Theorem baibd 908

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 299	. . 3 |- ( ch -> ( th <-> ( ch /\ th ) ) )
3	2	bicomd 140	. 2 |- ( ch -> ( ( ch /\ th ) <-> th ) )
4	1, 3	sylan9bb 457	1 |- ( ( ph /\ ch ) -> ( ps <-> th ) )

Syntax hints:   -> wi 4   /\ wa 103   <-> wb 104

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  eluz  9332  elicc4  9716  divalgmodcl  11614  iscld2  12262  cncnp2m  12389  cnnei  12390  reopnap  12696  cnlimc  12799