Metamath Proof Explorer


Theorem animorrl

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

Ref Expression
Assertion animorrl ( ( 𝜑 ∧ 𝜓 ) → ( 𝜓 ∨ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 simpr ⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜓 )
2 1 orcd ⊢ ( ( 𝜑 ∧ 𝜓 ) → ( 𝜓 ∨ 𝜒 ) )