Theorem ancomsd

Description: Deduction commuting conjunction in antecedent. (Contributed by NM, 12-Dec-2004)

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

Proof

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