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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 ) )