Metamath Proof Explorer


Theorem 3jaodan

Description: Disjunction of three antecedents (deduction). (Contributed by NM, 14-Oct-2005)

Ref Expression
Hypotheses 3jaodan.1
|- ( ( ph /\ ps ) -> ch )
3jaodan.2
|- ( ( ph /\ th ) -> ch )
3jaodan.3
|- ( ( ph /\ ta ) -> ch )
Assertion 3jaodan
|- ( ( ph /\ ( ps \/ th \/ ta ) ) -> ch )

Proof

Step Hyp Ref Expression
1 3jaodan.1
 |-  ( ( ph /\ ps ) -> ch )
2 3jaodan.2
 |-  ( ( ph /\ th ) -> ch )
3 3jaodan.3
 |-  ( ( ph /\ ta ) -> ch )
4 1 ex
 |-  ( ph -> ( ps -> ch ) )
5 2 ex
 |-  ( ph -> ( th -> ch ) )
6 3 ex
 |-  ( ph -> ( ta -> ch ) )
7 4 5 6 3jaod
 |-  ( ph -> ( ( ps \/ th \/ ta ) -> ch ) )
8 7 imp
 |-  ( ( ph /\ ( ps \/ th \/ ta ) ) -> ch )