Metamath Proof Explorer


Theorem nppcand

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 )
Assertion nppcand
|- ( ph -> ( ( ( A - B ) + C ) + B ) = ( A + 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 nppcan
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( A - B ) + C ) + B ) = ( A + C ) )
5 1 2 3 4 syl3anc
 |-  ( ph -> ( ( ( A - B ) + C ) + B ) = ( A + C ) )