Theorem baibd 925

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  6931  eluz  9661  elicc4  10062  s111  11085  divalgmodcl  12239  eqglact  13561  eqgid  13562  iscrng2  13777  issubrg3  14009  iscld2  14576  cncnp2m  14703  cnnei  14704  reopnap  15018  cnlimc  15144  2omap  15932