Theorem baibr 888

Description: Move conjunction outside of biconditional. (Contributed by NM, 11-Jul-1994.)

Hypothesis

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

Assertion

Ref	Expression
baibr	|- ( ps -> ( ch <-> ph ) )

Proof of Theorem baibr

Step	Hyp	Ref	Expression
1		baib.1	. . 3 |- ( ph <-> ( ps /\ ch ) )
2	1	baib 887	. 2 |- ( ps -> ( ph <-> ch ) )
3	2	bicomd 140	1 |- ( ps -> ( ch <-> ph ) )

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

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

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  rbaibr  890  pm5.44  893  exmoeu2  2023  r19.9rmv  3420  dfopg  3669  brinxp  4567  elioo5  9609  prmind2  11647