Theorem divcan6d

Description: Cancellation of inverted fractions. (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	divcan6d	|- ( ph -> ( ( A / B ) x. ( B / A ) ) = 1 )

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		divcan6	|- ( ( ( A e. CC /\ A =/= 0 ) /\ ( B e. CC /\ B =/= 0 ) ) -> ( ( A / B ) x. ( B / A ) ) = 1 )
6	1 3 2 4 5	syl22anc	|- ( ph -> ( ( A / B ) x. ( B / A ) ) = 1 )