\begin{frame}{Alternative Translations}
  Another example:
  \begin{talign}
    \sentence{Jan does not know a logician.}
  \end{talign}
  \pause
  Again two possibilities: 
  %
  \begin{talign}
    \mpause[1]{\parbox{0.55\textwidth}{\sentence{There is not anybody Jan knows}\\\sentence{who is a logician.}} & :\;\;}
    \mpause{ \formula{\lognot{\existsst{x}{(\logand{\binpred{K}{\const{j}}{x}}{\unpred{L}{x}})}}} }
    \\  
    \mpause{\sentence{Everybody Jan knows is not a logician.}  & ;\;\;}
    \mpause{ \formula{\forallst{x}{(\logimp{\binpred{K}{\const{j}}{x}}{\lognot{\unpred{L}{x}}})}} }
  \end{talign}
  \updatepause
  
  Using the equivalence 
  \formula{
  $ \lognot{\existsst{x}{}} 
      \logequiv
    \forallst{x}{\lognot{}}$},
  % Op grond van de equivalentie  $\niet \eris x  \,\equiv\, \alle x \niet $ kun je   
  we can transform
  \begin{center}
  $\formula{\lognot{\existsst{x}{(\logand{\binpred{K}{\const{j}}{x}}{\unpred{L}{x}})}}}$
   % $\niet \eris x (Kjx \en Lx) $
  \mbox{ } into \mbox{ }
  $\formula{\forallst{x}{\,\lognot{(\logand{\binpred{K}{\const{j}}{x}}{\unpred{L}{x}})}}}$.
   % $\alle x \niet (Kjx \en Lx)$, 
  \end{center}\pause{}
  From propositional logic we know that 
  \begin{center}
  $\formula{\lognot{(\logand{\binpred{K}{\const{j}}{x}}
                            {\unpred{L}{x}})}}
     \logequiv
   \formula{\logimp{\binpred{K}{\const{j}}{x}}{\lognot{\unpred{L}{x}}}}$   
  \end{center}\pause{}                       
   % $ \niet (Kjx \en Lx) \:\equiv \: Kjx \dan \niet Lx$
  \medskip
  
  Combining these two transformations we obtain in fact:
  \begin{equation*}
    \formula{\lognot{\existsst{x}{(\logimp{\binpred{K}{\const{j}}{x}}
                                          {\unpred{L}{x})}}}}
      \logequiv
    \formula{\forallst{x}{(\logimp{\binpred{K}{\const{j}}{x}}{\lognot{\unpred{L}{x}}})}}                      
    %$$\niet \eris x (Kjx \en Lx) \;\equiv\;\alle x  (Kjx \en \niet Lx)$$  
  \end{equation*}
\end{frame}

\theme{Interplay between Quantifiers and Connectives}