Theorem divrec2d

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

	Ref	Expression
Hypotheses	div1d.1	|- ( ph -> A e. CC )
	divcld.2	|- ( ph -> B e. CC )
	divcld.3	|- ( ph -> B =/= 0 )
Assertion	divrec2d	|- ( ph -> ( A / B ) = ( ( 1 / B ) x. A ) )

Proof

Step	Hyp	Ref	Expression
1		div1d.1	|- ( ph -> A e. CC )
2		divcld.2	|- ( ph -> B e. CC )
3		divcld.3	|- ( ph -> B =/= 0 )
4		divrec2	|- ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> ( A / B ) = ( ( 1 / B ) x. A ) )
5	1 2 3 4	syl3anc	|- ( ph -> ( A / B ) = ( ( 1 / B ) x. A ) )