Theorem recdiv2d

Description: Division into a reciprocal. (Contributed by Mario Carneiro, 27-May-2016)

	Ref	Expression
Hypotheses	div1d.1	|- ( ph -> A e. CC )
	divcld.2	|- ( ph -> B e. CC )
	divne0d.3	|- ( ph -> A =/= 0 )
	divne0d.4	|- ( ph -> B =/= 0 )
Assertion	recdiv2d	|- ( ph -> ( ( 1 / A ) / B ) = ( 1 / ( A x. B ) ) )

Proof

Step	Hyp	Ref	Expression
1		div1d.1	|- ( ph -> A e. CC )
2		divcld.2	|- ( ph -> B e. CC )
3		divne0d.3	|- ( ph -> A =/= 0 )
4		divne0d.4	|- ( ph -> B =/= 0 )
5		recdiv2	|- ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( 1 / A ) / B ) = ( 1 / ( A x. B ) ) )
6	1 3 2 4 5	syl22anc	|- ( ph -> ( ( 1 / A ) / B ) = ( 1 / ( A x. B ) ) )