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Metamath Proof Explorer

Theorem pm2.61dan

Description: Elimination of an antecedent. (Contributed by NM, 1-Jan-2005)

Ref Expression
Hypotheses pm2.61dan.1 
|- ( ( ph /\ ps ) -> ch )
pm2.61dan.2 
|- ( ( ph /\ -. ps ) -> ch )
Assertion pm2.61dan 
|- ( ph -> ch )

Proof

Step Hyp Ref Expression
1 pm2.61dan.1 
 |-  ( ( ph /\ ps ) -> ch )
2 pm2.61dan.2 
 |-  ( ( ph /\ -. ps ) -> ch )
3 1 ex 
 |-  ( ph -> ( ps -> ch ) )
4 2 ex 
 |-  ( ph -> ( -. ps -> ch ) )
5 3 4 pm2.61d 
 |-  ( ph -> ch )