Theorem baibr 925

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 924	. 2 |- ( ps -> ( ph <-> ch ) )
3	2	bicomd 141	1 |- ( ps -> ( ch <-> ph ) )

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:  rbaibr  927  pm5.44  930  exmoeu2  2126  r19.9rmv  3583  dfopg  3855  brinxp  4787  infidc  7101  elioo5  10129  prmind2  12642  eulerthlemfi  12750  phisum  12763  pcelnn  12844  bj-charfundcALT  16172