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\title{\huge \bfseries Discrete Mathematics - Assignment 2}
\author{\Large Teddy van Jerry}
\date{\today}
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\chead{Discrete Mathematics - Assignment 2}
\begin{document}
\maketitle
{\Large Lecture 1: 2021/03/05
Exercises for Section 1.2 and 1.3}
\section{Section 1.2}
\subsection{Question 3}
You can graduate only if you have completed the requirements
of your major and you do not owe money to the
university and you do not have an overdue library book.
Express your answer in terms of $g$: ``You can graduate,"
$m$: ``You owe money to the university," $r$: ``You have completed
the requirements of your major," and $b$: ``You have
an overdue library book."
\Solution
$$g\ra \neg m \wedge r \wedge \neg b$$
\subsection{Question 9}
Are these system specifications consistent? ``The system
is in multiuser state if and only if it is operating normally.
If the system is operating normally, the kernel is functioning.
The kernel is not functioning or the system is in
interrupt mode. If the system is not in multiuser state,
then it is in interrupt mode. The system is not in interrupt
mode."
\Solution
Let's denote
$a$ as ``the system is in multiuser state",
$b$ as ``it is operating normally",
$c$ as ``the kernel is functioning",
$d$ as ``the system is in interrupt mode",
We know that
\begin{equation}
a\leftrightarrow b
\label{l1}
\end{equation}
\begin{equation}
b\ra c
\label{l2}
\end{equation}
\begin{equation}
\neg(\neg c\oplus d)
\label{l3}
\end{equation}
\begin{equation}
\neg a\ra d
\label{l4}
\end{equation}
\begin{equation}
\neg d
\label{l5}
\end{equation}
From equation \ref{l5}, we know
\begin{equation}\label{l6}
d\equiv \mathrm{\mathbf{F}}
\end{equation}
Consider \ref{l4}, we have
\begin{equation}\label{l7}
\neg a\equiv \mathrm{\mathbf{F}}
\end{equation}
Therefore,
\begin{equation}\label{l8}
a\equiv \mathrm{\mathbf{T}}
\end{equation}
and consider \ref{l1}
\begin{equation}\label{l9}
b\equiv \mathrm{\mathbf{T}}
\end{equation}
Consider \ref{l2},
\begin{equation}\label{key}
c\equiv\mathrm{\mathbf{T}}
\end{equation}
However, \ref{l3} is false then.
Thus these system specifications are \textbf{not} consistent.
\subsection{Question 21}
When three professors are seated in a restaurant, the hostess
asks them: ``Does everyone want coffee?" The first
professor says ``I do not know." The second professor then
says ``I do not know." Finally, the third professor says
``No, not everyone wants coffee." The hostess comes back
and gives coffee to the professors who want it. How did
she figure out who wanted coffee?
\Solution
Since the third professor says No, he is sure at least one of them does not need coffee, but the first two professors do not give their choice clearly, we can conclude that the third professor do not need coffee.
Conversely, if the first two professors do not want coffee, they will answer No, so they both want coffee.
\subsection{Question 45}
Find the output of each of these combinatorial circuits.
\begin{figure}[hbtp]
\centering
\includegraphics[width = .5\linewidth]{1_2-45(b).jpg}
\caption{Question 45 (b)}
\end{figure}
\Solution
$$(\neg p\wedge\neg q)\vee(p\wedge r)$$
\section{Section 1.3}
\subsection{Question 5}
Use a truth table to verify the distributive law
$$p\wedge (q\vee r) \equiv (p \wedge q) \vee (p \wedge r).$$
\begin{proof}
We can construct the truth tables as below:
\begin{table}[htbp]
\centering
\caption{Truth Table for $p\wedge (q\vee r)$ and $(p \wedge q) \vee (p \wedge r)$}
\begin{tabular}{ccccc}
\toprule
$p$ & $q$ & $r$ & $q\vee r$ & $p\wedge (q\vee r)$ \\
\midrule
T & T & T & T & T \\
T & T & F & T & T \\
T & F & T & T & T \\
T & F & F & F & F \\
F & T & T & T & F \\
F & T & F & T & F \\
F & F & T & T & F \\
F & F & F & F & F \\
\bottomrule
\end{tabular}\qquad
\begin{tabular}{cccccc}
\toprule
$p$ & $q$ & $r$ & $q\wedge q$ & $p\wedge r$ & $(p \wedge q) \vee (p \wedge r)$\\
\midrule
T & T & T & T & T & T \\
T & T & F & T & F & T \\
T & F & T & F & T & T \\
T & F & F & F & F & F \\
F & T & T & F & F & F \\
F & T & F & F & F & F \\
F & F & T & F & F & F \\
F & F & F & F & F & F \\
\bottomrule
\end{tabular}%
\label{Table_11}%
\end{table}
\end{proof}
\subsection{Question 7}
Use De Morgan's laws to find the negation of each of the
following statements.
d) Ibrahim is smart and hard working.
\Solution
``Ibrahim is not smart or is lazy."
\subsection{Question 9}
For each of these compound propositions, use the
conditional-disjunction equivalence (Example 3) to find
an equivalent compound proposition that does not involve
conditionals.
c) $(\neg q\ra p)\ra(p\ra\neg q)$
\subsection{Question 11}
Show that each of these conditional statements is a tautology
by using truth tables.
d) $(p\wedge q)\ra(p\ra q)$
f) $\neg(p\ra q)\ra\neg q$
\begin{proof}
We can construct the truth tables as below:
\begin{table}[htbp]
\centering
\caption{Truth Table for $(p\wedge q)\ra(p\ra q)$ and $\neg(p\ra q)\ra\neg q$}
\begin{tabular}{ccccc}
\toprule
$p$ & $q$ & $p\wedge q$ & $q\ra q$ & $(p\wedge q)\ra(p\ra q)$ \\
\midrule
T & T & T & T & T \\
T & F & F & F & T \\
F & T & F & T & T \\
F & F & F & T & T \\
\bottomrule
\end{tabular}\qquad
\begin{tabular}{cccccc}
\toprule
$p$ & $q$ & $p\ra q$ & $\neg(p\ra q)$ & $\neg q$ & $\neg(p\ra q)\ra\neg q$\\
\midrule
T & T & T & F & F & T \\
T & F & F & T & T & T \\
F & T & T & F & F & T \\
F & F & T & F & T & T \\
\bottomrule
\end{tabular}%
\label{Table_21}%
\end{table}
\end{proof}
\subsection{Question 13}
Show that each conditional statement in Exercise 11 is
a tautology using the fact that a conditional statement is
false exactly when the hypothesis is true and the conclusion
is false. (Do not use truth tables.)
\begin{proof}[Proof for (d)]
Case $(p\wedge q)$ is false, the proposition is certain to be true.
Case $(p\wedge q)$ is true, $p$ and $q$ are both true, thus $p\ra q$ is true, the proposition is true.
\end{proof}
\begin{proof}[Proof for (f)]
Case $\neg(p\ra q)$ is false, the proposition is certain to be true.
Case $\neg(p\ra q)$ is true, i.e. $(p\ra q)$ is false, thus $p$ is true and $q$ is false. Therefore $\neg q$ is true, the proposition is true.
\end{proof}
\subsection{Question 15}
Show that each conditional statement in Exercise 11 is a
tautology by applying a chain of logical identities as in
Example 8. (Do not use truth tables.)
\begin{proof}[Proof for (d)]
$$
\begin{aligned}
(p\wedge q)\ra(p\ra q)\equiv & (p\wedge q) \ra (\neg p\vee q) \\
\equiv & \neg(p\wedge q)\vee(\neg p\vee q) \\
\equiv & \neg p \vee \neg q \vee \neg p \vee q \\
\equiv & (q\vee \neg q) \vee \neg p \\
\equiv & \mathrm{\mathbf{T}} \vee \neg p \\
\equiv & \mathrm{\mathbf{T}}
\end{aligned}
$$
\end{proof}
\begin{proof}[Proof for (f)]
$$
\begin{aligned}
\neg(p\ra q)\ra\neg q \equiv & \neg\neg(p\vee q)\vee \neg q \\
\equiv & (p\vee q)\vee \neg q \\
\equiv & p\vee (q\vee\neg q) \\
\equiv & p\vee \mathrm{\mathbf{T}} \\
\equiv & \mathrm{\mathbf{T}}
\end{aligned}
$$
\end{proof}
\subsection{Question 19}
Determine whether $(\neg q\wedge(p\ra q))\ra\neg p$ is a tautology.
Each of Exercises 20–32 asks you to show that two compound
propositions are logically equivalent. To do this, either show
that both sides are true, or that both sides are false, for exactly
the same combinations of truth values of the propositional
variables in these expressions (whichever is easier).
\Solution
$$
\begin{aligned}
(\neg q\wedge(p\ra q))\ra\neg p \equiv & \neg(\neg q\wedge(\neg p\vee q))\vee\neg p \\
\equiv & \neg(\neg p\wedge(\neg p\vee q)\wedge p) \\
\equiv & \neg((\neg p\wedge p)\wedge(\neg p\vee q)) \\
\equiv & \neg(\mathrm{\mathbf{F}}\wedge(\neg p\vee q)) \\
\equiv & \neg\mathrm{\mathbf{F}} \\
\equiv & \mathrm{\mathbf{T}}
\end{aligned}
$$
So it is a tautology.
\subsection{Question 29}
Show that $(p\ra r)\vee(q\ra r)$ and $(p\wedge q)\ra r$ are logically
equivalent.
\begin{proof}
On one hand,
$$
\begin{aligned}
(p\ra r)\vee(q\ra r) \equiv & (\neg p\vee r)\vee(\neg q\vee r) \\
\equiv & \neg p\vee\neg q\vee r
\end{aligned}
$$
On the other hand,
$$
\begin{aligned}
(p\wedge q)\ra r \equiv & \neg(p\wedge q)\vee r \\
\equiv & \neg p\vee\neg q\vee r
\end{aligned}
$$
\end{proof}
\subsection{Question 35}
Show that $(p\ra q)\ra r$ and $p\ra(q\ra r)$ are not logically equivalent.
\begin{proof}
Let $p, q, r$ all be false, then $(p\ra q)\ra r$ is false while $p\ra(q\ra r)$ is true. It is a contradiction.
$\therefore$ $(p\ra q)\ra r$ and $p\ra(q\ra r)$ are not logically equivalent.
\end{proof}
\subsection{Question 65}
Determine whether each of these compound propositions
is satisfiable.
a) $(p\vee\neg q)\wedge(\neg p\vee q)\wedge(\neg p\vee q)$
b) $(p\ra q)\wedge(p\ra\neg q)\wedge(\neg p\ra q)\wedge(\neg p\ra\neg q)$
c) $(p\leftrightarrow q)\wedge(\neg p\leftrightarrow q)$
\Solution
a) is satisfiable.
\begin{proof}
Let $p, q$ be false, the proposition is true.
\end{proof}
b) is not satisfiable.
\begin{proof}
$$
\begin{aligned}
(p\ra q)\wedge(p\ra\neg q)\wedge(\neg p\ra q)\wedge(\neg p\ra\neg q) \equiv & (\neg p\vee q)\wedge(p\vee\neg q)\wedge(p\vee q)\wedge(p\wedge\neg q) \\
\equiv & \neg p\wedge(q\vee \neg q)\wedge p\wedge(q\vee\neg q) \\
\equiv & \neg p\wedge \mathrm{\mathbf{T}} \wedge p\wedge \mathrm{\mathbf{T}} \\
\equiv & (\neg p\wedge p)\wedge \mathrm{\mathbf{T}} \\
\equiv & \mathrm{\mathbf{F}} \wedge \mathrm{\mathbf{T}} \\
\equiv & \mathrm{\mathbf{F}}
\end{aligned}
$$
\end{proof}
c) is not satisfiable.
\begin{proof}
$$
\begin{aligned}
(p\leftrightarrow q)\wedge(\neg p\leftrightarrow q) & \equiv (p\vee q)\wedge(\neg p\vee\neg q)\wedge(\neg p\vee q)\wedge(p\wedge\neg q) \\
\equiv & \mathrm{\mathbf{F}} \text{ (shown in the proof of (b))}
\end{aligned}
$$
\end{proof}
\section*{Appendix}
~
This assignment is written in \LaTeX.
\LaTeX\ code: \url{https://blog.csdn.net/weixin_50012998/article/details/114463071}\\
Lecture Note: \url{https://blog.csdn.net/weixin_50012998/article/details/114383689}
\end{document}
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Teddy van Jerry’s CSDN Homepage
Teddy van Jerry’s GitHub Homepage
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