${\Sigma }_2^{\mu }$ is decidable for ${\Pi }_2^{\mu }$

Abstract

Given a ${\Pi }_2^{\mu }$ formula of the modal $\mu$ calculus, it is decidable whether it is equivalent to a ${\Sigma }_2^{\mu }$ formula.

This is the extended pre-print of a paper of the same name accepted to Computability in Europe 2017