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Metamath Proof Explorer

Theorem nnncan1

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005) (Proof shortened by Andrew Salmon, 19-Nov-2011)

Ref Expression
Assertion nnncan1 
|- ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - B ) - ( A - C ) ) = ( C - B ) )

Proof

Step Hyp Ref Expression
1 subcl 
 |-  ( ( A e. CC /\ C e. CC ) -> ( A - C ) e. CC )
2 1 3adant2 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - C ) e. CC )
3 sub32 
 |-  ( ( A e. CC /\ B e. CC /\ ( A - C ) e. CC ) -> ( ( A - B ) - ( A - C ) ) = ( ( A - ( A - C ) ) - B ) )
4 2 3 syld3an3 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - B ) - ( A - C ) ) = ( ( A - ( A - C ) ) - B ) )
5 nncan 
 |-  ( ( A e. CC /\ C e. CC ) -> ( A - ( A - C ) ) = C )
6 5 3adant2 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( A - ( A - C ) ) = C )
7 6 oveq1d 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - ( A - C ) ) - B ) = ( C - B ) )
8 4 7 eqtrd 
 |-  ( ( A e. CC /\ B e. CC /\ C e. CC ) -> ( ( A - B ) - ( A - C ) ) = ( C - B ) )