Theorem rbaib

Description: Move conjunction outside of biconditional. (Contributed by Mario Carneiro, 11-Sep-2015) (Proof shortened by Wolf Lammen, 19-Jan-2020)

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

Proof

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