Metamath Proof Explorer


Theorem subcand

Description: Cancellation law for subtraction. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses negidd.1
|- ( ph -> A e. CC )
pncand.2
|- ( ph -> B e. CC )
subaddd.3
|- ( ph -> C e. CC )
subcand.4
|- ( ph -> ( A - B ) = ( A - C ) )
Assertion subcand
|- ( ph -> B = C )

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 subcand.4
 |-  ( ph -> ( A - B ) = ( A - C ) )
5 subcan
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - B ) = ( A - C ) <-> B = C ) )
6 1 2 3 5 syl3anc
 |-  ( ph -> ( ( A - B ) = ( A - C ) <-> B = C ) )
7 4 6 mpbid
 |-  ( ph -> B = C )