Theorem divcan7d

Description: Cancel equal divisors in a division. (Contributed by Mario Carneiro, 27-May-2016)

	Ref	Expression
Hypotheses	div1d.1	|- ( ph -> A e. CC )
	divcld.2	|- ( ph -> B e. CC )
	divmuld.3	|- ( ph -> C e. CC )
	divmuld.4	|- ( ph -> B =/= 0 )
	divdiv23d.5	|- ( ph -> C =/= 0 )
Assertion	divcan7d	|- ( ph -> ( ( A / C ) / ( B / C ) ) = ( A / B ) )

Proof

Step	Hyp	Ref	Expression
1		div1d.1	|- ( ph -> A e. CC )
2		divcld.2	|- ( ph -> B e. CC )
3		divmuld.3	|- ( ph -> C e. CC )
4		divmuld.4	|- ( ph -> B =/= 0 )
5		divdiv23d.5	|- ( ph -> C =/= 0 )
6		divcan7	|- ( ( A e. CC /\ ( B e. CC /\ B =/= 0 ) /\ ( C e. CC /\ C =/= 0 ) ) -> ( ( A / C ) / ( B / C ) ) = ( A / B ) )
7	1 2 4 3 5 6	syl122anc	|- ( ph -> ( ( A / C ) / ( B / C ) ) = ( A / B ) )