Theorem absdivd

Description: Absolute value distributes over division. (Contributed by Mario Carneiro, 29-May-2016)

	Ref	Expression
Hypotheses	abscld.1	|- ( ph -> A e. CC )
	abssubd.2	|- ( ph -> B e. CC )
	absdivd.2	|- ( ph -> B =/= 0 )
Assertion	absdivd	|- ( ph -> ( abs ` ( A / B ) ) = ( ( abs ` A ) / ( abs ` B ) ) )

Proof

Step	Hyp	Ref	Expression
1		abscld.1	|- ( ph -> A e. CC )
2		abssubd.2	|- ( ph -> B e. CC )
3		absdivd.2	|- ( ph -> B =/= 0 )
4		absdiv	|- ( ( A e. CC /\ B e. CC /\ B =/= 0 ) -> ( abs ` ( A / B ) ) = ( ( abs ` A ) / ( abs ` B ) ) )
5	1 2 3 4	syl3anc	|- ( ph -> ( abs ` ( A / B ) ) = ( ( abs ` A ) / ( abs ` B ) ) )