Theorem baibd 865

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

Hypothesis

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

Assertion

Ref	Expression
baibd	|- ( ( ph /\ ch ) -> ( ps <-> th ) )

Proof of Theorem baibd

Step	Hyp	Ref	Expression
1		baibd.1	. 2 |- ( ph -> ( ps <-> ( ch /\ th ) ) )
2		ibar 295	. . 3 |- ( ch -> ( th <-> ( ch /\ th ) ) )
3	2	bicomd 139	. 2 |- ( ch -> ( ( ch /\ th ) <-> th ) )
4	1, 3	sylan9bb 449	1 |- ( ( ph /\ ch ) -> ( ps <-> th ) )

Syntax hints:   -> wi 4   /\ wa 102   <-> wb 103

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  eluz  8632  elicc4  8963  divalgmodcl  10328