\documentclass[a4paper]{article} \usepackage{geometry} \usepackage[english]{babel} \usepackage{enumitem} \setenumerate[1]{label=\alph*)} \usepackage{fancyhdr} \renewcommand{\headrulewidth}{0pt} \renewcommand{\footrulewidth}{0pt} \fancyhead{} \fancyfoot[C]{Copyright {\textcopyright} 2016 Camil Staps} \pagestyle{fancy} \usepackage{sv} \title{Software Verification - assignment 2} \author{Camil Staps} \def\run#1{\mathit{run}_{#1}} \def\stop#1{\mathit{stop}_{#1}} \begin{document} \maketitle \thispagestyle{fancy} \setcounter{section}{2} \section{CTL-properties of processes} \begin{enumerate} \item $\invariantly(\lnot(\run1\land\run2))$. \item $\potentially(\run1\land\run2)$. \item $\inevitably\eventually(\run1)$. \item $\potentially\eventually(\run1)$. \item $(\potentially(\lnot\run1\until\run2)) \land (\potentially(\lnot\run2\until\run1))$. \item Impossible: `never' cannot be expressed in CTL: does it mean `in no case whatsoever can this happen' or `there is a path for which this never happens'? But in LTL: $(\always\lnot\run2)\then(\always\run1)$. \item For the same reason as above we cannot express this in CTL. We cannot express this in LTL either, because `may \dots forever' can only be expressed in CTL. \item $\invariantly\exists\next\stop{i}$ with $i\in\{1,2\}$. \item $\invariantly((\forall\lnot\stop1)\then\inevitably\run1)$. \item $\invariantly(\lnot\run1\then\potentially(\lnot\run1\untilW\run2))$. \end{enumerate} \end{document}