Theorem baibd

Description: Move conjunction outside of biconditional. (Contributed by Mario Carneiro, 11-Sep-2015)

	Ref	Expression
Hypothesis	baibd.1	|- ( ph -> ( ps <-> ( ch /\ th ) ) )
Assertion	baibd	|- ( ( ph /\ ch ) -> ( ps <-> th ) )

Proof

Step	Hyp	Ref	Expression
1		baibd.1	|- ( ph -> ( ps <-> ( ch /\ th ) ) )
2		ibar	|- ( ch -> ( th <-> ( ch /\ th ) ) )
3	2	bicomd	|- ( ch -> ( ( ch /\ th ) <-> th ) )
4	1 3	sylan9bb	|- ( ( ph /\ ch ) -> ( ps <-> th ) )