Metamath Proof Explorer

Theorem expcom

Description: Exportation inference with commuted antecedents. (Contributed by NM, 25-May-2005)

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

Proof

Step Hyp Ref Expression
1 ex.1 
 |-  ( ( ph /\ ps ) -> ch )
2 1 ex 
 |-  ( ph -> ( ps -> ch ) )
3 2 com12 
 |-  ( ps -> ( ph -> ch ) )