Theorem baib

Description: Move conjunction outside of biconditional. (Contributed by NM, 13-May-1999)

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

Proof

Step	Hyp	Ref	Expression
1		baib.1	|- ( ph <-> ( ps /\ ch ) )
2		ibar	|- ( ps -> ( ch <-> ( ps /\ ch ) ) )
3	2 1	syl6rbbr	|- ( ps -> ( ph <-> ch ) )