init

hw01/HW01DanielShchur.tex

+\documentclass[12pt,letterpaper]{article}
+
+\usepackage{caption}
+\usepackage[table]{xcolor}
+\usepackage{boldline}
+\usepackage{geometry}
+\usepackage{marginnote}
+\usepackage{bm}
+
+\newcommand{\problem}{\vspace{.5cm}\textbf{Problem:~~}}
+\newcommand{\answer}{\emph{Answer:~~}}
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{0cm}
+
+\title{CSCE 235H Homework 01}
+\author{Daniel Shchur}
+
+\begin{document}
+
+\maketitle
+
+\section{Book Work}
+
+\problem (Pg12 S1.1 \#2) Which of these sentences are propositions? What are the truth values of those  that are propositions?
+
+\begin{enumerate}
+
+\item[a)] Do not pass go.
+
+  \begin{enumerate}
+  \item[] \answer It is not a proposition (it's a command).
+  \end{enumerate}
+    
+\item[b)] What time is it?
+
+  \begin{enumerate}
+  \item[] \answer It is not a proposition (it's a question).
+  \end{enumerate}
+  
+\item[c)] There are no black flies in Maine.
+
+  \begin{enumerate}
+  \item[] \answer It is a false proposition assuming black flies have not gone \mbox{extinct}.
+  \end{enumerate}
+  
+\item[d)] $4 + 4 = 5$.
+
+  \begin{enumerate}
+  \item[] \answer It is not a proposition (contains a variable).
+  \end{enumerate}
+
+\item[e)] The moon is made of green cheese.
+
+  \begin{enumerate}
+  \item[] \answer It is a false proposition because the moon is most certainly not made of cheese nor is it green (although the former is true for Wallace and Gromit).
+  \end{enumerate}
+
+\item[f)] $2^n \geq 100$.
+
+  \begin{enumerate}
+  \item[] \answer It is not a proposition (contains a variable).
+  \end{enumerate}
+
+\end{enumerate}
+  
+\newpage
+
+\problem (Pg13 S1.1 \#14) Let $p$, $q$, and $r$ be the propositions\\
+\indent\hspace{0.75cm} $p$: You get an A on the final Exam.\\
+\indent\hspace{0.75cm} $q$: You do every exercise in this book.\\
+\indent\hspace{0.75cm} $r$: You get an A in this class.\\
+Write these propositions using $p$, $q$, and $r$ and logical connectives (including negations).
+
+\begin{enumerate}
+  
+\item[a)] You get an A in this class, but you do not do every exercise in this book
+
+  \begin{enumerate}
+  \item[] \answer $r \wedge \neg q$
+  \end{enumerate}
+    
+\item[b)] You get an A on the final, you do every exercise in this book, and you get an A  in this class.
+
+  \begin{enumerate}
+  \item[] \answer $p \wedge q \wedge r$
+  \end{enumerate}
+    
+\item[c)] To get an A in this class, it is necessary for you to get an A on the final.
+
+  \begin{enumerate}
+  \item[] \answer $p \rightarrow r$
+  \end{enumerate}
+    
+\item[d)] You get an A on the final, but you don't do every exercise in this book; nevertheless, you get an A in this class.
+
+  \begin{enumerate}
+  \item[] \answer $p \wedge \neg q \wedge r$
+  \end{enumerate}
+  
+\item[e)] Getting an A on the final and doing every exercise in this book is sufficient for getting an A in this class.
+
+  \begin{enumerate}
+  \item[] \answer $(p \wedge q) \rightarrow r$
+  \end{enumerate}
+  
+\item[f)] You will get an A in this class if and only if you either do every exercise in this book or you get an A on the final.
+
+  \begin{enumerate}
+  \item[] \answer $(p \vee q) \leftrightarrow r$
+  \end{enumerate}
+  
+\end{enumerate}
+
+\newpage
+
+\problem (Pg14 S1.1 \#18) Determine whether each of these conditional statements is true or false.
+
+\begin{enumerate}
+  
+\item[a)] If $1+1=3$, then unicorns exist.
+
+  \begin{enumerate}
+  \item[] \answer True
+  \end{enumerate}
+  
+\item[b)] If $1+1=3$, then dogs can fly.
+
+  \begin{enumerate}
+  \item[] \answer True
+  \end{enumerate}
+  
+\item[c)] If $1+1=2$, then dogs can fly.
+
+  \begin{enumerate}
+  \item[] \answer False
+  \end{enumerate}
+    
+\item[d)] If $2+2=4$, then $1+2=3$.
+
+  \begin{enumerate}
+  \item[] \answer True
+  \end{enumerate}
+  
+\end{enumerate}
+
+\newpage
+
+\problem (Pg15 S1.1 \#38) Construct a truth table for $((p \rightarrow q) \rightarrow r) \rightarrow s$.
+
+\indent\hspace{0.5cm}\answer
+
+  \begin{table}[htb]
+    \caption*{Problem \#38}
+    \begin{center}
+      \rowcolors{2}{lightgray}{white}
+      \begin{tabular}{V{3} c | c | c | c V{3} c | c | c V{3}}
+      \hlineB{3}
+      $p$ & $q$ & $r$ & $s$ & $p \rightarrow q$ & $(p \rightarrow q) \rightarrow r$ & $((p \rightarrow q) \rightarrow r) \rightarrow s$\\
+      \hlineB{3}
+      0 & 0 & 0 & 0 & 1 & 0 & 1\\
+      \hline
+      1 & 0 & 0 & 0 & 0 & 1 & 0\\
+      \hline
+      0 & 1 & 0 & 0 & 1 & 0 & 1\\
+      \hline
+      0 & 0 & 1 & 0 & 1 & 1 & 0\\
+      \hline
+      0 & 0 & 0 & 1 & 1 & 0 & 1\\
+      \hline
+      1 & 1 & 0 & 0 & 1 & 0 & 1\\
+      \hline
+      1 & 0 & 1 & 0 & 0 & 1 & 0\\
+      \hline
+      1 & 0 & 0 & 1 & 0 & 1 & 1\\
+      \hline
+      0 & 1 & 1 & 0 & 1 & 1 & 0\\
+      \hline
+      0 & 1 & 0 & 1 & 1 & 0 & 1\\
+      \hline
+      0 & 0 & 1 & 1 & 1 & 1 & 1\\
+      \hline
+      1 & 1 & 1 & 0 & 1 & 1 & 0\\
+      \hline
+      1 & 1 & 0 & 1 & 1 & 0 & 1\\
+      \hline
+      1 & 0 & 1 & 1 & 0 & 1 & 1\\
+      \hline
+      0 & 1 & 1 & 1 & 1 & 1 & 1\\
+      \hline
+      1 & 1 & 1 & 1 & 1 & 1 & 1\\
+      \hlineB{3}
+    \end{tabular}
+    \end{center}
+  \end{table}
+
+  \newpage
+
+  \newgeometry{left=1.9in, right=1in}  
+  \reversemarginpar
+  \marginparwidth=1in
+
+  \section{Problem A}
+  
+  \marginnote{\color{cyan} \textbf{Example 11}}
+  
+  Construct the truth table of the compound proposition
+  
+  \vspace{0.5cm}
+  \indent\hspace{0.5cm}$(p \vee \neg q) \rightarrow (p \wedge q)$
+  \vspace{0.5cm}
+  
+  \color{cyan}\emph{Solution: }\color{black}Because this truth table involves two propositional variables $p$ and $q$, there are four rows in this truth table, one for each of the pairs of truth values TT, TF, FT, and FF. The first two columns are used for the truth values of p and q, respectively. In the third column we find the truth value of $\neg q$, needed to find the truth value of $p \vee \neg q$, found in the fourth column. The fifth column gives the truth value of $p \wedge q$. Finally, the truth value of $(p \vee \neg q) \rightarrow (p \wedge q)$ is found in the last column. The resulting truth table is shown in Table 7.
+  
+  \begin{table}[htb]
+      \setlength{\tabcolsep}{11pt}
+      \renewcommand{\arraystretch}{1.5}
+      \arrayrulecolor{cyan}
+      \arrayrulewidth=.75pt
+      \begin{tabular}{|c c|c|c|c|c|}
+        \hline
+        \rowcolor{cyan!13}\multicolumn{6}{|l|}{\color{cyan} \begin{large}\textbf{TABLE 7} \end{large} \color{black} \textbf{The Truth Table of} \bm{$(p \vee \neg q) \rightarrow (p \wedge q).$}}\\
+        \hline
+        \bm{$p$} & \bm{$q$} & \bm{$\neg q$} & \bm{$p \vee \neg q$} & \bm{$p \wedge q$} & \bm{$(p \vee \neg q) \rightarrow (p \wedge q)$}\\
+        \hline
+        T & T & F & T & T & T\\
+        T & F & T & T & F & F\\
+        F & T & F & F & F & T\\
+        F & F & T & T & F & F\\
+        \hline
+      \end{tabular}
+  \end{table}
+      
+\end{document}
\ No newline at end of file

hw02/HW02DanielShchur.tex

+\documentclass[12pt,letterpaper]{article}
+
+\usepackage[table]{xcolor}
+\usepackage{boldline}
+\usepackage{bm}
+\usepackage{forest}
+\usepackage[letterpaper, margin=.75in]{geometry}
+\usepackage{adjustbox}
+
+\newcommand{\problem}{\vspace{.5cm}\textbf{Problem:~~}}
+\newcommand{\answer}{\emph{Answer:~~}}
+\newcommand{\ansindent}{\indent\hspace{0.5cm}}
+\newcommand{\ansend}{\vspace{0.25cm}\\}
+\newcommand{\rot}[1]{\multicolumn{1}{c}{\adjustbox{angle=90}{#1}}‌​}
+\newcommand{\rotend}[1]{\multicolumn{1}{c V{3}}{\adjustbox{angle=90}{#1}}‌​}
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{0cm}
+\setlength{\intextsep}{.25cm}
+
+\title{CSCE 235H Homework 02}
+\author{Daniel Shchur}
+
+\begin{document}
+
+\maketitle
+
+\problem (Pg15 S1.1 \#24) Write each if these statements in the form ``if $p$, then $q$'' in English. [\emph{Hint:} Refer to the list of common ways to express conditional statements provided in this section.]
+
+\begin{enumerate}
+
+\item[b)] To be a citizen of this country it is sufficient that you were born in the United States.
+
+  \begin{enumerate}
+  \item[] \answer If you were born in the United States, then you will be a citizen.
+  \end{enumerate}
+
+\item[f)] The beach erodes whenever there is a storm.
+  
+  \begin{enumerate}
+  \item[] \answer If there is a storm, the beach will erode.
+  \end{enumerate}
+
+\item[g)] It is necessary to have a valid password to log on to the server.
+
+  \begin{enumerate}
+  \item[] \answer If you have a valid password, you will log onto the server.
+  \end{enumerate}
+
+\item[h)] You will reach the summit unless you begin your climb too late.
+
+  \begin{enumerate}
+  \item[] \answer If you do not begin your climb too late, then you will reach the summit.
+  \end{enumerate}
+
+\end{enumerate}
+  
+\problem (Pg15 S1.1 \#28) State the converse, contrapositive, and inverse of each of these conditional statements.
+
+\begin{enumerate}
+
+\item[a)] If it snows tonight, then I will stay at home.
+
+  \begin{enumerate}
+  \item[] \answer\\
+    \ansindent Converse: If I stay at home, it will snow tonight.\ansend
+    \ansindent Contrapositive: If I will not stay at home, it will not snow tonight.\ansend
+    \ansindent Inverse: If it does not snow tonight, then I will not stay at home.
+  \end{enumerate}
+
+\item[b)] I go to the beach whenever it is a sunny summer day.
+
+  \begin{enumerate}
+  \item[] \answer\\
+    \ansindent Converse: It is a sunny summer day whenever I go to the beach.\ansend
+    \ansindent Contrapositive: It is not a sunny summer day whenever I do not go to the beach.\ansend
+    \ansindent Inverse: I do not go to the beach whenever it is not a sunny summer day.
+  \end{enumerate}
+
+\item[c)] When I stay up late, it is necessary that I sleep until noon.
+
+  \begin{enumerate}
+  \item[] \answer\\
+    \ansindent Converse: When I sleep until noon, it is necessary that I stay up late.\ansend
+    \ansindent Contrapositive: When I do not sleep until noon, it is not necessary that I stay up late.\ansend
+    \ansindent Inverse: When I do not stay up late, it is not necessary that I sleep until noon.
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (Pg22 S1.2 \#8) Express these system specifications using the propositions $p$ ``The user enters a valid password,'' $q$ ``Access is granted,'' and $r$ ``The user has paid the subscription fee'' and logical connectives (including negations).
+
+\begin{enumerate}
+
+\item[a)] ``The user has paid the subscription fee, but does not enter a valid password.''
+
+  \begin{enumerate}
+  \item[] \answer $r \wedge \neg p$
+  \end{enumerate}
+
+\item[b)] ``Access is granted whenever the user has paid the subscription fee and enters a valid \mbox{password}.''
+
+  \begin{enumerate}
+  \item[] \answer $(r \wedge p) \rightarrow q$
+  \end{enumerate}
+
+\item[c)] ``Access is denied if the user has not paid the subscription fee.''
+
+  \begin{enumerate}
+  \item[] \answer $\neg r \rightarrow \neg q$
+  \end{enumerate}
+
+\item[d)] ``If the user has not entered a valid password but has paid the subscription fee, then access is granted.''
+
+  \begin{enumerate}
+  \item[] \answer $(\neg p \wedge r) \rightarrow q$
+  \end{enumerate}
+
+\end{enumerate}
+
+\newpage
+
+\problem (Pg23 S1.2 \#12) Are these system specifications consistent? ``If the file system is not locked, then new messages will be queued. If the file system is not locked, then the system is functioning normally, and conversely. If new messages are not queued, then they will be sent to the message buffer. If the file system is not locked, then new messages will be sent to the message buffer. New messages will not be sent to the message buffer.''\vspace{0.25cm}
+
+\ansindent\answer Yes, if the file system is locked, messages are queued, and the system isn't functioning\\\ansindent normally, then messages will not be sent to the message buffer. $(l = 1 \wedge q = 1 \wedge s = 0 \wedge b = 0)$ \ansend
+
+\ansindent Let:\ansend
+\ansindent\ansindent $l$: File system is locked\ansend
+\ansindent\ansindent $q$: Messages are queued\ansend
+\ansindent\ansindent $s$: System is functioning normally\ansend
+\ansindent\ansindent $b$: Messages are sent to file buffer\ansend
+\ansindent And let:\ansend
+\ansindent\ansindent $(\neg l \rightarrow q)$: ``If the file system is not locked, then new messages will be queued.''\ansend
+\ansindent\ansindent $(\neg l \leftrightarrow s)$: ``If the file system is not locked, then the system is functioning normally, and\\\ansindent\ansindent conversely.''\ansend
+\ansindent\ansindent $(\neg q \rightarrow b)$: ``If new messages are not queued, then they will be sent to the message buffer.''\ansend
+\ansindent\ansindent $(\neg l \rightarrow b)$: ``If the file system is not locked, then new messages will be sent to the message\\\ansindent\ansindent buffer.''\ansend
+\ansindent\ansindent $\neg b$: ``New messages will not be sent to the message buffer.''\ansend
+\ansindent Where:\ansend
+
+\begin{center}
+  Problem \#12 S1.2 Pg23\\
+  \begin{forest}
+    for tree={draw}
+    [{$(\neg l \rightarrow q) \wedge (\neg l \leftrightarrow s) \wedge (\neg q \rightarrow b) \wedge (\neg l \rightarrow b) \wedge \neg b$}
+      [{$l=0$}
+        [{$q=0$}, color={red}]
+        [{$q=1$}
+          [{$s=0$}, color={red}]
+          [{$s=1$}
+            [{$b=0$}, color={red}]
+            [{$b=1$}, color={red}]
+          ]
+        ]
+      ]
+      [{$l=1$}
+        [{$q=0$}
+          [{$s=0$}
+            [{$b=0$}, color={red}]
+            [{$b=1$}, color={red}]
+          ]
+          [{$s=1$}, color={red}]  
+        ]
+        [{$q=1$}
+          [{$s=0$}
+            [{$\bm{b=0}$}, color={green}]
+            [{$b=1$}, color={red}]
+          ]
+          [{$s=1$}, color={red}]
+        ]
+      ]  
+    ]
+    \end{forest}
+\end{center}
+
+\newpage
+
+\problem (Pg24 S1.2 \#34) Five friends have access to a chat room. Is it possible to determine who is chatting if the following information is known? Either Kevin or Heather, or both, are chatting. Either Randy or Vijay, but not both, are chatting. If Abby is chatting, so is Randy. Vijay and Kevin are either both chatting or neither is. If Heather is chatting, then so are Abby and Kevin. Explain your reasoning.\vspace{0.25cm}
+
+\ansindent \answer Yes, both Kevin and Vijay are chatting. See Table 1 on next page.\ansend
+
+\ansindent Let:\ansend
+\ansindent \ansindent $k$: Kevin is chatting\ansend
+\ansindent \ansindent $h$: Heather is chatting\ansend
+\ansindent \ansindent $r$: Randy is chatting\ansend
+\ansindent \ansindent $v$: Vijay is chatting\ansend
+\ansindent \ansindent $a$: Abby is chatting\ansend
+\ansindent And let:\ansend
+\ansindent \ansindent $(k \vee h)$: ``Either Kevin or Heather, or both, are chatting.''\ansend
+\ansindent \ansindent $(r \oplus v)$: ``Either Randy or Vijay, but not both, are chatting.''\ansend
+\ansindent \ansindent $a \rightarrow r$: ``If Abby is chatting, so is Randy.''\ansend
+\ansindent \ansindent $v \leftrightarrow k$: ``Vijay and Kevin are either both chatting or neither is.''\ansend
+\ansindent \ansindent $h \rightarrow (a \wedge k)$: ``If Heather is chatting, then so are Abby and Kevin.''\ansend
+\ansindent Where:\ansend
+\ansindent\ansindent $(k \vee h) \wedge (r \oplus v) \wedge (a \rightarrow r) \wedge (v \leftrightarrow k) \wedge (h \rightarrow (a \wedge k))$
+
+\newpage
+
+\begin{table}
+  \caption{Problem \#34 S1.2 Pg24}
+  \centering
+  \rowcolors{2}{red!20}{red!20}
+  \begin{tabular}{V{3} r V{3} c | c | c | c | c V{3} c | c | c | c | c V{3}}
+    \hlineB{3}
+    \# & $k$ & $h$ & $r$ & $v$ & $a$ & $(k \vee h)$ & $(r \oplus v)$ & $(a \rightarrow r)$ & $(v \leftrightarrow k)$ & $(h \rightarrow (a \wedge k))$\\
+    \hlineB{3}
+    1.  & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1\\
+    2.  & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1\\
+    3.  & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 1\\
+    4.  & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1\\
+    5.  & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\
+    6.  & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1\\
+    7.  & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 1\\
+    8.  & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 1\\
+    9.  & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0\\
+    10. & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0\\
+    11. & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 0 & 0\\
+    12. & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0\\
+    13. & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 0\\
+    14. & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 0\\
+    15. & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0\\
+    16. & 0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0\\
+    17. & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1\\
+    18. & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1\\\rowcolor{green!20}
+    19. & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\
+    20. & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1\\
+    21. & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1\\
+    22. & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1\\
+    23. & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 & 1\\
+    24. & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1\\
+    25. & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0\\
+    26. & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1\\
+    27. & 1 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 0\\
+    28. & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1\\
+    29. & 1 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 0\\
+    30. & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1\\
+    31. & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 1 & 0\\
+    32. & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1\\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+\clearpage
+
+\problem (Pg34 S1.3 \#6) Use a truth table to verify the first De Morgan law $\neg(p \wedge q) \equiv \neg p \vee \neg q$. \vspace{0.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#6 S1.3 Pg34}
+  \centering
+  \rowcolors{2}{lightgray!50}{white}
+  \begin{tabular}{V{3} c | c | c | c V{3} c | c | c V{3}}
+    \hlineB{3}
+    $p$ & $q$ & $\neg p$ & $\neg q$ & $p \wedge q$ & $\neg (p \wedge q)$ & $\neg p \vee \neg q$ \\
+    \hlineB{3}
+    0 & 0 & 1 & 1 & 0 & 1 & 1 \\
+    0 & 1 & 1 & 0 & 0 & 1 & 1 \\
+    1 & 0 & 0 & 1 & 0 & 1 & 1 \\
+    1 & 1 & 0 & 0 & 1 & 0 & 0 \\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+
+\problem (Pg35 S1.3 \#14) Determine whether $(\neg p \wedge (p \rightarrow q)) \rightarrow \neg q$ is a tautology.\vspace{0.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#14.1 S1.3 Pg35}
+  \centering
+  \begin{tabular}{l l | l}
+    \hline
+    1. & $(\neg p \wedge (p \rightarrow q)) \rightarrow \neg q$ & Original Clause\\
+    2. & $\equiv \neg(\neg p \wedge (p \rightarrow q)) \vee \neg q$ & Implication Rule\\
+    3. & $\equiv \neg(\neg p \wedge (\neg p \vee q)) \vee \neg q$ & Implication Rule\\
+    4. & $\equiv (p \vee \neg(\neg p \vee q)) \vee \neg q$ & DeMorgan's Law\\
+    5. & $\equiv (p \vee (p \wedge \neg q)) \vee \neg q$ & DeMorgan's Law\\
+    6. & $\equiv \neg q \vee \neg q$ & Absorption Law\\
+    7. & $\equiv \neg q$ & Idempotent Law\\
+    \hline
+    \multicolumn{3}{c}{Not a tautology}\\
+    \hline
+\end{tabular}
+\end{table}
+
+\begin{table}[ht]
+  \caption{Problem \#14.2 S1.3 Pg35}
+  \centering
+  \rowcolors{2}{lightgray!50}{white}
+  \begin{tabular}{V{3} c | c | c | c V{3} c | c V{3}
+    c V{3}}
+    \hlineB{3}
+    $p$ & $q$ & $\neg p$ & $\neg q$ & $p \rightarrow q$ & $\neg p \wedge (p \rightarrow q)$ & $(\neg p \wedge (p \rightarrow q)) \rightarrow \neg q $ \\
+    \hlineB{3}
+    0 & 0 & 1 & 1 & 1 & 1 & 1\\
+    0 & 1 & 1 & 0 & 1 & 1 & 0\\
+    1 & 0 & 0 & 1 & 0 & 0 & 1\\
+    1 & 1 & 0 & 0 & 1 & 0 & 1\\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+\newpage
+
+\problem (Pg35 S1.3 \#16) Show that $p \leftrightarrow q$ and $(p \wedge q) \vee (\neg p \wedge \neg q)$ are logically equivalent.\vspace{.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#16 S1.3 Pg35}
+  \centering
+  \begin{tabular}{l l | l}
+    \hline
+    1. & $p \leftrightarrow q$ & Original Clause\\
+    2. & $\equiv (p \rightarrow q) \wedge (q \rightarrow p)$ & Biconditional\\
+    3. & $\equiv (\neg p \vee q) \wedge (\neg q \vee p)$ & Implication\\
+    4. & $\equiv ((\neg p \vee q) \wedge \neg q) \vee ((\neg p \vee q) \wedge p)$ & Distributive Law\\
+    5. & $\equiv (\neg q \wedge (q \vee \neg p)) \vee (p \wedge (\neg p \vee q))$ & Commutative Law\\
+    6. & $\equiv ((\neg q \wedge q) \vee (\neg q \wedge \neg p)) \vee ((p \wedge \neg p) \vee (p \wedge q))$ & Distributive Law\\
+    7. & $\equiv (0 \vee (\neg q \wedge \neg p)) \vee (0 \vee (p \wedge q))$ & Negation Law\\
+    8. & $\equiv (\neg q \wedge \neg p) \vee (p \wedge q)$ & Identity Law\\
+    8. & $\equiv (p \wedge q) \vee (\neg q \wedge \neg p)$ & Commutative Law\\
+    \hline
+    \multicolumn{3}{c}{$p \leftrightarrow q \equiv (p \wedge q) \vee (\neg p \wedge \neg q)$}\\
+    \hline
+\end{tabular}
+\end{table}
+
+\problem (Pg35 S1.3 \#20) Show that $\neg (p \oplus q)$ and $p \leftrightarrow q$ are logically equivalent.\vspace{.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#14 S1.3 Pg35}
+  \centering
+  \rowcolors{2}{lightgray!50}{white}
+  \begin{tabular}{V{3} c | c V{3} c | c | c V{3} c | c V{3}}
+    \hlineB{3}
+    $p$ & $q$ & $p \rightarrow q$ & $q \rightarrow p$ & $p \oplus q$ & $\neg (p \oplus q)$ & $p \leftrightarrow q \equiv (p \rightarrow q) \wedge (q \rightarrow p)$\\
+    \hlineB{3}
+    0 & 0 & 1 & 1 & 0 & 1 & 1\\
+    0 & 1 & 1 & 0 & 1 & 0 & 0\\
+    1 & 0 & 0 & 1 & 1 & 0 & 0\\
+    1 & 1 & 1 & 1 & 0 & 1 & 1\\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+\problem (Pg35 S1.3 \#26) Show that $\neg p \rightarrow (q \rightarrow r)$ and $q \rightarrow (p \wedge r)$ are logically equivalent.\vspace{.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#16 S1.3 Pg35}
+  \centering
+  \begin{tabular}{l l | l}
+    \hline
+    1. & $\neg p \rightarrow (q \rightarrow r)$ & Original Clause\\
+    2. & $\equiv \neg p \rightarrow (\neg q \vee r)$ & Implication\\
+    3. & $\equiv p \vee (\neg q \vee r)$ & Implication\\
+    4. & $\equiv \neg q \vee (p \vee r)$ & Associative Law\\
+    5. & $\equiv q \rightarrow (p \vee r)$ & Implication\\
+    \hline
+    \multicolumn{3}{c}{$\neg p \rightarrow (q \rightarrow r) \equiv q \rightarrow (p \wedge r)$}\\
+    \hline
+\end{tabular}
+\end{table}
+
+\newpage
+
+\problem (Pg35 S1.3 \#28) Show that $p \leftrightarrow q$ and $\neg p \leftrightarrow \neg q$ are logically equivalent.\vspace{0.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#28 S1.3 Pg35}
+  \centering
+  \begin{tabular}{l l | l}
+    \hline
+    1. & $p \leftrightarrow q$ & Original Clause\\
+    2. & $\equiv (p \rightarrow q) \wedge (q \rightarrow p)$ & Biconditional\\
+    3. & $\equiv (\neg p \rightarrow \neg q) \wedge (\neg q \rightarrow \neg p)$ & Contrapositive\\
+    4. & $\equiv \neg p \leftrightarrow \neg q$ & Biconditional\\
+    \hline
+    \multicolumn{3}{c}{$p \leftrightarrow q \equiv \neg p \leftrightarrow \neg q$}\\
+    \hline
+\end{tabular}
+\end{table}
+
+\problem (Pg36 S1.3 \#46) Construct a truth table for the logical operator \emph{NAND}.\vspace{.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#46 S1.3 Pg36}
+  \centering
+  \rowcolors{2}{lightgray!50}{white}
+  \begin{tabular}{V{3} c | c V{3} c | c V{3}}
+    \hlineB{3}
+    $p$ & $q$ & $p \wedge q$ & $\neg(p \wedge q) \equiv p \uparrow q$\\
+    \hlineB{3}
+    0 & 0 & 0 & 1\\
+    0 & 1 & 0 & 1\\
+    1 & 0 & 0 & 1\\
+    1 & 1 & 1 & 0\\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+\problem (Pg36 S1.3 \#48) Construct a truth table for the logical operator \emph{NOR}.\vspace{.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#48 S1.3 Pg36}
+  \centering
+  \rowcolors{2}{lightgray!50}{white}
+  \begin{tabular}{V{3} c | c V{3} c | c V{3}}
+    \hlineB{3}
+    $p$ & $q$ & $p \vee q$ & $\neg(p \vee q) \equiv p \downarrow q$\\
+    \hlineB{3}
+    0 & 0 & 0 & 1\\
+    0 & 1 & 1 & 0\\
+    1 & 0 & 1 & 0\\
+    1 & 1 & 1 & 0\\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+\newpage
+
+\problem (Pg36 S1.3 \#62) Determine whether each of these compound propositions is satisfiable.
+
+\begin{enumerate}
+
+\item[a)] $(p \vee q \vee \neg r) \wedge (p \vee \neg q \vee \neg s) \wedge (p \vee \neg r \vee \neg s) \wedge (\neg p \vee \neg q \vee \neg s) \wedge (p \vee q \vee \neg s)$\vspace{.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#62.a S1.3 Pg36}
+  \centering
+  \rowcolors{2}{red!20}{red!20}
+  \begin{tabular}{V{3} c | c | c | c V{3} c | c | c | c | c V{3}}
+    \hlineB{3}
+    $p$ & $q$ & $r$ & $s$ & $(p \vee q \vee \neg r)$ & $(p \vee \neg q \vee \neg s)$ & $(p \vee \neg r \vee \neg s)$ & $(\neg p \vee \neg q \vee \neg s)$ & $(p \vee q \vee \neg s)$\\
+    \hlineB{3} \rowcolor{green!20}
+    0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\
+    0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 0\\
+    0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1\\
+    0 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0\\\rowcolor{green!20}
+    0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\
+    0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 1\\\rowcolor{green!20}
+    0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\
+    0 & 1 & 1 & 1 & 1 & 0 & 0 & 1 & 1\\\rowcolor{green!20}
+    1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\
+    1 & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1\\\rowcolor{green!20}
+    1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\
+    1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1\\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+\newpage
+
+\item[b)] $(\neg p \vee \neg q \vee r) \wedge (\neg p \vee q \vee \neg s) \wedge (p \vee \neg q \vee \neg s) \wedge (\neg p \vee \neg r \vee \neg s) \wedge (p \vee q \vee \neg r) \wedge (p \vee \neg r \vee \neg s)$\vspace{.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#62.b S1.3 Pg36}
+  \centering
+  \rowcolors{2}{red!20}{red!20}
+  \tabcolsep=.11cm
+  \begin{tabular}{V{3} c | c | c | c V{3} c | c | c | c | c | c V{3}}
+    \hlineB{3}
+    $p$ & $q$ & $r$ & $s$ & $(\neg p \vee \neg q \vee r)$ & $(\neg p \vee q \vee \neg s)$ & $(p \vee \neg q \vee \neg s)$ & $(\neg p \vee \neg r \vee \neg s)$ & $(p \vee q \vee \neg r)$ & $(p \vee \neg r \vee \neg s)$\\
+    \hlineB{3} \rowcolor{green!20}
+    0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
+    0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1\\
+    0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0\\\rowcolor{green!20}
+    0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\
+    0 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1\\\rowcolor{green!20}
+    0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\
+    0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0\\\rowcolor{green!20}
+    1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\
+    1 & 0 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\
+    1 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 1\\
+    1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\
+    1 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\
+    1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1\\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+\newpage
+
+\item[c)] $(p \vee q \vee r) \wedge (p \vee \neg q \vee \neg s) \wedge (q \vee \neg r \vee s) \wedge (\neg p \vee r \vee s) \wedge (\neg p \vee q \vee \neg s) \wedge (p \vee \neg q \vee \neg r) \wedge (\neg p \vee \neg q \vee s) \wedge (\neg p \vee \neg r \vee \neg s)$\vspace{.25cm}
+
+\ansindent \answer
+
+\begin{table}[ht]
+  \caption{Problem \#62.c S1.3 Pg36}
+  \centering
+  \rowcolors{2}{red!20}{red!20}
+  \begin{tabular}{V{3} c | c | c | c V{3} c | c | c | c | c | c | c | c V{3}}
+    \hlineB{3}
+    \multicolumn{1}{V{3} c}{$p$} & \multicolumn{1}{c}{$q$} & \multicolumn{1}{c}{$r$} & $s$ & \rot{$(p \vee q \vee r)$} & \rot{$(p \vee \neg q \vee \neg s)$} & \rot{$(q \vee \neg r \vee s)$} & \rot{$(\neg p \vee r \vee s)$} & \rot{$(\neg p \vee q \vee \neg s)$} & \rot{$(p \vee \neg q \vee \neg r)$} & \rot{$(\neg p \vee \neg q \vee s)$} & \rotend{$(\neg p \vee \neg r \vee \neg s)$}\\
+    \hlineB{3}
+    0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
+    0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
+    0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\\rowcolor{green!20}
+    0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
+    0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\
+    0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1\\
+    0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 1\\
+    1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1\\
+    1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1\\
+    1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\
+    1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0\\
+    1 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 1\\\rowcolor{green!20}
+    1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
+    1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1\\
+    1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0\\
+    \hlineB{3}
+  \end{tabular}
+\end{table}
+
+\end {enumerate}
+
+\end{document}
\ No newline at end of file

hw03/HW03DanielShchur.aux

+\relax 

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hw03/HW03DanielShchur.tex

+\documentclass[12pt,letterpaper]{article}
+
+\usepackage{bm}
+\usepackage{forest}
+\usepackage[letterpaper, margin=.75in]{geometry}
+
+\newcommand{\problem}{\vspace{.5cm}\textbf{Problem:~~}}
+\newcommand{\answer}{\emph{Answer:~~}}
+\newcommand{\ansindent}{\indent\hspace{0.5cm}}
+\newcommand{\answerindent}{\hangindent=.5cm\hangafter=0\answer}
+\newcommand{\ansend}{\vspace{0.25cm}\\}
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{0cm}
+\setlength{\intextsep}{.25cm}
+
+\title{CSCE 235H Homework 03}
+\author{Daniel Shchur}
+
+\begin{document}
+
+\maketitle
+
+\problem (\#10 S1.4 Pg53) Let $C(x)$ be the statement ``$x$ has a cat,'' let $D(x)$ be the statement ``$x$ has a dog,'' and let $F(x)$ be the statement ``$x$ has a ferret.'' Express each of these statements in terms of $C(x), D(x), F(x)$, quantifiers, and logical connectives. Let the domain consist of all students in your class.
+
+\begin{enumerate}
+
+\item[a)]A student in your class has a cat, a dog, and a ferret.
+  \begin{enumerate}
+  \item[]\answer $\exists x (C(x) \wedge D(x) \wedge F(x))$
+  \end{enumerate}
+
+\item[b)]All students in your class have a cat, a dog, or a ferret.
+  \begin{enumerate}
+  \item[]\answer $\forall x (C(x) \wedge D(x) \wedge F(x))$
+  \end{enumerate}
+  
+\item[c)]Some student in your class has a cat and a ferret, but not a dog.
+  \begin{enumerate}
+  \item[]\answer $\exists x (C(x) \wedge F(x) \wedge \neg D(x))$
+  \end{enumerate}
+
+\item[d)]No student in your class has a cat, a dog, and a ferret.
+  \begin{enumerate}
+  \item[]\answer $\neg \exists x (C(x) \wedge D(x) \wedge F(x))$
+  \end{enumerate}
+
+\item[e)]For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.
+  \begin{enumerate}
+  \item[]\answer$\exists x,y,z (C(x) \wedge D(y) \wedge F(z))$
+  \end{enumerate}
+
+\end{enumerate}
+
+
+\problem (\#14 S1.4 Pg53) Determine the truth value of each of these statements if the domain consists of all real numbers.
+
+\begin{enumerate}
+
+\item[a)]$\exists x (x^3 = -1)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[b)]$\exists x (x^4 < x^2)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\item[c)]$\forall x ((-x)^2 = x^2)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[d)]$\forall x (2x>x)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (\#28 S1.4 Pg54) Translate each of these statements into logical expressions using predicates, quantifiers, and logical connectives.
+
+\begin{enumerate}
+
+\item[a)]Something is not in the correct place.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[b)]All tools are in the correct place and are in excellent condition.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\item[c)]Everything is in the correct place and in excellent condition.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[d)]Nothing is in the correct place and is in excellent condition.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[e)]One of your tools is not in the correct place, but it is in excellent condition.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (\#32 S1.4 Pg55) Express each of these statements using quantifiers. Then form the negation of the statement so that no negation is to the left of a quantifier. Next, express the negation in simple English. (Do not simply use the phrase ``It is not the case that.'')
+
+\begin{enumerate}
+
+\item[a)]All dogs have fleas.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[b)]There is a horse that can add.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\item[c)]Every koala can climb.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[d)]No monkey can speak French.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[e)]There exists a pig that can swim and catch fish.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+
+\problem (\#34 S1.4 Pg55) Express the negation of these propositions using quantifiers, and then express the negation in English.
+
+\begin{enumerate}
+
+\item[b)]All Swedish movies are serious.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[c)]No one can keep a secret.
+  \begin{enumerate}
+  \item[]\answer 
+  \end{enumerate}
+  
+\item[d)]There is someone in this class who does not have a good attitude.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (\#36 S1.4 Pg55) Find a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all real numbers.
+
+\begin{enumerate}
+
+\item[a)]$\forall x (x^2 \neq x)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[b)]$\forall x (x^2 \neq 2)$
+  \begin{enumerate}
+  \item[]\answer 
+  \end{enumerate}
+  
+\item[c)]$\forall x (|x| > 0)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (\#38 S1.4 Pg55) Translate these system specifications into English where the predicate $S(x, y)$ is ``$x$ is in state $y$'' and where the domain for $x$ and $y$ consists of all systems and all possible states, respectively.
+
+\begin{enumerate}
+
+\item[b)]$\forall x(S(x$, malfunctioning$) \vee S(x$, diagnostic))
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[e)]$\forall x \neg S(x$, working)
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\end{enumerate}
+
+\problem (\#6 S1.5 Pg65) Let $C(x, y)$ mean that student $x$ is enrolled in class $y$, where the domain for $x$ consists of all students in your school and the domain for $y$ consists of all classes being given at your school. Express each of these statements by a simple English sentence.
+
+\begin{enumerate}
+
+\item[a)]$C$(Randy Goldberg, CS 252)
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[b)]$\exists x C(x$, Math 695)
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\item[d)]$\exists x (C(x$, Math 222$) \wedge C(x$, CS 252)
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (\#10 S1.5 Pg65) Let $F(x, y)$ be the statement ``$x$ can fool $y$,'' where the domain consists of all people in the world. Use quantifiers to express each of these statements.
+
+\begin{enumerate}
+
+\item[e)]Everyone can be fooled by somebody.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[g)]Nancy can fool exactly two people.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\item[h)]There is exactly one person whom everybody can fool.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[i)]No one can fool himself or herself.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[j)]There is someone who can fool exactly one person besides himself or herself.
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (\#28 S1.5 Pg67) Determine the truth value of each of these statements if the domain of each variable consists of all real numbers.
+
+\begin{enumerate}
+
+\item[a)]$\forall x \exists y (x^2 = y)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[c)]$\exists x \forall y (xy = 0)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\item[e)]$\forall x(x \neq 0 \rightarrow \exists y(xy = 1))$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[g)]$\forall x \exists y (x + y = 1)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[i)]$\forall x \exists y (x + y = 2 \wedge 2x - y = 1)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (\#30 S1.5 Pg67) Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logical connectives).
+
+\begin{enumerate}
+
+\item[b)]$\neg \forall x \exists P(x, y)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[d)]$\neg \exists y (\exists x R(x, y) \vee \forall x S(x, y))$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\item[e)]$\neg \exists y (\forall x \exists z T(x, y, z) \vee \exists x \forall z U(x, y, z))$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\end{enumerate}
+
+\problem (\#40 S1.5 Pg68) Find a counterexample, if possible, to these universally quantified statements, where the domain for all variables consists of all integers.
+
+\begin{enumerate}
+
+\item[a)]$\forall x \exists y(x = 1/y)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[b)]$\forall x \exists y(y^2 - x < 100)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[c)]$\forall x \forall y(x^2 \neq y^3)$
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+\problem (\#48 S1.5 Pg68) Show that $\forall x P(x) \vee \forall x Q(x)$ and $\forall x \forall y (P(x) \vee Q(y))$, where all quantifiers have the same nonempty domain, are logically equivalent. (the new variable $y$ is used to combine the quantifications correctly.)\vspace{.25cm}
+
+\answerindent
+
+\end{document}
\ No newline at end of file

hw03/HW03DanielShchur.tex~

+\documentclass[12pt,letterpaper]{article}
+
+\usepackage{bm}
+\usepackage{forest}
+\usepackage[letterpaper, margin=.75in]{geometry}
+
+\newcommand{\problem}{\vspace{.5cm}\textbf{Problem:~~}}
+\newcommand{\answer}{\emph{Answer:~~}}
+\newcommand{\ansindent}{\indent\hspace{0.5cm}}
+\newcommand{\answerindent}{\hangindent=.5cm\hangafter=0\answer}
+\newcommand{\ansend}{\vspace{0.25cm}\\}
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{0cm}
+\setlength{\intextsep}{.25cm}
+
+\title{CSCE 235H Homework 03}
+\author{Daniel Shchur}
+
+\begin{document}
+
+\maketitle
+
+\problem (\#10 S1.4 Pg53)
+
+\problem (\#14 S1.4 Pg53)
+
+\problem (\#28 S1.4 Pg54)
+
+\problem (\#32 S1.4 Pg55)
+
+\problem (\#34 S1.4 Pg55)
+
+\ansindent\answer
+
+\begin{enumerate}
+
+\item[b]
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\item[c]
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+  
+\item[d]
+  \begin{enumerate}
+  \item[]\answer
+  \end{enumerate}
+
+\end{enumerate}
+
+
+\end{document}
\ No newline at end of file

hwSat01/HWSAT01DanielShchur.aux

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hwSat01/HWSAT01DanielShchur.tex

+\documentclass[12pt,letterpaper]{article}
+
+\usepackage{bm}
+\usepackage{forest}
+\usepackage[letterpaper, margin=.75in]{geometry}
+
+\newcommand{\problem}[1]{\vspace{.5cm}\textbf{Problem{ #1}:~~}}
+\newcommand{\answer}{\emph{Answer:~~}}
+\newcommand{\ansindent}{\indent\hspace{0.5cm}}
+\newcommand{\answerindent}{\hangindent=.5cm\hangafter=0\answer}
+\newcommand{\ansend}{\vspace{0.25cm}\\}
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{0cm}
+\setlength{\intextsep}{.25cm}
+
+\title{CSCE 235H SAT Homework 01}
+\author{Daniel Shchur}
+
+\begin{document}
+
+\maketitle
+
+\problem{A} Consider the following CNF Formula:
+
+\begin{center}
+  \begin{tabular}{c c l}
+    $s1 =$ &          & $(a \vee b \vee c)$\\
+           & $\wedge$ & $(a \vee \neg b)$\\
+           & $\wedge$ & $(a \vee \neg c)$\\
+           & $\wedge$ & $(c \vee \neg a)$\\
+           & $\wedge$ & $(\neg a)$\\
+           & $\wedge$ & $(b)$\\
+  \end{tabular}
+\end{center}
+
+Is it satisfiable? If so, provide a solution and show how \emph{each} clause is satisfied in that solution. If not, identify clause(s) that cannot possibly be satisfied and explain why.\vspace{0.25cm}
+
+\answerindent The statement is not satisfiable because if $a=1$, the clause $(\neg a)$ is not satisfied, and if $a=0$ and $b=0$, then the clause $(b)$ is not satisfied, while if $b=1$, then the clause $(a \vee \neg b)$ is not satisfied. (See tree in Problem D)
+
+\problem{B} Consider the following CNF formula:
+
+\begin{center}
+  \begin{tabular}{c c l}
+    $s2 =$ &          & $(a \vee b)$\\
+           & $\wedge$ & $(\neg a \vee b)$\\
+           & $\wedge$ & $(a \vee c)$\\
+           & $\wedge$ & $(\neg b \vee \neg c)$\\
+  \end{tabular}
+\end{center}
+
+Is it satisfiable? If so, provide a solution and show how \emph{each} clause is satisfied in the solution. If not, identify clause(s) that cannot possibly be satisfied and explain why.\vspace{.25cm}
+
+\answerindent The statement is satisfiable by one solution, where $a=1$, $b=1$, and $c=0$. When those are as stated, $(a \vee b) \equiv (1 \vee 1)$, $(\neg a \vee b) \equiv (0 \vee 1)$, $(a \vee c) \equiv (1 \vee 0)$, and $(\neg b \vee \neg c) \equiv (0 \vee 1)$, which are all equivalent to true. (See tree in Problem D)
+
+\newpage
+
+\problem{C} Consider the following satisfiable CNF formula:
+
+\begin{center}
+  \begin{tabular}{c c l}
+    $s3 =$ &          & $(a \vee b \vee c \vee d)$\\
+           & $\wedge$ & $(\neg a \vee \neg b)$\\
+           & $\wedge$ & $(\neg c \vee \neg d)$\\
+           & $\wedge$ & $(a)$\\
+  \end{tabular}
+\end{center}
+
+How many unique solutions does it have? Explain your reasoning.\vspace{.25cm}
+
+\answerindent There are 3 unique solutions. This is easily visible with a model such as the tree for Problem C in Problem D.
+
+\problem{D} For each of the three CNF formulas from the previous problems, draw the binary tree of all possible combinations while pruning any branch as soon as it cannot be extended in a satisfiable manner.
+
+\begin{center}
+  Problem A\ansend
+  \begin{forest}
+    for tree=draw
+    [{$(a \vee b \vee c) \wedge (a \vee \neg b) \wedge (b \vee \neg c) \wedge (c \vee \neg a) \wedge (\neg a) \wedge (b)$}
+      [{$a=0$}
+        [{$b=0$}, color=red]
+        [{$b=1$}, color=red]
+      ]
+      [{$a=1$}, color=red]
+    ]
+  \end{forest}
+\end{center}
+
+\begin{center}
+  Problem B\ansend
+  \begin{forest}
+    for tree=draw
+    [{$(a \vee b) \wedge (\neg a \vee b) \wedge (a \vee c) \wedge (\neg b \vee \neg c)$}
+      [{$a=0$}
+        [{$b=0$}, color=red]
+        [{$b=1$}
+          [{$c=0$}, color=red]
+          [{$c=1$}, color=red]
+        ]
+      ]
+      [{$a=1$}
+        [{$b=0$}, color=red]
+        [{$b=1$}
+          [{$c=0$}, color={green!75!black}]
+          [{$c=1$}, color={red}]
+        ]
+      ]
+    ]
+  \end{forest}
+\end{center}
+
+\begin{center}
+  Problem C\ansend
+  \begin{forest}
+    for tree=draw
+    [{$(a \vee b \vee c \vee d) \wedge (\neg a \vee \neg b) \wedge (\neg c \vee \neg d) \wedge (a)$}
+      [{$a=0$}, color=red]
+      [{$a=1$}
+        [{$b=0$}
+          [{$c=0$}
+            [{$d=0$}, color={green!75!black}]
+            [{$d=1$}, color={green!75!black}]
+          ]
+          [{$c=1$}
+            [{$d=0$}, color={green!75!black}]
+            [{$d=1$}, color=red]
+          ]
+        ]
+        [{$b=1$}, color=red]
+      ]
+    ]
+  \end{forest}
+\end{center}
+
+\end{document}
\ No newline at end of file

hwSat01/HWSAT01DanielShchur.tex~

+\documentclass[12pt,letterpaper]{article}
+
+\usepackage[table]{xcolor}
+\usepackage{boldline}
+\usepackage{bm}
+\usepackage{forest}
+\usepackage[letterpaper, margin=.75in]{geometry}
+\usepackage{adjustbox}
+
+\newcommand{\problem}{\vspace{.5cm}\textbf{Problem:~~}}
+\newcommand{\answer}{\emph{Answer:~~}}
+\newcommand{\ansindent}{\indent\hspace{0.5cm}}
+\newcommand{\ansend}{\vspace{0.25cm}\\}
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+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{0cm}
+\setlength{\intextsep}{.25cm}
+
+\title{CSCE 235H SAT Homework 01}
+\author{Daniel Shchur}