Suppose $\Pi$ is a decidable decision problem.Does $\Pi\not \in NP$ imply $\Pi$ is $NP$-Hard?Edit: if we assume there exists $\Pi\in coNP\setminus NP$ then we are done. Can we refute the claim without any unknown assumptions?
Is $NP$ minimal, i.e. does $\Pi\notin NP$ imply $\Pi$ is $NP$-hard?
complexity theory;np hard;complexity classes;np