Metamath Proof Explorer


Theorem line2ylem

Description: Lemma for line2y . This proof is based on counterexamples for the following cases: 1. C =/= 0 : p = (0,0) (LHS of bicondional is false, RHS is true); 2. C = 0 /\ B =/= 0 : p = (1,-A/B) (LHS of bicondional is true, RHS is false); 3. A = B = C = 0 : p = (1,1) (LHS of bicondional is true, RHS is false). (Contributed by AV, 4-Feb-2023)

Ref Expression
Hypotheses line2ylem.i I=12
line2ylem.p P=I
Assertion line2ylem ABCpPAp1+Bp2=Cp1=0A0B=0C=0

Proof

Step Hyp Ref Expression
1 line2ylem.i I=12
2 line2ylem.p P=I
3 ianor ¬A0B=0C=0¬A0B=0¬C=0
4 df-ne C0¬C=0
5 0re 0
6 1 2 prelrrx2 001020P
7 5 5 6 mp2an 1020P
8 eqneqall C=0C0¬0=0
9 8 com12 C0C=0¬0=0
10 eqid 0=0
11 10 pm2.24i ¬0=0C=0
12 9 11 impbid1 C0C=0¬0=0
13 12 adantl ABCC0C=0¬0=0
14 xor3 ¬C=00=0C=0¬0=0
15 13 14 sylibr ABCC0¬C=00=0
16 simp1 ABCA
17 16 recnd ABCA
18 17 mul01d ABCA0=0
19 simp2 ABCB
20 19 recnd ABCB
21 20 mul01d ABCB0=0
22 18 21 oveq12d ABCA0+B0=0+0
23 00id 0+0=0
24 22 23 eqtrdi ABCA0+B0=0
25 24 eqeq1d ABCA0+B0=C0=C
26 eqcom 0=CC=0
27 25 26 bitrdi ABCA0+B0=CC=0
28 27 adantr ABCC0A0+B0=CC=0
29 28 bibi1d ABCC0A0+B0=C0=0C=00=0
30 15 29 mtbird ABCC0¬A0+B0=C0=0
31 fveq1 p=1020p1=10201
32 1ex 1V
33 c0ex 0V
34 1ne2 12
35 fvpr1g 1V0V1210201=0
36 32 33 34 35 mp3an 10201=0
37 31 36 eqtrdi p=1020p1=0
38 37 oveq2d p=1020Ap1=A0
39 fveq1 p=1020p2=10202
40 2ex 2V
41 fvpr2g 2V0V1210202=0
42 40 33 34 41 mp3an 10202=0
43 39 42 eqtrdi p=1020p2=0
44 43 oveq2d p=1020Bp2=B0
45 38 44 oveq12d p=1020Ap1+Bp2=A0+B0
46 45 eqeq1d p=1020Ap1+Bp2=CA0+B0=C
47 37 eqeq1d p=1020p1=00=0
48 46 47 bibi12d p=1020Ap1+Bp2=Cp1=0A0+B0=C0=0
49 48 notbid p=1020¬Ap1+Bp2=Cp1=0¬A0+B0=C0=0
50 49 rspcev 1020P¬A0+B0=C0=0pP¬Ap1+Bp2=Cp1=0
51 7 30 50 sylancr ABCC0pP¬Ap1+Bp2=Cp1=0
52 51 expcom C0ABCpP¬Ap1+Bp2=Cp1=0
53 4 52 sylbir ¬C=0ABCpP¬Ap1+Bp2=Cp1=0
54 notnotb C=0¬¬C=0
55 ianor ¬A0B=0¬A0¬B=0
56 df-ne B0¬B=0
57 1red ABCB0C=01
58 16 adantr ABCB0C=0A
59 58 renegcld ABCB0C=0A
60 19 adantr ABCB0C=0B
61 simprl ABCB0C=0B0
62 59 60 61 redivcld ABCB0C=0AB
63 1 2 prelrrx2 1AB112ABP
64 57 62 63 syl2anc ABCB0C=0112ABP
65 ax-1ne0 10
66 65 neii ¬1=0
67 10 66 2th 0=0¬1=0
68 xor3 ¬0=01=00=0¬1=0
69 67 68 mpbir ¬0=01=0
70 17 mulridd ABCA1=A
71 70 adantr ABCB0C=0A1=A
72 17 negcld ABCA
73 72 adantr ABCB0C=0A
74 20 adantr ABCB0C=0B
75 73 74 61 divcan2d ABCB0C=0BAB=A
76 71 75 oveq12d ABCB0C=0A1+BAB=A+A
77 17 negidd ABCA+A=0
78 77 adantr ABCB0C=0A+A=0
79 76 78 eqtrd ABCB0C=0A1+BAB=0
80 simprr ABCB0C=0C=0
81 79 80 eqeq12d ABCB0C=0A1+BAB=C0=0
82 81 bibi1d ABCB0C=0A1+BAB=C1=00=01=0
83 69 82 mtbiri ABCB0C=0¬A1+BAB=C1=0
84 fveq1 p=112ABp1=112AB1
85 fvpr1g 1V1V12112AB1=1
86 32 32 34 85 mp3an 112AB1=1
87 84 86 eqtrdi p=112ABp1=1
88 87 oveq2d p=112ABAp1=A1
89 fveq1 p=112ABp2=112AB2
90 ovex ABV
91 fvpr2g 2VABV12112AB2=AB
92 40 90 34 91 mp3an 112AB2=AB
93 89 92 eqtrdi p=112ABp2=AB
94 93 oveq2d p=112ABBp2=BAB
95 88 94 oveq12d p=112ABAp1+Bp2=A1+BAB
96 95 eqeq1d p=112ABAp1+Bp2=CA1+BAB=C
97 87 eqeq1d p=112ABp1=01=0
98 96 97 bibi12d p=112ABAp1+Bp2=Cp1=0A1+BAB=C1=0
99 98 notbid p=112AB¬Ap1+Bp2=Cp1=0¬A1+BAB=C1=0
100 99 rspcev 112ABP¬A1+BAB=C1=0pP¬Ap1+Bp2=Cp1=0
101 64 83 100 syl2anc ABCB0C=0pP¬Ap1+Bp2=Cp1=0
102 101 expcom B0C=0ABCpP¬Ap1+Bp2=Cp1=0
103 102 ex B0C=0ABCpP¬Ap1+Bp2=Cp1=0
104 56 103 sylbir ¬B=0C=0ABCpP¬Ap1+Bp2=Cp1=0
105 notnotb B=0¬¬B=0
106 nne ¬A0A=0
107 106 bicomi A=0¬A0
108 1re 1
109 1 2 prelrrx2 111121P
110 108 108 109 mp2an 1121P
111 oveq1 A=0A1=01
112 111 adantl B=0A=0A1=01
113 ax-1cn 1
114 113 mul02i 01=0
115 112 114 eqtrdi B=0A=0A1=0
116 oveq1 B=0B1=01
117 116 adantr B=0A=0B1=01
118 117 114 eqtrdi B=0A=0B1=0
119 115 118 oveq12d B=0A=0A1+B1=0+0
120 119 23 eqtrdi B=0A=0A1+B1=0
121 id C=0C=0
122 120 121 eqeqan12d B=0A=0C=0A1+B1=C0=0
123 122 bibi1d B=0A=0C=0A1+B1=C1=00=01=0
124 69 123 mtbiri B=0A=0C=0¬A1+B1=C1=0
125 fveq1 p=1121p1=11211
126 fvpr1g 1V1V1211211=1
127 32 32 34 126 mp3an 11211=1
128 125 127 eqtrdi p=1121p1=1
129 128 oveq2d p=1121Ap1=A1
130 fveq1 p=1121p2=11212
131 fvpr2g 2V1V1211212=1
132 40 32 34 131 mp3an 11212=1
133 130 132 eqtrdi p=1121p2=1
134 133 oveq2d p=1121Bp2=B1
135 129 134 oveq12d p=1121Ap1+Bp2=A1+B1
136 135 eqeq1d p=1121Ap1+Bp2=CA1+B1=C
137 128 eqeq1d p=1121p1=01=0
138 136 137 bibi12d p=1121Ap1+Bp2=Cp1=0A1+B1=C1=0
139 138 notbid p=1121¬Ap1+Bp2=Cp1=0¬A1+B1=C1=0
140 139 rspcev 1121P¬A1+B1=C1=0pP¬Ap1+Bp2=Cp1=0
141 110 124 140 sylancr B=0A=0C=0pP¬Ap1+Bp2=Cp1=0
142 141 a1d B=0A=0C=0ABCpP¬Ap1+Bp2=Cp1=0
143 142 ex B=0A=0C=0ABCpP¬Ap1+Bp2=Cp1=0
144 105 107 143 syl2anbr ¬¬B=0¬A0C=0ABCpP¬Ap1+Bp2=Cp1=0
145 104 144 jaoi3 ¬B=0¬A0C=0ABCpP¬Ap1+Bp2=Cp1=0
146 145 orcoms ¬A0¬B=0C=0ABCpP¬Ap1+Bp2=Cp1=0
147 55 146 sylbi ¬A0B=0C=0ABCpP¬Ap1+Bp2=Cp1=0
148 147 com12 C=0¬A0B=0ABCpP¬Ap1+Bp2=Cp1=0
149 54 148 sylbir ¬¬C=0¬A0B=0ABCpP¬Ap1+Bp2=Cp1=0
150 149 imp ¬¬C=0¬A0B=0ABCpP¬Ap1+Bp2=Cp1=0
151 53 150 jaoi3 ¬C=0¬A0B=0ABCpP¬Ap1+Bp2=Cp1=0
152 151 orcoms ¬A0B=0¬C=0ABCpP¬Ap1+Bp2=Cp1=0
153 152 com12 ABC¬A0B=0¬C=0pP¬Ap1+Bp2=Cp1=0
154 3 153 biimtrid ABC¬A0B=0C=0pP¬Ap1+Bp2=Cp1=0
155 rexnal pP¬Ap1+Bp2=Cp1=0¬pPAp1+Bp2=Cp1=0
156 154 155 imbitrdi ABC¬A0B=0C=0¬pPAp1+Bp2=Cp1=0
157 156 con4d ABCpPAp1+Bp2=Cp1=0A0B=0C=0
158 df-3an A0B=0C=0A0B=0C=0
159 157 158 syl6ibr ABCpPAp1+Bp2=Cp1=0A0B=0C=0