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Metamath Proof Explorer

Theorem nppcan

Description: Cancellation law for subtraction. (Contributed by NM, 1-Sep-2005)

Ref Expression
Assertion nppcan 
|- ( ( 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 simp3 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> C e. CC )
4 simp2 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> B e. CC )
5 2 3 4 add32d 
 |-  ( ( 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 oveq1d 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( ( A - B ) + B ) + C ) = ( A + C ) )
8 7 3adant3 
 |-  ( ( 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 ) )