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Metamath Proof Explorer

Theorem nnpcan

Description: Cancellation law for subtraction: ((a-b)-c)+b = a-c holds for complex numbers a,b,c. (Contributed by Alexander van der Vekens, 24-Mar-2018)

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

Proof

Step Hyp Ref Expression
1 subcl 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A - B ) e. CC )
2 1 3adant3 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - B ) e. CC )
3 addsub 
 |-  ( ( ( A - B ) e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( A - B ) + B ) - C ) = ( ( ( A - B ) - C ) + B ) )
4 3 eqcomd 
 |-  ( ( ( A - B ) e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( A - B ) - C ) + B ) = ( ( ( A - B ) + B ) - C ) )
5 2 4 syld3an1 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( A - B ) - C ) + B ) = ( ( ( A - B ) + B ) - C ) )
6 npcan 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( A - B ) + B ) = A )
7 6 3adant3 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - B ) + B ) = A )
8 7 oveq1d 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( A - B ) + B ) - C ) = ( A - C ) )
9 5 8 eqtrd 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( ( A - B ) - C ) + B ) = ( A - C ) )