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Metamath Proof Explorer

Theorem expd

Description: Exportation deduction. (Contributed by NM, 20-Aug-1993) (Proof shortened by Wolf Lammen, 28-Jul-2022)

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

Proof

Step Hyp Ref Expression
1 expd.1 
 |-  ( ph -> ( ( ps /\ ch ) -> th ) )
2 1 expdcom 
 |-  ( ps -> ( ch -> ( ph -> th ) ) )
3 2 com3r 
 |-  ( ph -> ( ps -> ( ch -> th ) ) )