Theorem rbaibr 874

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 437	. . 3 |- ((ps /\ ch) <-> (ch /\ ps))
3	1, 2	bitri 240	. 2 |- (ph <-> (ch /\ ps))
4	3	baibr 872	1 |- (ch -> (ps <-> ph))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-an 360

This theorem is referenced by:  ssunsn2  3865