Theorem baibd 928

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 301	. . 3 |- ( ch -> ( th <-> ( ch /\ th ) ) )
3	2	bicomd 141	. 2 |- ( ch -> ( ( ch /\ th ) <-> th ) )
4	1, 3	sylan9bb 462	1 |- ( ( ph /\ ch ) -> ( ps <-> th ) )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  pw2f1odclem  7003  eluz  9747  elicc4  10148  s111  11179  divalgmodcl  12455  eqglact  13778  eqgid  13779  iscrng2  13994  issubrg3  14227  iscld2  14794  cncnp2m  14921  cnnei  14922  reopnap  15236  cnlimc  15362  2omap  16446  pw1map  16448