Theorem expcomd

Description: Deduction form of expcom . (Contributed by Alan Sare, 22-Jul-2012)

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

Proof

Step	Hyp	Ref	Expression
1		expcomd.1	|- ( ph -> ( ( ps /\ ch ) -> th ) )
2	1	expd	|- ( ph -> ( ps -> ( ch -> th ) ) )
3	2	com23	|- ( ph -> ( ch -> ( ps -> th ) ) )