\begin{frame}
  \frametitle{Wherever you put the $q$'s, $q \to \all  \some  q$ always holds!}
 
  \begin{minipage}{0.53\textwidth}
    \begin{center}
    \begin{tikzpicture}[
        default,
        point/.style={circle, draw=blue, thick, inner sep=3pt, minimum size=9mm},
        node distance=20mm]
  
      \node (3) [point] {?};
      \node [ao=3] {$w_3$};
      
      \node (2) [point, below right of=3,yshift=-3mm] {?};
      \node [aro=2] {$w_2$};

      \node (1) [point, below left of=3,yshift=-3mm] {?};
      \node [alo=1] {$w_1$};

      \begin{scope}[shorten <= 1mm, shorten >= 1mm, very thick,>=stealth]
        \draw [->] (1) to[bend left=15] (2);
        \draw [->] (2) to[bend left=15] (1);
        \draw [->] (2) to[bend left=15] (3);
        \draw [->] (3) to[bend left=15] (2);
        \draw [->] (3) to[bend left=15] (1);
        \draw [->] (1) to[bend left=15] (3);
      \end{scope}

      \draw [rounded corners=2mm, dashed] (-25mm,-23mm) rectangle (25mm,9mm);
      \node at (-25mm,8mm) [anchor=north east,inner sep=2mm] {$\mathcal{M}_?$};
    \end{tikzpicture}
    \end{center}
  \end{minipage}~%
  \begin{minipage}{0.46\textwidth}
    \begin{itemize}  
      \item  $\BLUE{W}= \{\;w_1,w_2,w_3\;\}$
      \item  $R = \{\pair{w_1}{w_2},\; \pair{w_2}{w_1},$ \\
             \;\hfill $\pair{w_2}{w_3},\; \pair{w_3}{w_2}, $ \\
             \;\hfill $\pair{w_1}{w_3},\; \pair{w_3}{w_1}\;\}$
      \item $\M_? = (W,R,L_?)$
    \end{itemize}
  \end{minipage}
  \smallskip\pause

  \begin{exampleblock}{We check one world ($w_1$), with and without $q$}
    \begin{itemize}  
      \pause
      \item Assume that $q \not\in L_?(w_1)$, \pause \\
            then $w_1 \fc q \to\all\some  q$ \tabto{4.5cm} since $\;w_1 \not\fc q$
      \pause
      \item Assume that $q \in L_?(w_1) $,  \\
            \mpause[4]{then $w_1 \fc q \to \all\some q$ 
            \tabto{4.5cm} since}\mpause[3]{ $\;w_1 \fc \all\some q$
            \tabto{4.5cm} since}\mpause[2]{ $\;w_2 \fc \some q$ and $\;w_3 \fc \some q$
            \tabto{4.5cm} since}\mpause[1]{ $\;w_1\fc q$ }
    \end{itemize}    
  \end{exampleblock}
  \pause\pause\pause\pause\pause
  \emph{Because of the arrow configuration} (always back and forth), 
  $q \to \all\some q$ is \emph{always valid} wherever you put the $q$'s.
  \bigskip
\end{frame}