Metamath Proof Explorer

Theorem div11d

Description: One-to-one relationship for division. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 
|- ( ph -> A e. CC )
divcld.2 
|- ( ph -> B e. CC )
divmuld.3 
|- ( ph -> C e. CC )
divassd.4 
|- ( ph -> C =/= 0 )
div11d.5 
|- ( ph -> ( A / C ) = ( B / C ) )
Assertion div11d 
|- ( ph -> A = B )

Proof

Step Hyp Ref Expression
1 div1d.1 
 |-  ( ph -> A e. CC )
2 divcld.2 
 |-  ( ph -> B e. CC )
3 divmuld.3 
 |-  ( ph -> C e. CC )
4 divassd.4 
 |-  ( ph -> C =/= 0 )
5 div11d.5 
 |-  ( ph -> ( A / C ) = ( B / C ) )
6 div11 
 |-  ( ( A e. CC /\ B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( A / C ) = ( B / C ) <-> A = B ) )
7 1 2 3 4 6 syl112anc 
 |-  ( ph -> ( ( A / C ) = ( B / C ) <-> A = B ) )
8 5 7 mpbid 
 |-  ( ph -> A = B )