\begin{frame} \frametitle{Reflexive Frames: $\all p \to p$} Recall that \emph{reflexivity} means: $\myall{x}{R(x,x)}$. \smallskip \begin{goal}{Theorem} \begin{malign} \F \models \all p \to p \quad\iff\quad \text{the frame $\F$ is reflexive} \end{malign} \end{goal} \pause\medskip \alt<-8>{ \begin{block}{Proof} \begin{itemize} \item [$\Leftarrow$] Assume that $\F = \pair{W}{R}$ is reflexive. \smallskip\pause We show that $\F \models \all p \to p$. \smallskip\pause Let $L$ be an arbitrary labelling and $\M = \langle W,R,L \rangle$. \smallskip\pause We show for every world $x$:\; $\M,x \fc \all p \to p$. \smallskip\pause If \;\;$x \fc \all p$\;\;, then \pause \;\;$x \fc p$\;\; since $R(x,x)$. \smallskip\pause Hence $\M,x \fc \all p \to p$. \end{itemize} \end{block} }{ \pause[9] \begin{block}{Proof} \begin{itemize} \item [$\Rightarrow$] Assume that $\F \models \all p \to p$. \pause We show that $\F$ is reflexive. \smallskip\pause Assume, for a contradiction, $\F$ was not reflexive. \smallskip\pause Then there is a world $a$ with $\neg R(a,a)$. \smallskip\pause Let $\M = \langle W,R,\lab{L} \rangle$ where $\lab{L}$ is given by:\pause \begin{talign} \lab{L}(a) &= \emptyset &&& \lab{L}(w) &= \{\;p\;\} \quad \text{for every world $w \neq a$} \end{talign} \pause Then $a \fc \all p$ since $p$ holds in all worlds $\ne a$ \;and\; $\neg R(a,a)$.\\ \smallskip\pause But $a \not\fc p$. \pause Thus $a \not\fc \all p \to p$.\\ \smallskip\pause Thus $\F \not\fc \all p \to p$\pause; a contradiction! \pause Hence $\F$ is reflexive. \end{itemize} \end{block} } \vspace{5cm} \onslide<19>{} \end{frame}