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@@ -1335,4 +1335,43 @@ hypothèses $P_1,\ldots,P_r$, on a $Q$ ».
\end{frame}
%
+\begin{frame}
+\frametitle{Le calcul propositionnel intuitionniste : survol des règles}
+
+Axiomes : si $Q \in \{P_1,\ldots,P_r\}$ alors $P_1,\ldots,P_r \vdash Q$.
+
+\bigskip
+
+Introduction et élimination des connecteurs (style « déduction naturelle ») :
+
+\smallskip
+
+\begin{tabular}{c|c|c}
+&Intro&Élim\\\hline
+$\Rightarrow$
+&$\inferrule{\Gamma,P\vdash Q}{\Gamma\vdash P\Rightarrow Q}$
+&$\inferrule{\Gamma\vdash P\Rightarrow Q\\\Gamma\vdash P}{\Gamma\vdash Q}$
+\\\hline
+$\land$
+&$\inferrule{\Gamma\vdash Q_1\\\Gamma\vdash Q_2}{\Gamma\vdash Q_1\land Q_2}$
+&$\inferrule{\Gamma\vdash Q_1\land Q_2}{\Gamma\vdash Q_1}$
+\quad$\inferrule{\Gamma\vdash Q_1\land Q_2}{\Gamma\vdash Q_2}$
+\\\hline
+$\lor$
+&$\inferrule{\Gamma\vdash Q_1}{\Gamma\vdash Q_1\lor Q_2}$
+\quad$\inferrule{\Gamma\vdash Q_2}{\Gamma\vdash Q_1\lor Q_2}$
+&$\inferrule{\Gamma\vdash P_1\lor P_2\\\Gamma,P_1\vdash Q\\\Gamma,P_2\vdash Q}{\Gamma\vdash Q}$
+\\\hline
+$\top$
+&$\inferrule{\strut}{\Gamma\vdash \top}$
+&(néant)
+\\\hline
+$\bot$
+&(néant)
+&$\inferrule{\Gamma\vdash \bot}{\Gamma\vdash Q}$
+\\
+\end{tabular}
+
+\end{frame}
+%
\end{document}