Metamath Proof Explorer

Theorem npncan2

Description: Cancellation law for subtraction. (Contributed by Scott Fenton, 21-Jun-2013)

Ref Expression
Assertion npncan2 
|- ( ( A e. CC /\ B e. CC ) -> ( ( A - B ) + ( B - A ) ) = 0 )

Proof

Step Hyp Ref Expression
1 npncan 
 |-  ( ( A e. CC /\ B e. CC /\ A e. CC ) -> ( ( A - B ) + ( B - A ) ) = ( A - A ) )
2 1 3anidm13 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( A - B ) + ( B - A ) ) = ( A - A ) )
3 subid 
 |-  ( A e. CC -> ( A - A ) = 0 )
4 3 adantr 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A - A ) = 0 )
5 2 4 eqtrd 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( A - B ) + ( B - A ) ) = 0 )