Metamath Proof Explorer


Theorem subcan2ad

Description: Cancellation law for subtraction. Deduction form of subcan2 . Generalization of subcan2d . (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses negidd.1
|- ( ph -> A e. CC )
pncand.2
|- ( ph -> B e. CC )
subaddd.3
|- ( ph -> C e. CC )
Assertion subcan2ad
|- ( ph -> ( ( A - C ) = ( B - C ) <-> A = B ) )

Proof

Step Hyp Ref Expression
1 negidd.1
 |-  ( ph -> A e. CC )
2 pncand.2
 |-  ( ph -> B e. CC )
3 subaddd.3
 |-  ( ph -> 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 syl3anc
 |-  ( ph -> ( ( A - C ) = ( B - C ) <-> A = B ) )