Metamath Proof Explorer


Theorem exp4b

Description: An exportation inference. (Contributed by NM, 26-Apr-1994) (Proof shortened by Wolf Lammen, 23-Nov-2012) Shorten exp4a . (Revised by Wolf Lammen, 20-Jul-2021)

Ref Expression
Hypothesis exp4b.1
|- ( ( ph /\ ps ) -> ( ( ch /\ th ) -> ta ) )
Assertion exp4b
|- ( ph -> ( ps -> ( ch -> ( th -> ta ) ) ) )

Proof

Step Hyp Ref Expression
1 exp4b.1
 |-  ( ( ph /\ ps ) -> ( ( ch /\ th ) -> ta ) )
2 1 expd
 |-  ( ( ph /\ ps ) -> ( ch -> ( th -> ta ) ) )
3 2 ex
 |-  ( ph -> ( ps -> ( ch -> ( th -> ta ) ) ) )