Theorem subrecd

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

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

Proof

Step	Hyp	Ref	Expression
1		subrecd.1	|- ( ph -> A e. CC )
2		subrecd.2	|- ( ph -> B e. CC )
3		subrecd.3	|- ( ph -> A =/= 0 )
4		subrecd.4	|- ( ph -> 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	syl22anc	|- ( ph -> ( ( 1 / A ) - ( 1 / B ) ) = ( ( B - A ) / ( A x. B ) ) )