Metamath Proof Explorer


Theorem subcan2i

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

Ref Expression
Hypotheses negidi.1
|- A e. CC
pncan3i.2
|- B e. CC
subadd.3
|- C e. CC
Assertion subcan2i
|- ( ( A - C ) = ( B - C ) <-> A = B )

Proof

Step Hyp Ref Expression
1 negidi.1
 |-  A e. CC
2 pncan3i.2
 |-  B e. CC
3 subadd.3
 |-  C e. CC
4 subcan2
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - C ) = ( B - C ) <-> A = B ) )
5 1 2 3 4 mp3an
 |-  ( ( A - C ) = ( B - C ) <-> A = B )