Metamath Proof Explorer


Theorem divne1d

Description: If two complex numbers are unequal, their quotient is not one. Contrapositive of diveq1d . (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses div1d.1
|- ( ph -> A e. CC )
divcld.2
|- ( ph -> B e. CC )
divcld.3
|- ( ph -> B =/= 0 )
divne1d.4
|- ( ph -> A =/= B )
Assertion divne1d
|- ( ph -> ( A / B ) =/= 1 )

Proof

Step Hyp Ref Expression
1 div1d.1
 |-  ( ph -> A e. CC )
2 divcld.2
 |-  ( ph -> B e. CC )
3 divcld.3
 |-  ( ph -> B =/= 0 )
4 divne1d.4
 |-  ( ph -> A =/= B )
5 1 2 3 diveq1ad
 |-  ( ph -> ( ( A / B ) = 1 <-> A = B ) )
6 5 necon3bid
 |-  ( ph -> ( ( A / B ) =/= 1 <-> A =/= B ) )
7 4 6 mpbird
 |-  ( ph -> ( A / B ) =/= 1 )