Metamath Proof Explorer


Theorem expimpd

Description: Exportation followed by a deduction version of importation. (Contributed by NM, 6-Sep-2008)

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

Proof

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