MA0301/exercise2/main.tex

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\documentclass[12pt]{article}
\usepackage{ntnu}
\usepackage{ntnu-math}
\author{Øystein Tveit}
\title{MA0301 Exercise 2}
\begin{document}
\ntnuTitle{}
\break{}
\begin{excs}
\exc{}
\begin{truthtable}
{e|c|c|e}
{$p$ & $q$ & $p \wedge q$ & $p \vee (p \wedge q)$}
\T & \T & \T & \T \\
\T & \F & \F & \T \\
\F & \T & \F & \F \\
\F & \F & \F & \F \\
\end{truthtable}
\exc{}
(DL1)
\begin{truthtable}
{c|c|c|c|e|c|c|e}
{$\alpha$ & $\beta$ & $\gamma$ & $(\beta \wedge \gamma)$ & $\alpha \vee (\beta \wedge \gamma)$ & $\alpha \vee \beta$ & $\alpha \vee \gamma$ & $(\alpha \vee \gamma) \wedge (\alpha \vee \gamma)$}
\T & \T & \T & \T & \T & \T & \T & \T \\
\T & \T & \F & \F & \T & \T & \T & \T \\
\T & \F & \T & \F & \T & \T & \T & \T \\
\T & \F & \F & \F & \T & \T & \T & \T \\
\F & \T & \T & \T & \T & \T & \T & \T \\
\F & \T & \F & \F & \F & \T & \F & \F \\
\F & \F & \T & \F & \F & \F & \T & \F \\
\F & \F & \F & \F & \F & \F & \F & \F \\
\end{truthtable}
(DL2)
\begin{truthtable}
{c|c|c|c|e|c|c|e}
{$\alpha$ & $\beta$ & $\gamma$ & $(\beta \vee \gamma)$ & $\alpha \wedge (\beta \vee \gamma)$ & $\alpha \wedge \beta$ & $\alpha \wedge \gamma$ & $(\alpha \wedge \gamma) \vee (\alpha \wedge \gamma)$}
\T & \T & \T & \T & \T & \T & \T & \T \\
\T & \T & \F & \T & \T & \T & \F & \T \\
\T & \F & \T & \T & \T & \F & \T & \T \\
\T & \F & \F & \F & \F & \F & \F & \F \\
\F & \T & \T & \T & \F & \F & \F & \F \\
\F & \T & \F & \T & \F & \F & \F & \F \\
\F & \F & \T & \T & \F & \F & \F & \F \\
\F & \F & \F & \F & \F & \F & \F & \F \\
\end{truthtable}
\exc{}
\begin{align*}
p \Rightarrow (q \vee r) &\equiv (p \wedge \neg q) \Rightarrow r \\
&\equiv \neg (p \wedge \neg q) \vee r \\
&\equiv (\neg p \vee \neg\neg q) \vee r \\
&\equiv (\neg p \vee q) \vee r \\
&\equiv \neg p \vee (q \vee r) \\
&\equiv p \Rightarrow (q \vee r)
\end{align*}
\exc{}
\begin{align*}
[(q \wedge p) \vee q] \wedge \neg(\neg q \vee p) &\equiv q \wedge \neg p \\
q \wedge \neg p &\equiv [(q \wedge p) \vee q] \wedge \neg(\neg q \vee p) \\
&\equiv [q] \wedge (\neg\neg q \wedge \neg p) \\
&\equiv q \wedge (q \wedge \neg p) \\
&\equiv (q \wedge q) \wedge \neg p \\
&\equiv q \wedge \neg p
\end{align*}
\exc{}
\begin{subexcs}
\subexc{}
$\forall S(x)[H(x)]$
\subexc{}
$\exists S(x)[\neg H(x)]$
\subexc{}
$\forall S(x)[\neg H(x)]$
\subexc{}
$\forall \neg H(x) \exists S(x)$
\end{subexcs}
\exc{}
\begin{subexcs}
\subexc{}
The formula is true because of the case where $x < z < y$ which would mean that $x < y \wedge z < y \wedge x < z \wedge \neg (z < x)$
\subexc{}
The formula is false because $p(z, y)$ and $\neg p(z, x)$ cannot be fulfilled at the same time. $z \geq 0 \wedge \neg (z \geq 0) \equiv F$
\setsubexc{4}
\subexc{}
\fbox{see comment in ovsys}
\end{subexcs}
\exc{}
\begin{align*}
\neg (\forall x [p(x) \wedge q(x)]) &\equiv \exists x [\neg(p(x) \wedge q(x))] \\
&\equiv \exists x [\neg p(x) \vee \neg q(x)]
\end{align*}
\exc{}
\begin{align*}
\neg (\exists x \forall y [p(y) \vee \neg q(x,y)]) &\equiv \forall x \neg(\forall y [p(y) \vee \neg q(x,y)]) \\
&\equiv \forall x \exists y \neg[p(y) \vee \neg q(x,y)] \\
&\equiv \forall x \exists y [\neg p(y) \wedge \neg\neg q(x,y)] \\
&\equiv \forall x \exists y [\neg p(y) \wedge q(x,y)]
\end{align*}
\end{excs}
\end{document}