Metamath Proof Explorer


Theorem isosctrlem2

Description: Lemma for isosctr . Corresponds to the case where one vertex is at 0, another at 1 and the third lies on the unit circle. (Contributed by Saveliy Skresanov, 31-Dec-2016)

Ref Expression
Assertion isosctrlem2 A A = 1 ¬ 1 = A log 1 A = log A 1 A

Proof

Step Hyp Ref Expression
1 1cnd A A = 1 ¬ 1 = A A 1 A = 0 1
2 simpl1 A A = 1 ¬ 1 = A A 1 A = 0 A
3 1 2 negsubd A A = 1 ¬ 1 = A A 1 A = 0 1 + A = 1 A
4 1rp 1 +
5 4 a1i A A = 1 ¬ 1 = A A 1 A = 0 1 +
6 simpl3 A A = 1 ¬ 1 = A A 1 A = 0 ¬ 1 = A
7 simpl2 A A = 1 ¬ 1 = A A 1 A = 0 A = 1
8 1 2 1 sub32d A A = 1 ¬ 1 = A A 1 A = 0 1 - A - 1 = 1 - 1 - A
9 1m1e0 1 1 = 0
10 9 oveq1i 1 - 1 - A = 0 A
11 df-neg A = 0 A
12 10 11 eqtr4i 1 - 1 - A = A
13 8 12 eqtrdi A A = 1 ¬ 1 = A A 1 A = 0 1 - A - 1 = A
14 1cnd A A = 1 ¬ 1 = A 1
15 simp1 A A = 1 ¬ 1 = A A
16 14 15 subcld A A = 1 ¬ 1 = A 1 A
17 16 adantr A A = 1 ¬ 1 = A A 1 A = 0 1 A
18 ax-1cn 1
19 subeq0 1 A 1 A = 0 1 = A
20 18 19 mpan A 1 A = 0 1 = A
21 20 biimpd A 1 A = 0 1 = A
22 21 con3dimp A ¬ 1 = A ¬ 1 A = 0
23 22 neqned A ¬ 1 = A 1 A 0
24 23 3adant2 A A = 1 ¬ 1 = A 1 A 0
25 24 adantr A A = 1 ¬ 1 = A A 1 A = 0 1 A 0
26 17 25 recrecd A A = 1 ¬ 1 = A A 1 A = 0 1 1 1 A = 1 A
27 14 16 24 div2negd A A = 1 ¬ 1 = A 1 1 A = 1 1 A
28 27 adantr A A = 1 ¬ 1 = A A 1 A = 0 1 1 A = 1 1 A
29 15 negcld A A = 1 ¬ 1 = A A
30 29 16 24 cjdivd A A = 1 ¬ 1 = A A 1 A = A 1 A
31 15 cjnegd A A = 1 ¬ 1 = A A = A
32 fveq2 A = 0 A = 0
33 abs0 0 = 0
34 32 33 eqtrdi A = 0 A = 0
35 eqtr2 A = 1 A = 0 1 = 0
36 34 35 sylan2 A = 1 A = 0 1 = 0
37 ax-1ne0 1 0
38 neneq 1 0 ¬ 1 = 0
39 37 38 mp1i A = 1 A = 0 ¬ 1 = 0
40 36 39 pm2.65da A = 1 ¬ A = 0
41 40 adantl A A = 1 ¬ A = 0
42 df-ne A 0 ¬ A = 0
43 oveq1 A = 1 A 2 = 1 2
44 sq1 1 2 = 1
45 43 44 eqtrdi A = 1 A 2 = 1
46 45 adantl A A = 1 A 2 = 1
47 absvalsq A A 2 = A A
48 47 adantr A A = 1 A 2 = A A
49 46 48 eqtr3d A A = 1 1 = A A
50 49 3adant3 A A = 1 A 0 1 = A A
51 50 oveq1d A A = 1 A 0 1 A = A A A
52 simp1 A A = 1 A 0 A
53 52 cjcld A A = 1 A 0 A
54 simp3 A A = 1 A 0 A 0
55 53 52 54 divcan3d A A = 1 A 0 A A A = A
56 51 55 eqtrd A A = 1 A 0 1 A = A
57 42 56 syl3an3br A A = 1 ¬ A = 0 1 A = A
58 41 57 mpd3an3 A A = 1 1 A = A
59 58 eqcomd A A = 1 A = 1 A
60 59 3adant3 A A = 1 ¬ 1 = A A = 1 A
61 60 negeqd A A = 1 ¬ 1 = A A = 1 A
62 31 61 eqtrd A A = 1 ¬ 1 = A A = 1 A
63 62 oveq1d A A = 1 ¬ 1 = A A 1 A = 1 A 1 A
64 cjsub 1 A 1 A = 1 A
65 18 64 mpan A 1 A = 1 A
66 1red A 1
67 66 cjred A 1 = 1
68 67 oveq1d A 1 A = 1 A
69 65 68 eqtrd A 1 A = 1 A
70 69 adantr A A = 1 1 A = 1 A
71 59 oveq2d A A = 1 1 A = 1 1 A
72 70 71 eqtrd A A = 1 1 A = 1 1 A
73 72 3adant3 A A = 1 ¬ 1 = A 1 A = 1 1 A
74 73 oveq2d A A = 1 ¬ 1 = A 1 A 1 A = 1 A 1 1 A
75 30 63 74 3eqtrd A A = 1 ¬ 1 = A A 1 A = 1 A 1 1 A
76 40 3ad2ant2 A A = 1 ¬ 1 = A ¬ A = 0
77 76 neqned A A = 1 ¬ 1 = A A 0
78 1cnd A A 0 1
79 simpl A A 0 A
80 simpr A A 0 A 0
81 78 79 80 divnegd A A 0 1 A = 1 A
82 81 oveq1d A A 0 1 A 1 1 A = 1 A 1 1 A
83 15 77 82 syl2anc A A = 1 ¬ 1 = A 1 A 1 1 A = 1 A 1 1 A
84 14 negcld A A = 1 ¬ 1 = A 1
85 84 15 77 divcld A A = 1 ¬ 1 = A 1 A
86 15 77 reccld A A = 1 ¬ 1 = A 1 A
87 14 86 subcld A A = 1 ¬ 1 = A 1 1 A
88 16 24 cjne0d A A = 1 ¬ 1 = A 1 A 0
89 73 88 eqnetrrd A A = 1 ¬ 1 = A 1 1 A 0
90 85 87 15 89 77 divcan5d A A = 1 ¬ 1 = A A 1 A A 1 1 A = 1 A 1 1 A
91 84 15 77 divcan2d A A = 1 ¬ 1 = A A 1 A = 1
92 15 14 86 subdid A A = 1 ¬ 1 = A A 1 1 A = A 1 A 1 A
93 15 mulid1d A A = 1 ¬ 1 = A A 1 = A
94 15 77 recidd A A = 1 ¬ 1 = A A 1 A = 1
95 93 94 oveq12d A A = 1 ¬ 1 = A A 1 A 1 A = A 1
96 92 95 eqtrd A A = 1 ¬ 1 = A A 1 1 A = A 1
97 91 96 oveq12d A A = 1 ¬ 1 = A A 1 A A 1 1 A = 1 A 1
98 83 90 97 3eqtr2d A A = 1 ¬ 1 = A 1 A 1 1 A = 1 A 1
99 subcl A 1 A 1
100 99 negnegd A 1 A 1 = A 1
101 negsubdi2 A 1 A 1 = 1 A
102 101 negeqd A 1 A 1 = 1 A
103 100 102 eqtr3d A 1 A 1 = 1 A
104 15 14 103 syl2anc A A = 1 ¬ 1 = A A 1 = 1 A
105 104 oveq2d A A = 1 ¬ 1 = A 1 A 1 = 1 1 A
106 75 98 105 3eqtrd A A = 1 ¬ 1 = A A 1 A = 1 1 A
107 106 adantr A A = 1 ¬ 1 = A A 1 A = 0 A 1 A = 1 1 A
108 29 16 24 divcld A A = 1 ¬ 1 = A A 1 A
109 108 adantr A A = 1 ¬ 1 = A A 1 A = 0 A 1 A
110 simpr A A = 1 ¬ 1 = A A 1 A = 0 A 1 A = 0
111 109 110 reim0bd A A = 1 ¬ 1 = A A 1 A = 0 A 1 A
112 111 cjred A A = 1 ¬ 1 = A A 1 A = 0 A 1 A = A 1 A
113 112 111 eqeltrd A A = 1 ¬ 1 = A A 1 A = 0 A 1 A
114 107 113 eqeltrrd A A = 1 ¬ 1 = A A 1 A = 0 1 1 A
115 28 114 eqeltrrd A A = 1 ¬ 1 = A A 1 A = 0 1 1 A
116 16 24 recne0d A A = 1 ¬ 1 = A 1 1 A 0
117 116 adantr A A = 1 ¬ 1 = A A 1 A = 0 1 1 A 0
118 115 117 rereccld A A = 1 ¬ 1 = A A 1 A = 0 1 1 1 A
119 26 118 eqeltrrd A A = 1 ¬ 1 = A A 1 A = 0 1 A
120 1red A A = 1 ¬ 1 = A A 1 A = 0 1
121 119 120 resubcld A A = 1 ¬ 1 = A A 1 A = 0 1 - A - 1
122 13 121 eqeltrrd A A = 1 ¬ 1 = A A 1 A = 0 A
123 2 122 negrebd A A = 1 ¬ 1 = A A 1 A = 0 A
124 123 absord A A = 1 ¬ 1 = A A 1 A = 0 A = A A = A
125 eqeq1 A = 1 A = A 1 = A
126 125 biimpd A = 1 A = A 1 = A
127 eqeq1 A = 1 A = A 1 = A
128 127 biimpd A = 1 A = A 1 = A
129 126 128 orim12d A = 1 A = A A = A 1 = A 1 = A
130 7 124 129 sylc A A = 1 ¬ 1 = A A 1 A = 0 1 = A 1 = A
131 130 ord A A = 1 ¬ 1 = A A 1 A = 0 ¬ 1 = A 1 = A
132 6 131 mpd A A = 1 ¬ 1 = A A 1 A = 0 1 = A
133 132 5 eqeltrrd A A = 1 ¬ 1 = A A 1 A = 0 A +
134 5 133 rpaddcld A A = 1 ¬ 1 = A A 1 A = 0 1 + A +
135 3 134 eqeltrrd A A = 1 ¬ 1 = A A 1 A = 0 1 A +
136 135 relogcld A A = 1 ¬ 1 = A A 1 A = 0 log 1 A
137 136 reim0d A A = 1 ¬ 1 = A A 1 A = 0 log 1 A = 0
138 133 135 rpdivcld A A = 1 ¬ 1 = A A 1 A = 0 A 1 A +
139 138 relogcld A A = 1 ¬ 1 = A A 1 A = 0 log A 1 A
140 139 reim0d A A = 1 ¬ 1 = A A 1 A = 0 log A 1 A = 0
141 137 140 eqtr4d A A = 1 ¬ 1 = A A 1 A = 0 log 1 A = log A 1 A
142 16 24 logcld A A = 1 ¬ 1 = A log 1 A
143 142 adantr A A = 1 ¬ 1 = A A 1 A 0 log 1 A
144 143 imcld A A = 1 ¬ 1 = A A 1 A 0 log 1 A
145 144 recnd A A = 1 ¬ 1 = A A 1 A 0 log 1 A
146 108 adantr A A = 1 ¬ 1 = A A 1 A 0 A 1 A
147 15 77 negne0d A A = 1 ¬ 1 = A A 0
148 29 16 147 24 divne0d A A = 1 ¬ 1 = A A 1 A 0
149 148 adantr A A = 1 ¬ 1 = A A 1 A 0 A 1 A 0
150 146 149 logcld A A = 1 ¬ 1 = A A 1 A 0 log A 1 A
151 150 imcld A A = 1 ¬ 1 = A A 1 A 0 log A 1 A
152 151 recnd A A = 1 ¬ 1 = A A 1 A 0 log A 1 A
153 106 fveq2d A A = 1 ¬ 1 = A log A 1 A = log 1 1 A
154 153 adantr A A = 1 ¬ 1 = A A 1 A 0 log A 1 A = log 1 1 A
155 logcj A 1 A A 1 A 0 log A 1 A = log A 1 A
156 108 155 sylan A A = 1 ¬ 1 = A A 1 A 0 log A 1 A = log A 1 A
157 16 24 reccld A A = 1 ¬ 1 = A 1 1 A
158 157 116 logcld A A = 1 ¬ 1 = A log 1 1 A
159 158 negnegd A A = 1 ¬ 1 = A log 1 1 A = log 1 1 A
160 isosctrlem1 A A = 1 ¬ 1 = A log 1 A π
161 logrec 1 A 1 A 0 log 1 A π log 1 A = log 1 1 A
162 16 24 160 161 syl3anc A A = 1 ¬ 1 = A log 1 A = log 1 1 A
163 162 negeqd A A = 1 ¬ 1 = A log 1 A = log 1 1 A
164 27 fveq2d A A = 1 ¬ 1 = A log 1 1 A = log 1 1 A
165 159 163 164 3eqtr4rd A A = 1 ¬ 1 = A log 1 1 A = log 1 A
166 165 adantr A A = 1 ¬ 1 = A A 1 A 0 log 1 1 A = log 1 A
167 154 156 166 3eqtr3rd A A = 1 ¬ 1 = A A 1 A 0 log 1 A = log A 1 A
168 167 fveq2d A A = 1 ¬ 1 = A A 1 A 0 log 1 A = log A 1 A
169 143 imnegd A A = 1 ¬ 1 = A A 1 A 0 log 1 A = log 1 A
170 150 imcjd A A = 1 ¬ 1 = A A 1 A 0 log A 1 A = log A 1 A
171 168 169 170 3eqtr3d A A = 1 ¬ 1 = A A 1 A 0 log 1 A = log A 1 A
172 145 152 171 neg11d A A = 1 ¬ 1 = A A 1 A 0 log 1 A = log A 1 A
173 141 172 pm2.61dane A A = 1 ¬ 1 = A log 1 A = log A 1 A