomd P0=AP1=BP2=CP3=DABACBCBDCDP3P1
44 43 a1d P0=AP1=BP2=CP3=DABACBCBDCD31P3P1
45 simp3 ABACBCBDCDCD
46 45 necomd ABACBCBDCDDC
47 46 adantl P0=AP1=BP2=CP3=DABACBCBDCDDC
48 simpl P3=DP2=CP3=D
49 simpr P3=DP2=CP2=C
50 48 49 neeq12d P3=DP2=CP3P2DC
51 50 ancoms P2=CP3=DP3P2DC
52 51 adantl P0=AP1=BP2=CP3=DP3P2DC
53 52 adantr P0=AP1=BP2=CP3=DABACBCBDCDP3P2DC
54 47 53 mpbird P0=AP1=BP2=CP3=DABACBCBDCDP3P2
55 54 a1d P0=AP1=BP2=CP3=DABACBCBDCD32P3P2
56 44 55 jca P0=AP1=BP2=CP3=DABACBCBDCD31P3P132P3P2
57 33 36 56 jca31 P0=AP1=BP2=CP3=DABACBCBDCD21P2P122P2P231P3P132P3P2
58 22 31 57 jca31 P0=AP1=BP2=CP3=DABACBCBDCD01P0P102P0P211P1P112P1P221P2P122P2P231P3P132P3P2
59 5 4 58 syl2anc φ01P0P102P0P211P1P112P1P221P2P122P2P231P3P132P3P2
60 1 fveq2i P=⟨“ABCD”⟩
61 s4len ⟨“ABCD”⟩=4
62 60 61 eqtri P=4
63 62 oveq2i 0..^P=0..^4
64 fzo0to42pr 0..^4=0123
65 63 64 eqtri 0..^P=0123
66 65 raleqi k0..^Pk1PkP1k2PkP2k0123k1PkP1k2PkP2
67 ralunb k0123k1PkP1k2PkP2k01k1PkP1k2PkP2k23k1PkP1k2PkP2
68 c0ex 0V
69 1ex 1V
70 neeq1 k=0k101
71 fveq2 k=0Pk=P0
72 71 neeq1d k=0PkP1P0P1
73 70 72 imbi12d k=0k1PkP101P0P1
74 neeq1 k=0k202
75 71 neeq1d k=0PkP2P0P2
76 74 75 imbi12d k=0k2PkP202P0P2
77 73 76 anbi12d k=0k1PkP1k2PkP201P0P102P0P2
78 neeq1 k=1k111
79 fveq2 k=1Pk=P1
80 79 neeq1d k=1PkP1P1P1
81 78 80 imbi12d k=1k1PkP111P1P1
82 neeq1 k=1k212
83 79 neeq1d k=1PkP2P1P2
84 82 83 imbi12d k=1k2PkP212P1P2
85 81 84 anbi12d k=1k1PkP1k2PkP211P1P112P1P2
86 68 69 77 85 ralpr k01k1PkP1k2PkP201P0P102P0P211P1P112P1P2
87 2ex 2V
88 3ex 3V
89 neeq1 k=2k121
90 fveq2 k=2Pk=P2
91 90 neeq1d k=2PkP1P2P1
92 89 91 imbi12d k=2k1PkP121P2P1
93 neeq1 k=2k222
94 90 neeq1d k=2PkP2P2P2
95 93 94 imbi12d k=2k2PkP222P2P2
96 92 95 anbi12d k=2k1PkP1k2PkP221P2P122P2P2
97 neeq1 k=3k131
98 fveq2 k=3Pk=P3
99 98 neeq1d k=3PkP1P3P1
100 97 99 imbi12d k=3k1PkP131P3P1
101 neeq1 k=3k232
102 98 neeq1d k=3PkP2P3P2
103 101 102 imbi12d k=3k2PkP232P3P2
104 100 103 anbi12d k=3k1PkP1k2PkP231P3P132P3P2
105 87 88 96 104 ralpr k23k1PkP1k2PkP221P2P122P2P231P3P132P3P2
106 86 105 anbi12i k01k1PkP1k2PkP2k23k1PkP1k2PkP201P0P102P0P211P1P112P1P221P2P122P2P231P3P132P3P2
107 66 67 106 3bitri k0..^Pk1PkP1k2PkP201P0P102P0P211P1P112P1P221P2P122P2P231P3P132P3P2
108 59 107 sylibr φk0..^Pk1PkP1k2PkP2
109 2 fveq2i F=⟨“JKL”⟩
110 s3len ⟨“JKL”⟩=3
111 109 110 eqtri F=3
112 111 oveq2i 1..^F=1..^3
113 fzo13pr 1..^3=12
114 112 113 eqtri 1..^F=12
115 114 raleqi j1..^FkjPkPjj12kjPkPj
116 neeq2 j=1kjk1
117 fveq2 j=1Pj=P1
118 117 neeq2d j=1PkPjPkP1
119 116 118 imbi12d j=1kjPkPjk1PkP1
120 neeq2 j=2kjk2
121 fveq2 j=2Pj=P2
122 121 neeq2d j=2PkPjPkP2
123 120 122 imbi12d j=2kjPkPjk2PkP2
124 69 87 119 123 ralpr j12kjPkPjk1PkP1k2PkP2
125 115 124 bitri j1..^FkjPkPjk1PkP1k2PkP2
126 125 ralbii k0..^Pj1..^FkjPkPjk0..^Pk1PkP1k2PkP2
127 108 126 sylibr φk0..^Pj1..^FkjPkPj