Theorem imdistand

Description: Distribution of implication with conjunction (deduction form). (Contributed by NM, 27-Aug-2004)

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

Proof

Step	Hyp	Ref	Expression
1		imdistand.1	|- ( ph -> ( ps -> ( ch -> th ) ) )
2		imdistan	|- ( ( ps -> ( ch -> th ) ) <-> ( ( ps /\ ch ) -> ( ps /\ th ) ) )
3	1 2	sylib	|- ( ph -> ( ( ps /\ ch ) -> ( ps /\ th ) ) )