Theorem expcomdg

Description: Biconditional form of expcomd . (Contributed by Alan Sare, 22-Jul-2012) (New usage is discouraged.)

	Ref	Expression
Assertion	expcomdg	|- ( ( ph -> ( ( ps /\ ch ) -> th ) ) <-> ( ph -> ( ch -> ( ps -> th ) ) ) )

Proof

Step	Hyp	Ref	Expression
1		ancomst	|- ( ( ( ps /\ ch ) -> th ) <-> ( ( ch /\ ps ) -> th ) )
2		impexp	|- ( ( ( ch /\ ps ) -> th ) <-> ( ch -> ( ps -> th ) ) )
3	1 2	bitri	|- ( ( ( ps /\ ch ) -> th ) <-> ( ch -> ( ps -> th ) ) )
4	3	imbi2i	|- ( ( ph -> ( ( ps /\ ch ) -> th ) ) <-> ( ph -> ( ch -> ( ps -> th ) ) ) )