Metamath Proof Explorer


Theorem line2xlem

Description: Lemma for line2x . This proof is based on counterexamples for the following cases: 1. M =/= ( C / B ) : p = (0,C/B) (LHS of bicondional is true, RHS is false); 2. A =/= 0 /\ M = ( C / B ) : p = (1,C/B) (LHS of bicondional is false, RHS is true). (Contributed by AV, 4-Feb-2023)

Ref Expression
Hypotheses line2.i I=12
line2.e E=msup
line2.p P=I
line2.l L=LineME
line2.g G=pP|Ap1+Bp2=C
line2x.x X=102M
line2x.y Y=112M
Assertion line2xlem ABB0CMpPAp1+Bp2=Cp2=MA=0M=CB

Proof

Step Hyp Ref Expression
1 line2.i I=12
2 line2.e E=msup
3 line2.p P=I
4 line2.l L=LineME
5 line2.g G=pP|Ap1+Bp2=C
6 line2x.x X=102M
7 line2x.y Y=112M
8 ianor ¬A=0M=CB¬A=0¬M=CB
9 df-ne A0¬A=0
10 df-ne MCB¬M=CB
11 9 10 orbi12i A0MCB¬A=0¬M=CB
12 8 11 bitr4i ¬A=0M=CBA0MCB
13 0red MCBABB0CM0
14 simp3 ABB0CC
15 14 adantr ABB0CMC
16 simpl BB0B
17 16 3ad2ant2 ABB0CB
18 17 adantr ABB0CMB
19 simp2r ABB0CB0
20 19 adantr ABB0CMB0
21 15 18 20 redivcld ABB0CMCB
22 21 adantl MCBABB0CMCB
23 1 3 prelrrx2 0CB102CBP
24 13 22 23 syl2anc MCBABB0CM102CBP
25 id MCBMCB
26 25 necomd MCBCBM
27 26 neneqd MCB¬CB=M
28 27 a1d MCBC=C¬CB=M
29 eqidd ¬CB=MC=C
30 29 a1i MCB¬CB=MC=C
31 28 30 impbid MCBC=C¬CB=M
32 xor3 ¬C=CCB=MC=C¬CB=M
33 31 32 sylibr MCB¬C=CCB=M
34 33 adantr MCBABB0CM¬C=CCB=M
35 0red ABB0CM0
36 fv1prop 0102CB1=0
37 35 36 syl ABB0CM102CB1=0
38 37 oveq2d ABB0CMA102CB1=A0
39 recn AA
40 39 mul01d AA0=0
41 40 3ad2ant1 ABB0CA0=0
42 41 adantr ABB0CMA0=0
43 38 42 eqtrd ABB0CMA102CB1=0
44 ovexd ABB0CMCBV
45 fv2prop CBV102CB2=CB
46 44 45 syl ABB0CM102CB2=CB
47 46 oveq2d ABB0CMB102CB2=BCB
48 14 recnd ABB0CC
49 48 adantr ABB0CMC
50 16 recnd BB0B
51 50 3ad2ant2 ABB0CB
52 51 adantr ABB0CMB
53 49 52 20 divcan2d ABB0CMBCB=C
54 47 53 eqtrd ABB0CMB102CB2=C
55 43 54 oveq12d ABB0CMA102CB1+B102CB2=0+C
56 55 adantl MCBABB0CMA102CB1+B102CB2=0+C
57 48 addlidd ABB0C0+C=C
58 57 adantr ABB0CM0+C=C
59 58 adantl MCBABB0CM0+C=C
60 56 59 eqtrd MCBABB0CMA102CB1+B102CB2=C
61 60 eqeq1d MCBABB0CMA102CB1+B102CB2=CC=C
62 46 eqeq1d ABB0CM102CB2=MCB=M
63 62 adantl MCBABB0CM102CB2=MCB=M
64 61 63 bibi12d MCBABB0CMA102CB1+B102CB2=C102CB2=MC=CCB=M
65 34 64 mtbird MCBABB0CM¬A102CB1+B102CB2=C102CB2=M
66 fveq1 p=102CBp1=102CB1
67 66 oveq2d p=102CBAp1=A102CB1
68 fveq1 p=102CBp2=102CB2
69 68 oveq2d p=102CBBp2=B102CB2
70 67 69 oveq12d p=102CBAp1+Bp2=A102CB1+B102CB2
71 70 eqeq1d p=102CBAp1+Bp2=CA102CB1+B102CB2=C
72 68 eqeq1d p=102CBp2=M102CB2=M
73 71 72 bibi12d p=102CBAp1+Bp2=Cp2=MA102CB1+B102CB2=C102CB2=M
74 73 notbid p=102CB¬Ap1+Bp2=Cp2=M¬A102CB1+B102CB2=C102CB2=M
75 74 rspcev 102CBP¬A102CB1+B102CB2=C102CB2=MpP¬Ap1+Bp2=Cp2=M
76 24 65 75 syl2anc MCBABB0CMpP¬Ap1+Bp2=Cp2=M
77 76 ex MCBABB0CMpP¬Ap1+Bp2=Cp2=M
78 nne ¬MCBM=CB
79 1red ABB0C1
80 14 17 19 redivcld ABB0CCB
81 79 80 jca ABB0C1CB
82 81 adantr ABB0CM1CB
83 1 3 prelrrx2 1CB112CBP
84 82 83 syl ABB0CM112CBP
85 84 adantl M=CBA0ABB0CM112CBP
86 eqneqall A=0A0¬CB=M
87 86 com12 A0A=0¬CB=M
88 87 adantl M=CBA0A=0¬CB=M
89 pm2.24 CB=M¬CB=MA=0
90 89 eqcoms M=CB¬CB=MA=0
91 90 adantr M=CBA0¬CB=MA=0
92 88 91 impbid M=CBA0A=0¬CB=M
93 xor3 ¬A=0CB=MA=0¬CB=M
94 92 93 sylibr M=CBA0¬A=0CB=M
95 94 adantr M=CBA0ABB0CM¬A=0CB=M
96 simprl1 M=CBA0ABB0CMA
97 96 recnd M=CBA0ABB0CMA
98 15 adantl M=CBA0ABB0CMC
99 98 recnd M=CBA0ABB0CMC
100 97 99 addcomd M=CBA0ABB0CMA+C=C+A
101 100 eqeq1d M=CBA0ABB0CMA+C=CC+A=C
102 recn CC
103 39 102 anim12ci ACCA
104 103 3adant2 ABB0CCA
105 104 adantr ABB0CMCA
106 105 adantl M=CBA0ABB0CMCA
107 addid0 CAC+A=CA=0
108 106 107 syl M=CBA0ABB0CMC+A=CA=0
109 101 108 bitrd M=CBA0ABB0CMA+C=CA=0
110 109 bibi1d M=CBA0ABB0CMA+C=CCB=MA=0CB=M
111 95 110 mtbird M=CBA0ABB0CM¬A+C=CCB=M
112 1ex 1V
113 112 a1i ABB0CM1V
114 fv1prop 1V112CB1=1
115 113 114 syl ABB0CM112CB1=1
116 115 oveq2d ABB0CMA112CB1=A1
117 ax-1rid AA1=A
118 117 3ad2ant1 ABB0CA1=A
119 118 adantr ABB0CMA1=A
120 116 119 eqtrd ABB0CMA112CB1=A
121 fv2prop CBV112CB2=CB
122 44 121 syl ABB0CM112CB2=CB
123 122 oveq2d ABB0CMB112CB2=BCB
124 15 recnd ABB0CMC
125 124 52 20 divcan2d ABB0CMBCB=C
126 123 125 eqtrd ABB0CMB112CB2=C
127 120 126 oveq12d ABB0CMA112CB1+B112CB2=A+C
128 127 eqeq1d ABB0CMA112CB1+B112CB2=CA+C=C
129 122 eqeq1d ABB0CM112CB2=MCB=M
130 128 129 bibi12d ABB0CMA112CB1+B112CB2=C112CB2=MA+C=CCB=M
131 130 notbid ABB0CM¬A112CB1+B112CB2=C112CB2=M¬A+C=CCB=M
132 131 adantl M=CBA0ABB0CM¬A112CB1+B112CB2=C112CB2=M¬A+C=CCB=M
133 111 132 mpbird M=CBA0ABB0CM¬A112CB1+B112CB2=C112CB2=M
134 fveq1 p=112CBp1=112CB1
135 134 oveq2d p=112CBAp1=A112CB1
136 fveq1 p=112CBp2=112CB2
137 136 oveq2d p=112CBBp2=B112CB2
138 135 137 oveq12d p=112CBAp1+Bp2=A112CB1+B112CB2
139 138 eqeq1d p=112CBAp1+Bp2=CA112CB1+B112CB2=C
140 136 eqeq1d p=112CBp2=M112CB2=M
141 139 140 bibi12d p=112CBAp1+Bp2=Cp2=MA112CB1+B112CB2=C112CB2=M
142 141 notbid p=112CB¬Ap1+Bp2=Cp2=M¬A112CB1+B112CB2=C112CB2=M
143 142 rspcev 112CBP¬A112CB1+B112CB2=C112CB2=MpP¬Ap1+Bp2=Cp2=M
144 85 133 143 syl2anc M=CBA0ABB0CMpP¬Ap1+Bp2=Cp2=M
145 144 ex M=CBA0ABB0CMpP¬Ap1+Bp2=Cp2=M
146 78 145 sylanb ¬MCBA0ABB0CMpP¬Ap1+Bp2=Cp2=M
147 77 146 jaoi3 MCBA0ABB0CMpP¬Ap1+Bp2=Cp2=M
148 147 orcoms A0MCBABB0CMpP¬Ap1+Bp2=Cp2=M
149 148 com12 ABB0CMA0MCBpP¬Ap1+Bp2=Cp2=M
150 rexnal pP¬Ap1+Bp2=Cp2=M¬pPAp1+Bp2=Cp2=M
151 149 150 imbitrdi ABB0CMA0MCB¬pPAp1+Bp2=Cp2=M
152 12 151 biimtrid ABB0CM¬A=0M=CB¬pPAp1+Bp2=Cp2=M
153 152 con4d ABB0CMpPAp1+Bp2=Cp2=MA=0M=CB