Theorem rbaibr 912

Description: Move conjunction outside of biconditional. (Contributed by Mario Carneiro, 11-Sep-2015.)

Hypothesis

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

Assertion

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

Proof of Theorem rbaibr

Step	Hyp	Ref	Expression
1		baib.1	. . 3 |- ( ph <-> ( ps /\ ch ) )
2		ancom 264	. . 3 |- ( ( ps /\ ch ) <-> ( ch /\ ps ) )
3	1, 2	bitri 183	. 2 |- ( ph <-> ( ch /\ ps ) )
4	3	baibr 910	1 |- ( ch -> ( ps <-> ph ) )

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: (None)