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\begin{frame}
  \frametitle{Semantic Implication}
  
  \begin{goal}{Semantic Implication / Consequence}
    \vspace{-1ex}
    \begin{talign}
      \quad\quad \phi_1,\ldots,\phi_n \;\models \; \psi
    \end{talign}
    means
    \begin{center}
      Whenever $\phi_1,\ldots,\phi_n$ are all true, $\psi$ is also true.
    \end{center}
  \end{goal}
  \bigskip
  \pause
  
  \begin{exampleblock}{}
    Do we have \quad $q \;\models\; p \to q$ \quad ?
    \pause
    
    \begin{center}
    \begin{tabular}{|c|c|c|}
      \hline
      \thd $p$ & \thd $q$ & \thd $p \to q$ \\
      \hline
      $\F$ & $\F$ & $\T$\\
      \hline
      $\F$ & \malert{1}{4}{$\T$} & \malert{2}{3}{$\T$}\\
      \hline
      $\T$ & $\F$ & $\F$\\
      \hline
      $\T$ & \malert{1}{4}{$\T$} & \malert{2}{3}{$\T$}\\
      \hline
    \end{tabular}
    \end{center}
    \pause\pause\pause
    Whenever $q$ is $\T$ also $p \to q$ is $\T$. 
    \pause
    Hence: $q \;\models\; p \to q$\;.
  \end{exampleblock}
\end{frame}