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Metamath Proof Explorer

Theorem negsub

Description: Relationship between subtraction and negative. Theorem I.3 of Apostol p. 18. (Contributed by NM, 21-Jan-1997) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion negsub 
|- ( ( 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 2 a1i 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A + -u B ) = ( A + ( 0 - B ) ) )
4 0cn 
 |-  0 e. CC
5 addsubass 
 |-  ( ( A e. CC /\ 0 e. CC /\ B e. CC ) -> ( ( A + 0 ) - B ) = ( A + ( 0 - B ) ) )
6 4 5 mp3an2 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( A + 0 ) - B ) = ( A + ( 0 - B ) ) )
7 simpl 
 |-  ( ( A e. CC /\ B e. CC ) -> A e. CC )
8 7 addridd 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A + 0 ) = A )
9 8 oveq1d 
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( A + 0 ) - B ) = ( A - B ) )
10 3 6 9 3eqtr2d 
 |-  ( ( A e. CC /\ B e. CC ) -> ( A + -u B ) = ( A - B ) )