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\newcommand{\naam}          {Sam Titarsolej}
\newcommand{\studentNummer} {12206385}
\newcommand{\opdrachtNaam}  {Huiswerk}
\newcommand{\opdrachtNummer}{2}
\newcommand{\datum}         {13/11/2018}

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\begin{document}

\section*{Vraag 1}

a) $\begin{array}{c|c|c|c}
p & q & p \vee q & q \rightarrow p\\ \hline
1 & 1 & 1 & 1\\
1 & 0 & 1 & 1\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{array}$\\

$p \vee q, q \rightarrow p, \vDash p$ is een geldige redenering, omdat voor elke ware conlcusie, de premissen ook waar zijn.
\newline
\newline

b) $\begin{array}{c|c|c|c|c|c|c}
p & q & s & \neg p & \neg p \rightarrow q & q \vee s & \neg s\\ \hline
1 & 1 & 1 & 0 & 1 & 1 & 0\\
1 & 1 & 0 & 0 & 1 & 1 & 1\\
1 & 0 & 1 & 0 & 1 & 1 & 0\\
1 & 0 & 0 & 0 & 1 & 0 & 1\\
0 & 1 & 1 & 1 & 1 & 1 & 0\\
0 & 1 & 0 & 1 & 1 & 1 & 1\\
0 & 0 & 1 & 1 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 1\\
\end{array}$\\

$\neg p, \neg p \rightarrow q, q \vee s \vDash \neg s$ is geen geldige redenering. Tegenvoorbeeld: $V(p) = 0,\ V(q) = 1,\ V(s) = 1$.

\section*{Vraag 2}

a) $\begin{array}{c|c|c|c|c}
p & q & \neg p & p \rightarrow q & \neg p \rightarrow (p \rightarrow q)\\ \hline
1 & 1 & 0 & 1 & 1\\
1 & 0 & 0 & 0 & 1\\
0 & 1 & 1 & 1 & 1\\
0 & 0 & 1 & 1 & 1\\
\end{array}$\\

$\neg p \rightarrow (p \rightarrow q)$ is een tautologie.
\newline
\newline

b) $\begin{array}{c|c|c}
p & \neg p & p \leftrightarrow \neg p\\ \hline
1 & 0 & 0\\
0 & 1 & 0\\
\end{array}$\\

$p \leftrightarrow \neg p$ is een contradictie.
\newline
\newline

c) $\begin{array}{c|c|c|c|c}
p & q & \neg (p \wedge q) & \neg p \vee \neg q & \neg (p \wedge q) \leftrightarrow (\neg p \vee \neg q)\\ \hline
1 & 1 & 0 & 1 & 1\\
1 & 0 & 1 & 0 & 1\\
0 & 1 & 1 & 0 & 1\\
0 & 0 & 1 & 0 & 1\\
\end{array}$\\

$\neg (p \wedge q) \leftrightarrow (\neg p \vee \neg q$ is een tautologie.
\newline
\newline

d) $\begin{array}{c|c|c|c|c}
p & q & p \rightarrow q & p \wedge q & (p \rightarrow q) \rightarrow (p \wedge q)\\ \hline
1 & 1 & 1 & 1 & 1\\
1 & 0 & 0 & 0 & 1\\
0 & 1 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
\end{array}$\\

$(p \rightarrow q) \rightarrow (p \wedge q)$ is contingent.

\section*{Vraag 3}

$W \vee K, W \vDash \neg K\\\\
\begin{array}{c|c|c|c}
W & K & W \vee K & \neg K \\ \hline
1 & 1 & 1 & 0\\
1 & 0 & 1 & 1\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{array}$\\
\newline

De redenering is niet geldig. Tegenvoorbeeld: $V(W) = 1, V(K) = 1$ (Als Turing zowel wiskundige als KI'er was, dan klopt de conclusie niet).

\newpage
\section*{Vraag 4}

a)
\begin{fitch}
$p \rightarrow (q \rightarrow r)$ & \\
\fj $q$ & Ass. \\
\fa $q \rightarrow r$ & $E \rightarrow 1, 2$\\
\fa $q$ & $E \rightarrow 1, 2, 3$\\
\fa $r$ & $E \rightarrow 3, 4$\\
\fa $p \rightarrow r$ & $I \rightarrow 2 - 5$ \\
\fa $p \rightarrow (p \rightarrow r)$ & $I \rightarrow 2 - 7$ \\
\end{fitch}\\
\\

b)
\begin{fitch}
$p \rightarrow (q \rightarrow r)$ & \\
\fj $p \vee (p \wedge q)$ & Ass. \\
\fa $p$ & $Ass.$\\
\fa \fj $p \wedge q$ & $G \vee 1$\\
\fa \fa $q$ & $I \wedge 3$\\
\fa \fh $p$ & $E \vee 1$ \\
\fa \fa $p$ & $Herh.$ \\
\fa $p$ & $I \rightarrow 2 - 5$ \\
\end{fitch}\\

\section*{Vraag 5}

\begin{fitch}
\fa $p \rightarrow q$ \\
\fj $r \rightarrow s$ \\
\fa \fh $p$ & $Ass.$\\
\fa \fa $p \vee\ r$ & $I \vee 3$\\
\fa \fh $q$ & $E \rightarrow 1, 3$\\
\fa \fa $q \vee\ s$ & $I \vee 5$ \\
\fa $(p \vee r) \rightarrow (q \vee s)$ & $I \rightarrow 4, 6$ \\
\end{fitch}\\

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