Would anyone care to explain how can we dualize the Reflexive Transitive Closure in PDL from : $$\frac{(\phi\vee\langle\alpha\rangle\psi)\rightarrow\psi}{\langle\alpha^*\rangle\phi\rightarrow\psi}$$ to: $$\frac{\psi\rightarrow (\phi\vee [\alpha]\psi)}{\psi\rightarrow [\alpha^*]\phi}$$?
Where $\alpha$ is a program, $\phi,\psi$ are formulas and $\langle\alpha\rangle$ is the diamond from PDL.
$\def\<#1>{\langle#1\rangle}$We start from the relation between box and diamond:
$$ \<\alpha>p \equiv \neg [\alpha] \neg p \enspace. $$
Substitution in the reflexive transitive closure rule yields
$$ \frac{(\phi \vee \neg[\alpha]\neg \psi) \rightarrow \psi}{\neg[\alpha^*]\neg\phi \rightarrow \psi} \enspace. $$
Taking the contrapositive of both premise and conclusion,
$$ \frac{\neg \psi \rightarrow (\neg\phi \wedge [\alpha]\neg \psi)}{\neg\psi \rightarrow [\alpha^*]\neg\phi} \enspace. $$
We now rename, so that $\neg\phi$ is replaced by $\phi$ and $\neg\psi$ is replaced by $\psi$:
$$ \frac{\psi \rightarrow (\phi \wedge [\alpha]\psi)}{\psi \rightarrow [\alpha^*]\phi} \enspace. $$
Finally, if we wish to obtain the loop invariance rule, we substitute $\psi$ for $\phi$ and simplify:
$$ \frac{\psi \rightarrow [\alpha]\psi}{\psi \rightarrow [\alpha^*]\psi} \enspace, $$
having noted that $\psi \rightarrow (\psi \wedge [\alpha]\psi) \equiv \psi \rightarrow [\alpha]\psi$.