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Metamath Proof Explorer

Theorem subneg

Description: Relationship between subtraction and negative. (Contributed by NM, 10-May-2004) (Revised by Mario Carneiro, 27-May-2016)

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

Proof

Step Hyp Ref Expression
1 df-neg 
 |-  -u B = ( 0 - B )
2 1 oveq2i 
 |-  ( A - -u B ) = ( A - ( 0 - B ) )
3 0cn 
 |-  0 e. CC
4 subsub 
 |-  ( ( A e. CC /\ 0 e. CC /\ B e. CC ) -> ( A - ( 0 - B ) ) = ( ( A - 0 ) + B ) )
5 3 4 mp3an2 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A - ( 0 - B ) ) = ( ( A - 0 ) + B ) )
6 2 5 eqtrid 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A - -u B ) = ( ( A - 0 ) + B ) )
7 subid1 
 |-  ( A e. CC -> ( A - 0 ) = A )
8 7 adantr 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A - 0 ) = A )
9 8 oveq1d 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( A - 0 ) + B ) = ( A + B ) )
10 6 9 eqtrd 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A - -u B ) = ( A + B ) )