Metamath Proof Explorer


Theorem angpieqvdlem2

Description: Equivalence used in angpieqvd . (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses angpieqvd.angdef
|- F = ( x e. ( CC \ { 0 } ) , y e. ( CC \ { 0 } ) |-> ( Im ` ( log ` ( y / x ) ) ) )
angpieqvd.A
|- ( ph -> A e. CC )
angpieqvd.B
|- ( ph -> B e. CC )
angpieqvd.C
|- ( ph -> C e. CC )
angpieqvd.AneB
|- ( ph -> A =/= B )
angpieqvd.BneC
|- ( ph -> B =/= C )
Assertion angpieqvdlem2
|- ( ph -> ( -u ( ( C - B ) / ( A - B ) ) e. RR+ <-> ( ( A - B ) F ( C - B ) ) = _pi ) )

Proof

Step Hyp Ref Expression
1 angpieqvd.angdef
 |-  F = ( x e. ( CC \ { 0 } ) , y e. ( CC \ { 0 } ) |-> ( Im ` ( log ` ( y / x ) ) ) )
2 angpieqvd.A
 |-  ( ph -> A e. CC )
3 angpieqvd.B
 |-  ( ph -> B e. CC )
4 angpieqvd.C
 |-  ( ph -> C e. CC )
5 angpieqvd.AneB
 |-  ( ph -> A =/= B )
6 angpieqvd.BneC
 |-  ( ph -> B =/= C )
7 4 3 subcld
 |-  ( ph -> ( C - B ) e. CC )
8 2 3 subcld
 |-  ( ph -> ( A - B ) e. CC )
9 2 3 5 subne0d
 |-  ( ph -> ( A - B ) =/= 0 )
10 7 8 9 divcld
 |-  ( ph -> ( ( C - B ) / ( A - B ) ) e. CC )
11 6 necomd
 |-  ( ph -> C =/= B )
12 4 3 11 subne0d
 |-  ( ph -> ( C - B ) =/= 0 )
13 7 8 12 9 divne0d
 |-  ( ph -> ( ( C - B ) / ( A - B ) ) =/= 0 )
14 lognegb
 |-  ( ( ( ( C - B ) / ( A - B ) ) e. CC /\ ( ( C - B ) / ( A - B ) ) =/= 0 ) -> ( -u ( ( C - B ) / ( A - B ) ) e. RR+ <-> ( Im ` ( log ` ( ( C - B ) / ( A - B ) ) ) ) = _pi ) )
15 10 13 14 syl2anc
 |-  ( ph -> ( -u ( ( C - B ) / ( A - B ) ) e. RR+ <-> ( Im ` ( log ` ( ( C - B ) / ( A - B ) ) ) ) = _pi ) )
16 1 8 9 7 12 angvald
 |-  ( ph -> ( ( A - B ) F ( C - B ) ) = ( Im ` ( log ` ( ( C - B ) / ( A - B ) ) ) ) )
17 16 eqeq1d
 |-  ( ph -> ( ( ( A - B ) F ( C - B ) ) = _pi <-> ( Im ` ( log ` ( ( C - B ) / ( A - B ) ) ) ) = _pi ) )
18 15 17 bitr4d
 |-  ( ph -> ( -u ( ( C - B ) / ( A - B ) ) e. RR+ <-> ( ( A - B ) F ( C - B ) ) = _pi ) )