Metamath Proof Explorer


Theorem nnncan1d

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 nnncan1d
|- ( ph -> ( ( A - B ) - ( A - C ) ) = ( C - 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 nnncan1
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - B ) - ( A - C ) ) = ( C - B ) )
5 1 2 3 4 syl3anc
 |-  ( ph -> ( ( A - B ) - ( A - C ) ) = ( C - B ) )