Questionnaire-2024/natural_deduction_propositional_logic/1001_sol.tex

Translation: \\
$p:$ \quad I am ill. \\
$q:$ \quad I go to the doctor.

\begin{enumerate}
\item
    \begin{quote}
        \begin{tabbing}
            If I am ill, I go to the doctor. \quad \= $p \imp q$ \\
            I am ill. \> $p$ \\
            Therefore, I go to the doctor. \> $\vdash q$ \\
        \end{tabbing}
        \emph{Sequent:} \quad $p \imp q, p \vdash q$ \\

        \begin{logicproof}{1}
            p \imp q & \prem \\
            p & \prem \\
            q & $\impe 2,1$
        \end{logicproof}
    \end{quote}

\item
    \begin{quote}
        \begin{tabbing}
            If I am ill, I go to the doctor. \quad \= $p \imp q$ \\
            I go to the doctor. \> $q$ \\
            Therefore, I am ill. \> $\vdash p$ \\
        \end{tabbing}
        \emph{Sequent:} \quad $p \imp q, q \vdash p$ \\
        \vspace{0.2cm}
        This sequent is not provable.

        $\mathcal{M} : p = F, q = T$ \\
        $\mathcal{M} \models p \imp q, q$ \\
        $\mathcal{M} \nmodels p$
    \end{quote}
\item
    \begin{quote}
        \begin{tabbing}
            (Solve without using the Modus Tollens)\\
            If I am ill, I go to the doctor. \quad \= $p \imp q$ \\
            I did not go to the doctor. \> $\lnot q$ \\
            Therefore, I am not ill. \> $\lnot p$ \\
        \end{tabbing}
        \emph{Sequent:} \quad $p \imp q, \lnot q \vdash \lnot p$ \\
        \vspace{0.2cm}

        \begin{logicproof}{1}
            p \imp q & \prem \\
            \lnot q & \prem \\
            \begin{subproof}
                p & \assum \\
                q & $\impe 3,1$ \\
                \bot & $\nege 4,2$
            \end{subproof}
            \lnot p & $\negi 3-5$
        \end{logicproof}
    \end{quote}
\end{enumerate}