Metamath Proof Explorer


Theorem atandm2

Description: This form of atandm is a bit more useful for showing that the logarithms in df-atan are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015)

Ref Expression
Assertion atandm2 A dom arctan A 1 i A 0 1 + i A 0

Proof

Step Hyp Ref Expression
1 atandm A dom arctan A A i A i
2 3anass A 1 i A 0 1 + i A 0 A 1 i A 0 1 + i A 0
3 ax-1cn 1
4 ax-icn i
5 mulcl i A i A
6 4 5 mpan A i A
7 subeq0 1 i A 1 i A = 0 1 = i A
8 3 6 7 sylancr A 1 i A = 0 1 = i A
9 4 4 mulneg2i i i = i i
10 ixi i i = 1
11 10 negeqi i i = -1
12 negneg1e1 -1 = 1
13 9 11 12 3eqtri i i = 1
14 13 eqeq2i i A = i i i A = 1
15 eqcom i A = 1 1 = i A
16 14 15 bitri i A = i i 1 = i A
17 8 16 bitr4di A 1 i A = 0 i A = i i
18 id A A
19 4 negcli i
20 19 a1i A i
21 4 a1i A i
22 ine0 i 0
23 22 a1i A i 0
24 18 20 21 23 mulcand A i A = i i A = i
25 17 24 bitrd A 1 i A = 0 A = i
26 25 necon3bid A 1 i A 0 A i
27 addcom 1 i A 1 + i A = i A + 1
28 3 6 27 sylancr A 1 + i A = i A + 1
29 subneg i A 1 i A -1 = i A + 1
30 6 3 29 sylancl A i A -1 = i A + 1
31 28 30 eqtr4d A 1 + i A = i A -1
32 31 eqeq1d A 1 + i A = 0 i A -1 = 0
33 3 negcli 1
34 subeq0 i A 1 i A -1 = 0 i A = 1
35 6 33 34 sylancl A i A -1 = 0 i A = 1
36 32 35 bitrd A 1 + i A = 0 i A = 1
37 10 eqeq2i i A = i i i A = 1
38 36 37 bitr4di A 1 + i A = 0 i A = i i
39 18 21 21 23 mulcand A i A = i i A = i
40 38 39 bitrd A 1 + i A = 0 A = i
41 40 necon3bid A 1 + i A 0 A i
42 26 41 anbi12d A 1 i A 0 1 + i A 0 A i A i
43 42 pm5.32i A 1 i A 0 1 + i A 0 A A i A i
44 3anass A A i A i A A i A i
45 43 44 bitr4i A 1 i A 0 1 + i A 0 A A i A i
46 2 45 bitri A 1 i A 0 1 + i A 0 A A i A i
47 1 46 bitr4i A dom arctan A 1 i A 0 1 + i A 0