Metamath Proof Explorer

Theorem subcan2

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005)

Ref Expression
Assertion subcan2 
|- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - C ) = ( B - C ) <-> A = B ) )

Proof

Step Hyp Ref Expression
1 simp1 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> A e. CC )
2 simp3 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> C e. CC )
3 subcl 
 |-  ( ( B e. CC /\ C e. CC ) -> ( B - C ) e. CC )
4 3 3adant1 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( B - C ) e. CC )
5 subadd2 
 |-  ( ( A e. CC /\ C e. CC /\ ( B - C ) e. CC ) -> ( ( A - C ) = ( B - C ) <-> ( ( B - C ) + C ) = A ) )
6 1 2 4 5 syl3anc 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - C ) = ( B - C ) <-> ( ( B - C ) + C ) = A ) )
7 npcan 
 |-  ( ( B e. CC /\ C e. CC ) -> ( ( B - C ) + C ) = B )
8 7 3adant1 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( B - C ) + C ) = B )
9 8 eqeq1d 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( B - C ) + C ) = A <-> B = A ) )
10 eqcom 
 |-  ( B = A <-> A = B )
11 9 10 syl6bb 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( B - C ) + C ) = A <-> A = B ) )
12 6 11 bitrd 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - C ) = ( B - C ) <-> A = B ) )