Theorem neg2subd

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

	Ref	Expression
Hypotheses	negidd.1	|- ( ph -> A e. CC )
	pncand.2	|- ( ph -> B e. CC )
Assertion	neg2subd	|- ( ph -> ( -u A - -u B ) = ( B - A ) )

Proof

Step	Hyp	Ref	Expression
1		negidd.1	|- ( ph -> A e. CC )
2		pncand.2	|- ( ph -> B e. CC )
3		neg2sub	|- ( ( A e. CC /\ B e. CC ) -> ( -u A - -u B ) = ( B - A ) )
4	1 2 3	syl2anc	|- ( ph -> ( -u A - -u B ) = ( B - A ) )