Theorem expd

Description: Exportation deduction. (Contributed by NM, 20-Aug-1993) (Proof shortened by Wolf Lammen, 28-Jul-2022)

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

Proof

Step	Hyp	Ref	Expression
1		expd.1	|- ( ph -> ( ( ps /\ ch ) -> th ) )
2	1	expdcom	|- ( ps -> ( ch -> ( ph -> th ) ) )
3	2	com3r	|- ( ph -> ( ps -> ( ch -> th ) ) )