Metamath Proof Explorer

Theorem 3exp

Description: Exportation inference. (Contributed by NM, 30-May-1994) (Proof shortened by Wolf Lammen, 22-Jun-2022)

Ref Expression
Hypothesis 3exp.1 
|- ( ( ph /\ ps /\ ch ) -> th )
Assertion 3exp 
|- ( ph -> ( ps -> ( ch -> th ) ) )

Proof

Step Hyp Ref Expression
1 3exp.1 
 |-  ( ( ph /\ ps /\ ch ) -> th )
2 1 3expa 
 |-  ( ( ( ph /\ ps ) /\ ch ) -> th )
3 2 exp31 
 |-  ( ph -> ( ps -> ( ch -> th ) ) )