Metamath Proof Explorer


Theorem isosctrlem3

Description: Lemma for isosctr . Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017)

Ref Expression
Hypothesis isosctrlem3.1 F = x 0 , y 0 log y x
Assertion isosctrlem3 A B A 0 B 0 A B A = B A F B A = A B F B

Proof

Step Hyp Ref Expression
1 isosctrlem3.1 F = x 0 , y 0 log y x
2 simp1l A B A 0 B 0 A B A = B A
3 simp21 A B A 0 B 0 A B A = B A 0
4 simp1r A B A 0 B 0 A B A = B B
5 2 4 subcld A B A 0 B 0 A B A = B A B
6 simp23 A B A 0 B 0 A B A = B A B
7 2 4 6 subne0d A B A 0 B 0 A B A = B A B 0
8 1 angneg A A 0 A B A B 0 A F A B = A F A B
9 2 3 5 7 8 syl22anc A B A 0 B 0 A B A = B A F A B = A F A B
10 2 4 negsubdi2d A B A 0 B 0 A B A = B A B = B A
11 10 oveq2d A B A 0 B 0 A B A = B A F A B = A F B A
12 1cnd A B A 0 B 0 A B A = B 1
13 ax-1ne0 1 0
14 13 a1i A B A 0 B 0 A B A = B 1 0
15 4 2 3 divcld A B A 0 B 0 A B A = B B A
16 12 15 subcld A B A 0 B 0 A B A = B 1 B A
17 6 necomd A B A 0 B 0 A B A = B B A
18 4 2 3 17 divne1d A B A 0 B 0 A B A = B B A 1
19 18 necomd A B A 0 B 0 A B A = B 1 B A
20 12 15 19 subne0d A B A 0 B 0 A B A = B 1 B A 0
21 1 12 14 16 20 angvald A B A 0 B 0 A B A = B 1 F 1 B A = log 1 B A 1
22 16 div1d A B A 0 B 0 A B A = B 1 B A 1 = 1 B A
23 22 fveq2d A B A 0 B 0 A B A = B log 1 B A 1 = log 1 B A
24 23 fveq2d A B A 0 B 0 A B A = B log 1 B A 1 = log 1 B A
25 4 2 3 absdivd A B A 0 B 0 A B A = B B A = B A
26 simp3 A B A 0 B 0 A B A = B A = B
27 26 eqcomd A B A 0 B 0 A B A = B B = A
28 27 oveq1d A B A 0 B 0 A B A = B B A = A A
29 2 abscld A B A 0 B 0 A B A = B A
30 29 recnd A B A 0 B 0 A B A = B A
31 2 3 absne0d A B A 0 B 0 A B A = B A 0
32 30 31 dividd A B A 0 B 0 A B A = B A A = 1
33 25 28 32 3eqtrd A B A 0 B 0 A B A = B B A = 1
34 19 neneqd A B A 0 B 0 A B A = B ¬ 1 = B A
35 isosctrlem2 B A B A = 1 ¬ 1 = B A log 1 B A = log B A 1 B A
36 15 33 34 35 syl3anc A B A 0 B 0 A B A = B log 1 B A = log B A 1 B A
37 15 negcld A B A 0 B 0 A B A = B B A
38 simp22 A B A 0 B 0 A B A = B B 0
39 4 2 38 3 divne0d A B A 0 B 0 A B A = B B A 0
40 15 39 negne0d A B A 0 B 0 A B A = B B A 0
41 1 16 20 37 40 angvald A B A 0 B 0 A B A = B 1 B A F B A = log B A 1 B A
42 36 41 eqtr4d A B A 0 B 0 A B A = B log 1 B A = 1 B A F B A
43 21 24 42 3eqtrd A B A 0 B 0 A B A = B 1 F 1 B A = 1 B A F B A
44 2 mulid1d A B A 0 B 0 A B A = B A 1 = A
45 2 12 15 subdid A B A 0 B 0 A B A = B A 1 B A = A 1 A B A
46 4 2 3 divcan2d A B A 0 B 0 A B A = B A B A = B
47 44 46 oveq12d A B A 0 B 0 A B A = B A 1 A B A = A B
48 45 47 eqtrd A B A 0 B 0 A B A = B A 1 B A = A B
49 44 48 oveq12d A B A 0 B 0 A B A = B A 1 F A 1 B A = A F A B
50 1 angcan 1 1 0 1 B A 1 B A 0 A A 0 A 1 F A 1 B A = 1 F 1 B A
51 12 14 16 20 2 3 50 syl222anc A B A 0 B 0 A B A = B A 1 F A 1 B A = 1 F 1 B A
52 49 51 eqtr3d A B A 0 B 0 A B A = B A F A B = 1 F 1 B A
53 2 15 mulneg2d A B A 0 B 0 A B A = B A B A = A B A
54 46 negeqd A B A 0 B 0 A B A = B A B A = B
55 53 54 eqtrd A B A 0 B 0 A B A = B A B A = B
56 48 55 oveq12d A B A 0 B 0 A B A = B A 1 B A F A B A = A B F B
57 1 angcan 1 B A 1 B A 0 B A B A 0 A A 0 A 1 B A F A B A = 1 B A F B A
58 16 20 37 40 2 3 57 syl222anc A B A 0 B 0 A B A = B A 1 B A F A B A = 1 B A F B A
59 56 58 eqtr3d A B A 0 B 0 A B A = B A B F B = 1 B A F B A
60 43 52 59 3eqtr4d A B A 0 B 0 A B A = B A F A B = A B F B
61 9 11 60 3eqtr3d A B A 0 B 0 A B A = B A F B A = A B F B