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	addid1d	|- ( ( 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 ) )