\input{../settings/settings} \usepackage{amssymb}% \usepackage{MnSymbol}% %\usepackage{wasysym}% \begin{document} \klausur{Gdl-SaV-B (Logik)} {Prof. M. Mendler, Ph. D.} {Wintersemester 16/17} {90} {Wörterbuch (Englisch-Deutsch/Deutsch-Englisch)} \newpage \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline \textbf{(Sub-)Question} & 1 & 2a & 2b & 3a & 3b & 4a & 4b \\ \hline \textbf{Available Marks} & 12 & 12 & 20 & 10 & 12 & 8 & 16 \\ \hline $\Sigma$ & & & & & & & \\ \hline \end{tabular} \end{center} \newpage \textbf{Syntax} \begin{enumerate} \item Consider the formulas $\phi$ and $\psi$ defined as follows: $$ \phi =_{def} (P \supset \square R ) \wedge \neg P $$ $$ \psi =_{def} S \wedge \neg \diamondsuit Q $$ Use the Martelli-Montanari Algorithm (the algorithm is given in Apeendix I) to check whether there is a \textit{most general unifier} $\theta$ for $\phi$ and $\psi$, i.e., a solution for the unification problem $ E = \{\phi = \psi\}$. In case there is such a unifier, state it. Make clear how you obtain your results by providing \textbf{every single transformation step} of the algorithm, specifyfing the corresponding number of the rule which you apply. \textbf{Hilbert and Tableau Calculus} \item \begin{enumerate} \item With the model axiom $\Gamma = \{A\}$, find a suitable Hilbert deduction in the modal KD such that $$ KD;\Gamma;\vdash_{H} \lozenge(A \vee C).$$ In other words: prove that $\lozenge(A \vee C)$ holds if $A$ holds, using the modal Hilbert Calculus with the axiom schmees from KD.\\ For this, you need (beside your model axiom) the following: \begin{itemize} \item the rule of \textit{Modus Ponens,} \item the rule of \textit{Necessitation,} \item the propositional axiom scheme: $P \supset (P \vee Q),$ \item the modal axiom scheme (D): $\square P \supset \lozenge P$. \end{itemize} \textbf{Hint:} You may need to instantiate the variables $P$ and $Q$ of the axiom schemes appropriately to perform the proof. \item Consider the formula $\phi$ given as $$ ((\lozenge \square P) \wedge \lozenge Q \supset \square (P \wedge \lozenge Q). $$ Using the S5 tableau calculus (see Appendix II) show that $\phi$ is valid in all S5-frames. \end{enumerate} \item A traffic light can have four different states: \textit{red, red/yellow, green} and \textit{yellow}. In a working traffic light, these states change in the indicated order. Consider the set $Var =_{def} \{RED, YELLOW, GREEN\}$ of propositional variables with the following meanings: \begin{center} \begin{tabular}{|l|l|} \hline \textbf{Variable} & \textbf{Meaning} \\ \hline \hline $RED$ & \textit{The red light is on.} \\ \hline $YELLOW$ & \textit{The yellow light is on.} \\ \hline $GREEN$ & \textit{The green light is on.} \\ \hline \end{tabular} \end{center} In the \textit{red/yellow}, both $RED$ and $YELLOW$ hold. \begin{enumerate} \item Some broken traffic light has a malfunction and satisties the \textit{Propositional Temporal Logic (PLTL)} formula $\psi$ defined thus: $$ \psi =_{def} \diamondsuit RED \wedge \square (RED \supset X RED) $$ Draw a time-line representing a model $(\mathcal{T},V)$ with at least one world $t$ where $\psi$ holds. \textbf{Explain in detail} the properties of your model, which enforce that $\mathcal{T},V,t \models \psi$ and how they enforce it. \item Find a formula of PLTL that expresses the following statement: \begin{quote} \textit{Whenever the light is red, it becomes green eventually after being red/yellow for some time (i.e., red/yellow at least for one time instance).} Note that between states \textit{red} and \textit{green}, \textbf{only} \textit{red/yellow} is permitted. \end{quote} \end{enumerate} \textbf{Semantics and Correspondence Theory} \item \begin{enumerate} \item Consider a mono-modal frame $\mathcal{F}_{1} = (W_{1} \rightarrow_{1})$ with $W_{1} =_{def} \{w_{1},w_{2},w_{3},w_{4}\}$and $\rightarrow_{1}$ as indicated in the following figure. \image{0.5}{fig1.PNG}{}{} Copy the transition system of $\mathcal{F}_{1}$ to your exam answer paper and add further transitions to obtain an extended frame $\mathcal{F}_{2} = (W_{1},\rightarrow)$ such that the \textbf{frame axiom (5)} holds, i.e., such that $\mathcal{F}_{2} \models \lozenge P \supset \square \lozenge P.$ Add only a minimal amount of transitions that is necessary to make the statement become true. Do not remove any transition and do not change $W_{1}$! \item Prove formally that in \textbf{all} frames $\mathcal{F}$ (not only in $\mathcal{F}_{1}$ or $\mathcal{F}_{2}$) it holds: $$ \mathcal{F} \models \square (P \supset Q) \supset (\square P \supset \square Q). $$ Do this by arguing on the formal semantics of normal modal logics. In other words prove $$ \mathcal{F},V,w \models \square (P \supset Q) \supset (\square P \supset \square Q) $$ for an arbitrary frame $\mathcal{F} = (W,\rightarrow)$, valuation $V$ and world $w \in W$. Do \textbf{not} use Hilbert Calculus or a Tableau Calculus! \end{enumerate} \end{enumerate} \end{document}