Metamath Proof Explorer


Theorem jaod

Description: Deduction disjoining the antecedents of two implications. (Contributed by NM, 18-Aug-1994)

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

Proof

Step Hyp Ref Expression
1 jaod.1
 |-  ( ph -> ( ps -> ch ) )
2 jaod.2
 |-  ( ph -> ( th -> ch ) )
3 1 com12
 |-  ( ps -> ( ph -> ch ) )
4 2 com12
 |-  ( th -> ( ph -> ch ) )
5 3 4 jaoi
 |-  ( ( ps \/ th ) -> ( ph -> ch ) )
6 5 com12
 |-  ( ph -> ( ( ps \/ th ) -> ch ) )