Metamath Proof Explorer

Theorem div11

Description: One-to-one relationship for division. (Contributed by NM, 20-Apr-2006) (Proof shortened by Mario Carneiro, 27-May-2016) (Proof shortened by SN, 9-Jul-2025)

Ref Expression
Assertion div11 
|- ( ( A e. CC /\ B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( A / C ) = ( B / C ) <-> A = B ) )

Proof

Step Hyp Ref Expression
1 divcl 
 |-  ( ( B e. CC /\ C e. CC /\ C =/= 0 ) -> ( B / C ) e. CC )
2 1 3expb 
 |-  ( ( B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( B / C ) e. CC )
3 2 3adant1 
 |-  ( ( A e. CC /\ B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( B / C ) e. CC )
4 divmul3 
 |-  ( ( A e. CC /\ ( B / C ) e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( A / C ) = ( B / C ) <-> A = ( ( B / C ) x. C ) ) )
5 3 4 syld3an2 
 |-  ( ( A e. CC /\ B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( A / C ) = ( B / C ) <-> A = ( ( B / C ) x. C ) ) )
6 divcan1 
 |-  ( ( B e. CC /\ C e. CC /\ C =/= 0 ) -> ( ( B / C ) x. C ) = B )
7 6 3expb 
 |-  ( ( B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( B / C ) x. C ) = B )
8 7 3adant1 
 |-  ( ( A e. CC /\ B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( B / C ) x. C ) = B )
9 8 eqeq2d 
 |-  ( ( A e. CC /\ B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( A = ( ( B / C ) x. C ) <-> A = B ) )
10 5 9 bitrd 
 |-  ( ( A e. CC /\ B e. CC /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( A / C ) = ( B / C ) <-> A = B ) )