Theorem subreci

Description: Subtraction of reciprocals. (Contributed by Scott Fenton, 9-Jan-2017)

	Ref	Expression
Hypotheses	subreci.1	|- A e. CC
	subreci.2	|- B e. CC
	subreci.3	|- A =/= 0
	subreci.4	|- B =/= 0
Assertion	subreci	|- ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) )

Proof

Step	Hyp	Ref	Expression
1		subreci.1	|- A e. CC
2		subreci.2	|- B e. CC
3		subreci.3	|- A =/= 0
4		subreci.4	|- B =/= 0
5		subrec	|- ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) ) )
6	1 3 2 4 5	mp4an	|- ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) )