\begin{frame}
  \frametitle{Example from a Previous Exam}

  \begin{exampleblock}{}
    Show that \quad$\vdash\; \neg q \vee (p \to q)$\quad with PBC instead of LEM:
    \smallskip
    
    \begin{tikzpicture}
    \naturaldeduction{
      \mpause[1]{
        \proofbox{
          \mpause{
            \proofstep{$\neg (\neg q \vee (p \to q))$}{assumption};
          }
          \mpause{
            \proofbox{
              \mpause{
                \proofstep{$\neg q$}{assumption};
              }
              \mpause{
                \proofstep{$\neg q \vee (p\to q)$}{$\vee_{i_1}$ 2};
              }
              \mpause{
                \proofstep{$\bot$}{$\neg_e$ 3,1};
              }
            }
          }
          \mpause{
            \proofstep{$q$}{PBC 2--4};
          }
          \mpause{
            \proofbox{
              \mpause{
                \proofstep{$p$}{assumption};
              }
              \mpause{
                \proofstep{$q$}{copy 5};
              }
            }
          }
          \mpause{
            \proofstep{$p \to q$}{$\to_i$ 6--7};
          }
          \mpause{
            \proofstep{$\neg q \vee (p \to q)$}{$\vee_{i_2}$ 8};
          }
          \mpause{
            \proofstep{$\bot$}{$\neg_e$ 9,1};
          }
        }
      }
      \mpause{
        \proofstep{$\neg q \vee (p \to q)$}{PBC 1--11};
      }
    }
    \end{tikzpicture}
  \end{exampleblock}
\end{frame}