\begin{frame} \frametitle{Derivation with Natural Deduction} \begin{exampleblock}{} Can we derive \quad$\neg r \to \neg q$\quad from \quad$p \to (q \to r),\; p$\quad? \pause\medskip \begin{tikzpicture} \naturaldeduction{ \proofstep{$p \to (q \to r)$}{premise}; \mpause[1]{ \proofstep{$p$}{premise}; } \mpause{ \proofbox{ \mpause{ \proofstep{$\neg r$}{assumption}; } \mpause{ \proofstep{$q \to r$}{$\to_e$ 2,1}; } \mpause{ \proofstep{$\neg q$}{MT 4,3}; } } } \mpause{ \proofstep{$\neg r \to \neg q$}{$\to_i$ 3--5}; } } \end{tikzpicture} \pause\pause\pause\pause\pause\pause\pause \medskip Hence we have derived \begin{talign} p \to (q \to r),\; p \quad\vdash\quad \neg r \to \neg q \end{talign} \end{exampleblock} \end{frame}