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 bitrdi
 |-  ( ( 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 ) )