Metamath Proof Explorer


Theorem exp4d

Description: An exportation inference. (Contributed by NM, 26-Apr-1994)

Ref Expression
Hypothesis exp4d.1
|- ( ph -> ( ( ps /\ ( ch /\ th ) ) -> ta ) )
Assertion exp4d
|- ( ph -> ( ps -> ( ch -> ( th -> ta ) ) ) )

Proof

Step Hyp Ref Expression
1 exp4d.1
 |-  ( ph -> ( ( ps /\ ( ch /\ th ) ) -> ta ) )
2 1 expd
 |-  ( ph -> ( ps -> ( ( ch /\ th ) -> ta ) ) )
3 2 exp4a
 |-  ( ph -> ( ps -> ( ch -> ( th -> ta ) ) ) )