Metamath Proof Explorer


Theorem 3expia

Description: Exportation from triple conjunction. (Contributed by NM, 19-May-2007) (Proof shortened by Wolf Lammen, 22-Jun-2022)

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

Proof

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