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59 lines
1.7 KiB
59 lines
1.7 KiB
Translation: \\
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$p:$ \quad I am ill. \\
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$q:$ \quad I go to the doctor.
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\begin{enumerate}
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\item
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\begin{quote}
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\begin{tabbing}
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If I am ill, I go to the doctor. \quad \= $p \imp q$ \\
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I am ill. \> $p$ \\
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Therefore, I go to the doctor. \> $\vdash q$ \\
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\end{tabbing}
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\emph{Sequent:} \quad $p \imp q, p \vdash q$ \\
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\begin{logicproof}{1}
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p \imp q & \prem \\
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p & \prem \\
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q & $\impe 2,1$
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\end{logicproof}
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\end{quote}
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\item
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\begin{quote}
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\begin{tabbing}
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If I am ill, I go to the doctor. \quad \= $p \imp q$ \\
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I go to the doctor. \> $q$ \\
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Therefore, I am ill. \> $\vdash p$ \\
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\end{tabbing}
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\emph{Sequent:} \quad $p \imp q, q \vdash p$ \\
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\vspace{0.2cm}
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This sequent is not provable.
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$\mathcal{M} : p = F, q = T$ \\
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$\mathcal{M} \models p \imp q, q$ \\
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$\mathcal{M} \nmodels p$
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\end{quote}
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\item
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\begin{quote}
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\begin{tabbing}
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(Solve without using the Modus Tollens)\\
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If I am ill, I go to the doctor. \quad \= $p \imp q$ \\
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I did not go to the doctor. \> $\lnot q$ \\
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Therefore, I am not ill. \> $\lnot p$ \\
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\end{tabbing}
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\emph{Sequent:} \quad $p \imp q, \lnot q \vdash \lnot p$ \\
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\vspace{0.2cm}
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\begin{logicproof}{1}
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p \imp q & \prem \\
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\lnot q & \prem \\
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\begin{subproof}
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p & \assum \\
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q & $\impe 3,1$ \\
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\bot & $\nege 4,2$
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\end{subproof}
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\lnot p & $\negi 3-5$
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\end{logicproof}
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\end{quote}
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\end{enumerate}
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