Metamath Proof Explorer

Theorem animorlr

Description: Conjunction implies disjunction with one common formula (3/4). (Contributed by BJ, 4-Oct-2019)

Ref Expression
Assertion animorlr 
|- ( ( ph /\ ps ) -> ( ch \/ ph ) )

Proof

Step Hyp Ref Expression
1 simpl 
 |-  ( ( ph /\ ps ) -> ph )
2 1 olcd 
 |-  ( ( ph /\ ps ) -> ( ch \/ ph ) )