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Metamath Proof Explorer

Theorem nnncan

Description: Cancellation law for subtraction. (Contributed by NM, 4-Sep-2005)

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

Proof

Step Hyp Ref Expression
1 subcl 
 |-  ( ( B e. CC /\ C e. CC ) -> ( B - C ) e. CC )
2 1 3adant1 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( B - C ) e. CC )
3 subsub4 
 |-  ( ( A e. CC /\ ( B - C ) e. CC /\ C e. CC ) -> ( ( A - ( B - C ) ) - C ) = ( A - ( ( B - C ) + C ) ) )
4 2 3 syld3an2 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - ( B - C ) ) - C ) = ( A - ( ( B - C ) + C ) ) )
5 npcan 
 |-  ( ( B e. CC /\ C e. CC ) -> ( ( B - C ) + C ) = B )
6 5 oveq2d 
 |-  ( ( B e. CC /\ C e. CC ) -> ( A - ( ( B - C ) + C ) ) = ( A - B ) )
7 6 3adant1 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - ( ( B - C ) + C ) ) = ( A - B ) )
8 4 7 eqtrd 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - ( B - C ) ) - C ) = ( A - B ) )