new teachers, new work

18 files changed, 2084 insertions, 5 deletions

diff --git a/.gitignore b/.gitignore
index dedd592..3417ff3 100644
--- a/.gitignore
+++ b/.gitignore
@@ -1,5 +1,13 @@
 *.log
 *.pdf
+*.aux
+*.bbl
+*.blg
 *.ps
+*.out
+*.run.xml
+*.bcf
+
 li/output.txt
 li/insurance.csv
+stanzione/yang.jpg
diff --git a/README b/README
index bf457b0..3c5539c 100644
--- a/README
+++ b/README
@@ -8,6 +8,11 @@
 - Mayya Zhilova for MATH 4261: Math Statistics I
 - Martin M Jarrio for PHYS 2212: Intro Physics II
 
+- Ronnie Howard for CS 2050: Intro Discrete Math
+- Christopher Stanzione for PSYC 1101: Introduction to Psychology
+- Michelle Rosbruck for APPH 1040: Sci Foundation of Health
+- Neha Gupta for MATH 2106: Foundations of Math Proof
+
 ## Structure
 
 The growth of this repository will be pretty organic, but as a general
@@ -23,8 +28,3 @@ for when different events occur.
 
 I have chosen to release this writing under CC BY-SA 4.0.
 The full license is available in LICENSE.
-
-The license does not apply to images that are used in
-`hireme/final-pres.odp` (which are used under an exception to copyright)
-nor any PDF forms I have included (which are likely insufficiently
-original to warrant copyright protection).
diff --git a/gupta/Makefile b/gupta/Makefile
new file mode 100644
index 0000000..1188bc9
--- /dev/null
+++ b/gupta/Makefile
@@ -0,0 +1,18 @@
+.POSIX:
+.SUFFIXES: .tex .pdf
+
+PDFTEX = pdftex
+PDFLATEX = pdflatex
+
+.tex.pdf:
+	$(PDFTEX) $<
+
+all: hw1.pdf hw2.pdf hw3.pdf hw4.pdf hw5.pdf
+
+clean:
+	rm -f *.pdf *.log *.aux
+
+hw2.pdf: hw2.tex
+	$(PDFLATEX) hw2.tex
+	$(PDFLATEX) hw2.tex
+	$(PDFLATEX) hw2.tex
diff --git a/gupta/hw1.tex b/gupta/hw1.tex
new file mode 100644
index 0000000..5519891
--- /dev/null
+++ b/gupta/hw1.tex
@@ -0,0 +1,18 @@
+\noindent{\it (a)}
+
+I have access to eduroam and I live on campus, so my internet is stable.
+I have a scanner app for pictures of paper work, and I have a mic and
+camera for online lectures. I don't have any other technological
+concerns.
+
+\noindent{\it (b)}
+
+I'm a first-year Math and CS double major, and I'm taking this class
+because I'm interested in abstract algebra and analysis, which have this
+class as a prerequisite.
+
+\noindent{\it (c)}
+
+I don't have any other concerns about this class.
+
+\bye
diff --git a/gupta/hw2.tex b/gupta/hw2.tex
new file mode 100644
index 0000000..922703a
--- /dev/null
+++ b/gupta/hw2.tex
@@ -0,0 +1,386 @@
+\documentclass[10pt,twoside]{article}
+
+\usepackage{amssymb,amsmath,amsthm,amsfonts, epsfig, graphicx, dsfont,
+  bbm, bbold, url, color, setspace, multirow,pinlabel,tikz,pgfplots}
+\usepackage[all]{xy}
+
+\usepackage{fancyhdr} \setlength{\voffset}{-1in}
+\setlength{\topmargin}{0in} \setlength{\textheight}{9.5in}
+\setlength{\textwidth}{6.5in} \setlength{\hoffset}{0in}
+\setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in}
+\setlength{\marginparsep}{0in} \setlength{\marginparwidth}{0in}
+\setlength{\headsep}{0.25in} \setlength{\headheight}{0.5in}
+\pagestyle{fancy}
+
+\onehalfspace
+
+\fancyhead[LO,LE]{Math 2106 - Dr. Gupta} \fancyhead[RO,RE]{Due Thursday 1/20/2022 at 11:59 pm}
+\chead{\textbf{}} \cfoot{}
+\fancyfoot[LO,LE]{} \fancyfoot[RO,RE]{Page \thepage\ of
+  \pageref{LastPage}} \renewcommand{\footrulewidth}{0.5pt}
+\parindent 0in
+
+\makeatletter
+\def\old@comma{,}
+\catcode`\,=13
+\def,{%
+  \ifmmode%
+    \old@comma\discretionary{}{}{}%
+  \else%
+    \old@comma%
+  \fi%
+}
+\makeatother
+
+\let\implies\Rightarrow
+\def\lnot{{\sim}}
+
+%% ------------------------------------------------------%%
+%% -------------------Begin Document---------------------%%
+%% ------------------------------------------------------%%
+\begin{document}
+
+\begin{center}
+    \huge{\bf{Homework 2} - Holden Rohrer}
+\end{center}
+
+\medskip
+
+\noindent \large{\textbf{Collaborators:}}
+
+\medskip
+
+\begin{itemize}
+	\item Hammack 1.1: 16, 28, 52
+	\begin{itemize}
+	    \item[16.] Write the set $\{6a + 2b ~|~ a, b \in \mathbb Z\}$ by listing its elements between curly braces.
+	    \begin{proof}[Answer]
+            If and only if $x$ is in this set, $x+6$ and $x+2$ are in
+            this set. $0$ is in this set with $a=0$ and $b=0.$ This is
+            the set of even numbers $\{\ldots,-4,-2,0,2,4,\ldots\}.$
+	    \end{proof}
+        \item[28.] Write the set $\{\ldots, -\frac32, -\frac34, 0,
+            \frac34, \frac32, \frac94, 3, \frac{15}4, \frac92\}$ in
+            set-biulder notation.
+        \begin{proof}[Answer]
+            This set is $\{\frac34 x : x \in \mathbb Z\}.$
+        \end{proof}
+        \item[52.] Sketch the set of points $\{(x,y)\in\mathbb R^2 :
+            (y-x^2)(y+x^2) = 0\}.$
+        \begin{proof}[Answer]
+            This set is the union of the two graphs $y = x^2$ and $y =
+            -x^2.$
+            \begin{figure}[h!]
+            \centering
+            \begin{tikzpicture}[scale=3]
+                \draw[<->] (-1,0) -- (1,0) node[right] {$x$};
+                \draw[<->] (0,-1) -- (0,1) node[above] {$y$};
+                \draw (0,0) parabola (1,1);
+                \draw (0,0) parabola (-1,1);
+                \draw (0,0) parabola (1,-1);
+                \draw (0,0) parabola (-1,-1);
+            \end{tikzpicture}
+            \end{figure}
+            %%TO DRAW
+        \end{proof}
+	\end{itemize}
+	\item Hammack 1.2: 2
+    \begin{itemize}
+        \item[2.] Suppose $A = \{\pi,e,0\}$ and $B = \{0,1\}.$
+        \begin{itemize}
+            \item[(a)] Write out $A\times B.$
+            \begin{proof}[Answer]
+                $A\times B = \{(\pi,0), (\pi, 1), (e,0), (e,1), (0,0),
+                (1,0)\}.$
+            \end{proof}
+            \item[(b)] Write out $B\times A.$
+            \begin{proof}[Answer]
+                $B\times A = \{(0,\pi), (1,\pi), (0,e), (1,e), (0,0),
+                (1,0)\}.$
+            \end{proof}
+            \item[(c)] Write out $A\times A.$
+            \begin{proof}[Answer]
+                $A\times A = \{(\pi,\pi), (\pi,e), (\pi, 0), (e,\pi),
+                (e,e), (e,0), (0,\pi), (0,e), (0,0)\}.$
+            \end{proof}
+            \item[(d)] Write out $B\times B.$
+            \begin{proof}[Answer]
+                $B\times B = \{(0,0), (0,1), (1,0), (1,1)\}.$
+            \end{proof}
+            \item[(e)] Write out $A\times \emptyset.$
+            \begin{proof}[Answer]
+                $A\times\emptyset = \emptyset.$
+            \end{proof}
+            \item[(f)] Write out $(A\times B)\times B.$
+            \begin{proof}[Answer]
+                $(A\times B)\times B = \{((\pi,0),0), ((\pi, 1),0),
+                ((e,0),0), ((e,1),0), ((0,0),0), ((0,1),0),
+                ((\pi,0),1), ((\pi, 1),1),
+                ((e,0),1), ((e,1),1), ((0,0),1), ((0,1),1)\}.$
+            \end{proof}
+            \item[(g)] Write out $A\times(B\times B).$
+            \begin{proof}[Answer]
+                $A\times(B\times B) = \{(\pi,(0,0)), (\pi,(1,0)),
+                (e,(0,0)), (e,(1,0)), (0,(0,0)), (0,(1,0)),
+                (\pi,(0,1), (\pi,(1,1)),
+                (e,(0,1)), (e,(1,1)), (0,(0,1)), (0,(1,1))\}.$
+            \end{proof}
+            \item[(h)] Write out $A\times B\times B.$
+            \begin{proof}[Answer]
+                $A\times B\times B = \{(\pi,0,0), (\pi, 1,0),
+                (e,0,0), (e,1,0), (0,0,0), (0,1,0),
+                (\pi,0,1), (\pi, 1,1),
+                (e,0,1), (e,1,1), (0,0,1), (0,1,1)\}.$
+            \end{proof}
+        \end{itemize}
+    \end{itemize}
+	\item Hammack 1.3: 2, 8, 10, 16
+    \begin{itemize}
+        \item[2.] List all subsets of $\{1,2,\emptyset\}.$
+        \begin{proof}[Answer]
+            The subsets are $\emptyset, \{1\}, \{2\}, \{\emptyset\},
+            \{1,2\}, \{1,\emptyset\}, \{2,\emptyset\},
+            \{1,2,\emptyset\}.$
+        \end{proof}
+        \item[8.] List all subsets of $\{\{0,1\}, \{0,1,\{2\}\},
+            \{0\}\}.$
+        \begin{proof}[Answer]
+            The subsets are $\emptyset, \{\{0,1\}\}, \{\{0,1,\{2\}\}\},
+            \{\{0\}\}, \{\{0,1\},\{0\}\}, \{\{0,1\},\{0,1,\{2\}\}\},
+            \{\{0\},\{0,1,\{2\}\}\}, \{\{0,1\},\{0\},\{0,1,\{2\}\}\}.$
+        \end{proof}
+        \item[10.] Write out $\{X\subseteq N : |X| \leq 1\}.$
+        \begin{proof}[Answer]
+            This set is $\{-1,0,1\}.$
+        \end{proof}
+        \item[16.] Decide if $\{(x,y) : x^2 - x = 0\} \subseteq
+            \{(x,y) : x-1 = 0\}$ is true or false.
+        \begin{proof}[Answer]
+            False. $(0,0)$ is in the first set but not in the second
+            set, so it cannot be a subset.
+        \end{proof}
+    \end{itemize}
+	\item Hammack 1.4: 6, 14, 16, 18, 20
+    \begin{itemize}
+        \item[6.] Find $\mathcal P(\{1,2\})\times\mathcal P(\{3\}).$
+        \begin{proof}[Answer]
+            This is $\{(\emptyset,\emptyset), (\{1\},\emptyset),
+            (\{2\},\emptyset), (\{1,2\},\emptyset), (\emptyset, \{3\}),
+            (\{1\}, \{3\}), (\{2\}, \{3\}), (\{1,2\}, \{3\}).$
+        \end{proof}
+        \item[14.] Suppose $|A| = m$ and $|B| = n.$ Find
+            $|\mathcal P(\mathcal P(A))|.$
+        \begin{proof}[Answer]
+            $|\mathcal P(\mathcal P(A))| = 2^{|\mathcal P(A)|} =
+            2^{2^m}.$
+        \end{proof}
+        \item[16.] Suppose $|A| = m$ and $|B| = n.$ Find
+            $|\mathcal P(A)\times \mathcal P(B)|.$
+        \begin{proof}[Answer]
+            This is $|\mathcal P(A)|\cdot|\mathcal P(B)| = 2^{|\mathcal
+            P(A)|}2^{|\mathcal P(B)|} = 2^{nm}$
+        \end{proof}
+        \item[18.] Suppose $|A| = m$ and $|B| = n.$ Find
+            $|\mathcal P(A\times \mathcal P(B))|.$
+        \begin{proof}[Answer]
+            By similar techniques, this set has cardinality $2^{m2^n}.$
+        \end{proof}
+        \item[20.] Suppose $|A| = m$ and $|B| = n.$ Find $|\{X\subseteq\mathcal P(A) : |X|\leq 1\}|.$
+        \begin{proof}[Answer]
+            This is $\emptyset$ and every one-element subset of
+            $\mathcal P(A),$ so it has cardinality $2^m + 1.$
+        \end{proof}
+    \end{itemize}
+	\item Hammack 1.6: 2
+    \begin{itemize}
+        \item[2.] Let $A = \{0,2,4,6,8\}$ and $B = \{1,3,5,7\}$ have
+            universal set $U = \{0,1,2,\ldots,8\}.$ Find:
+        \begin{itemize}
+            \item[(a)] $\bar A = B.$
+            \item[(b)] $\bar B = A.$
+            \item[(c)] $A\cap\bar A = \emptyset.$
+            \item[(d)] $A\cup\bar A = U.$
+            \item[(e)] $A - \bar A = A.$
+            \item[(f)] $\bar{A\cup B} = \emptyset.$
+            \item[(g)] $\bar A \cap \bar B = \emptyset.$
+            \item[(h)] $\bar{A\cap B} = U.$
+            \item[(i)] $\bar A \times B = B^2.$
+        \end{itemize}
+        %\begin{proof}[Answer]
+        %\end{proof}
+    \end{itemize}
+	\item Hammack 1.8: 4, 10, 12, 14
+    \begin{itemize}
+        \item[4.] For each $n\in \mathbb N,$ let $A_n = \{-2n,0,2n\}.$
+        \begin{itemize}
+            \item[(a)] $\bigcup_{i\in N} A_i = \{2n:n\in\mathbb N\}$
+            \item[(b)] $\bigcap_{i\in N} A_i = \{0\}$
+        \end{itemize}
+        \item[10.]
+        \begin{itemize}
+            \item[(a)] $\bigcup_{i\in [0,1]} [x,1]\times[0,x^2] =
+                \{(x,y)\in \mathbb [0,1]^2 : y \leq x^2\}.$
+            \item[(b)] $\bigcap_{i\in [0,1]} [x,1]\times[0,x^2] =
+                \{(1,0)\},$ as can be seen from the intersection of the
+                $[0,1]\times[0,0]$ and $[1,1]\times[0,1]$ elements.
+        \end{itemize}
+        \item[12.] If $\bigcap_{\alpha\in I} A_\alpha =
+            \bigcup_{\alpha\in I} A_\alpha,$ what do you think can be
+            said about the relationships between the sets $A_{\alpha}?$
+        \begin{proof}[Answer]
+            For any $\alpha,\beta \in I,$ $A_\alpha = A_\beta.$
+        \end{proof}
+        \item[14.] If $J\neq\emptyset$ and $J\subseteq I,$ does it
+            follow that $\bigcap_{\alpha\in I} A_\alpha \subseteq
+            \bigcap_{\alpha\in J} A_\alpha?$ Explain.
+        \begin{proof}[Answer]
+            Yes, this does follow because
+            $\bigcap_{\alpha\in I} A_\alpha = \bigcap_{\alpha\in J}
+            A_\alpha \cap \bigcap_{\beta\in I - J} A_\beta
+            \subseteq \bigcap_{\alpha\in J} A_\alpha.$
+        \end{proof}
+    \end{itemize}
+	\item Hammack 2.3: 2, 6
+    \begin{itemize}
+        \item[2.] Convert the following sentence into the form ``{\em If
+            P, then Q}.'' ``For a function to be continuous, it is
+            sufficient that it is differentiable.''
+        \begin{proof}[Answer]
+            ``If a function is differentiable, that function is
+            continuous.''
+        \end{proof}
+        \item[6.] Convert the following sentence into the form ``{\em If
+            P, then Q}.'' ``Whenever a surface has only one side, it is
+            non-orientable.''
+        \begin{proof}[Answer]
+            ``If a surface has one side, that surface is
+            non-orientable.''
+        \end{proof}
+    \end{itemize}
+	\item Hammack 2.5: 10
+    \begin{itemize}
+        \item[10.] Suppose the statement $((P\land Q)\lor R) \implies
+            (R\lor S)$ is false. Find the truth values of $P$, $Q,$ $R,$
+            and $S.$
+        \begin{proof}[Answer]
+            $P=Q={\rm True}$ and $R=S={\rm False}$ make this statement
+            false.
+        \end{proof}
+    \end{itemize}
+	\item Hammack 2.6: 14
+    \begin{itemize}
+        \item[14.] Decide whether or not $P\land(Q\lor\lnot Q)$ and
+            $(\lnot P) \implies (Q\land\lnot Q)$ are logically
+            equivalent.
+        \begin{proof}[Answer]
+            These are logically equivalent. $Q\lor ~Q \equiv T$ and
+            $Q\land ~Q \equiv F,$ giving us simplified expressions
+            $P\land T \equiv P$ and $\lnot P \implies F \equiv P,$ so
+            these are logically equivalent expressions.
+        \end{proof}
+    \end{itemize}
+	\item Problems not from the textbook
+  	\begin{enumerate}
+         \item Use truth tables to prove that each of the following compound propositions is \emph{not} a tautology. These implication are common logical fallacies (errors in reasoning) since the conclusion does not follow from the hypotheses.
+         \begin{enumerate}
+              \item $[(P\implies Q)\land Q]\implies P$
+              \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \implies Q$ & $(P \implies Q)\land
+                      Q$ & $[(P\implies Q)\land Q] \implies P$ \\\hline
+
+                      T & T & T & T & T\\\hline
+                      T & F & F & F & T\\\hline
+                      F & T & T & T & F\\\hline
+                      F & F & T & F & T\\\hline
+                  \end{tabular}
+              \end{proof}
+  			  \item $[(P\implies Q)\land\lnot{P}]\implies\lnot{Q}$
+              \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \implies Q$ & $(P \implies Q)\land
+                      \lnot P$ & $[(P\implies Q)\land \lnot P] \implies
+                      \lnot Q$ \\\hline
+
+                      T & T & T & F & T\\\hline
+                      T & F & F & F & T\\\hline
+                      F & T & T & T & F\\\hline
+                      F & F & T & T & T\\\hline
+                  \end{tabular}
+              \end{proof}
+ 		\end{enumerate}
+		\item Use truth tables to prove that each of the following compound propositions is a tautology. These are the four most important ``rules of inference'' in propositional logic. Each rule gives a conclusion that follows from a set of hypotheses and thus give building blocks for correct proofs.
+ 		 \begin{enumerate}
+             \item $[P\land(P\implies Q)]\implies Q$
+             \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \implies Q$ &
+                      $P\land (P \implies Q)$ &
+                      $[P\land (P\implies Q)] \implies Q$ \\\hline
+
+                      T & T & T & T & T\\\hline
+                      T & F & F & F & T\\\hline
+                      F & T & T & F & T\\\hline
+                      F & F & T & F & T\\\hline
+                  \end{tabular}
+             \end{proof}
+			 \item $[\lnot Q\land(P\implies Q)]\implies\lnot{P} $
+             \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \implies Q$ & $(P \implies Q)\land
+                      \lnot P$ & $[(P\implies Q)\land \lnot P] \implies
+                      \lnot Q$ \\\hline
+
+                      T & T & T & F & T\\\hline
+                      T & F & F & F & T\\\hline
+                      F & T & T & T & F\\\hline
+                      F & F & T & T & T\\\hline
+                  \end{tabular}
+             \end{proof}
+			 \item $[(P\implies Q)\land(Q\implies R)]\implies(P\implies R)$
+             \begin{proof}[Answer]
+                 Let $(1)$ be $(P\implies Q)\land(Q\implies R).$
+
+                  \begin{tabular}{|c|c|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $R$ & $P \implies Q$ &
+                      $Q \implies R$ &
+                      $(1)$ &
+                      $P \implies R$ &
+                      $(1) \implies (P\implies R)$ \\\hline
+
+                      T & T & T & T & T & T & T & T\\\hline
+                      T & T & F & T & F & F & F & T\\\hline
+                      T & F & T & F & T & F & T & T\\\hline
+                      T & F & F & F & T & F & F & T\\\hline
+                      F & T & T & T & T & T & T & T\\\hline
+                      F & T & F & T & F & F & T & T\\\hline
+                      F & F & T & T & T & T & T & T\\\hline
+                      F & F & F & T & T & T & T & T\\\hline
+                      % to break up lol
+                  \end{tabular}
+             \end{proof}
+             \item $[(P\lor Q)\land\lnot{P}]\implies Q$
+             \begin{proof}[Answer]
+                  \begin{tabular}{|c|c|c|c|c|c|}
+                      \hline
+                      $P$ & $Q$ & $P \lor Q$ & $(P \lor Q)\land
+                      \lnot P$ & $[(P\lor Q)\land \lnot P] \implies Q$
+                      \\\hline
+
+                      T & T & T & F & T\\\hline
+                      T & F & T & F & T\\\hline
+                      F & T & T & T & T\\\hline
+                      F & F & F & F & T\\\hline
+                  \end{tabular}
+             \end{proof}
+		\end{enumerate}
+	\end{enumerate}
+\end{itemize}
+
+\label{LastPage}
+\end{document}
diff --git a/gupta/hw3.tex b/gupta/hw3.tex
new file mode 100644
index 0000000..1961ce3
--- /dev/null
+++ b/gupta/hw3.tex
@@ -0,0 +1,151 @@
+\newfam\bbold
+\def\bb#1{{\fam\bbold #1}}
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\font\bigbf=cmbx12 at 24pt
+
+\def\answer{\smallskip\bgroup}
+\def\endanswer{\egroup\medskip}
+\def\section#1{\medskip\goodbreak\noindent{\bf #1}}
+\let\impl\Rightarrow
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-01-27 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 3 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 2.7: 2, 9, 10}
+
+\item{2.} Write the following as an English sentence:
+$\forall x\in\bb R, \exists n\in \bb N, x^n\geq 0.$
+
+\answer
+For all real numbers $x,$ there is a natural number $n$ such that $x^n$
+is nonnegative.
+This statement is true because, for all real numbers, $x^2 \geq 0$ and
+$2\in\bb N.$
+\endanswer
+
+\item{9.} Write the following as an English sentence:
+$\forall n\in\bb Z, \exists m\in\bb Z, m = n+5.$
+
+\answer
+For all integers $n,$ there is an integer $m$ which is 5 greater than
+$n.$
+This statement is true because the integers are closed under addition.
+\endanswer
+
+\item{10.} Write the following as an English sentence:
+$\exists m\in\bb Z, \forall n\in\bb Z, m = n + 5.$
+
+\answer
+There is an integer $m$ such that for all integers $n,$ $m$ is 5 greater
+than $n.$
+This statement is false because $m$ cannot equal $0+5$ and $1+5$ at the
+same time.
+\endanswer
+
+\section{Hammack 2.9: 1, 7, 10}
+
+\item{1.} Translate the following sentence into symbolic logic: ``If $f$
+is a polynomial and its degree is greater than 2, then $f'$ is not
+constant.
+\answer
+Where $P$ is the set of polynomials, and $\mathop{\rm degree}(p)$ is the
+degree of a polynomial $p,$
+$$\forall p\in P, \left(\mathop{\rm degree}(p) > 2\right) \impl \exists
+a,b\in\bb R, f'(a) \neq f'(b).$$
+\endanswer
+
+\item{7.} Translate the following sentence into symbolic logic: ``There
+exists a real number $a$ for which $a+x = x$ for every real number $x.$
+\answer
+$$\exists a\in\bb R, \forall x\in\bb R, a+x = x.$$
+\endanswer
+
+\item{10.} Translate the following sentence into symbolic logic: ``If
+$\sin(x) < 0,$ then it is not the case that $0\leq x\leq\pi.$
+\answer
+$$\forall x\in\bb R, \sin(x) < 0 \impl \lnot(0\leq x\leq\pi).$$
+\endanswer
+
+\section{Hammack 2.10: 2, 5, 10}
+\item{2.} Negate the following sentence: ``If $x$ is prime, then $\sqrt
+x$ is not a rational number.''
+
+\answer
+There is a prime number $x$ such that $\sqrt x$ is a rational number.
+\endanswer
+
+\item{5.} Negate the following sentence: ``For every positive number
+$\epsilon,$ there is a positive number $M$ for which $|f(x)-b|<\epsilon$
+whenever $x > M.$
+
+\answer
+There is a positive number $\epsilon$ such that for all $M$ there is an
+$x > M$ such that $|f(x)-b|>\epsilon$
+\endanswer
+
+\item{10.} If $f$ is a polynomial and its degree is greater than 2, then
+$f'$ is not constant.
+
+\answer
+There is a polynomial with degree greater than 2 such that $f'$ is
+constant.
+\endanswer
+
+\section{Hammack 4: 4, 12, 20}
+
+\item{4.} Prove ``Suppose $x,y\in\bb Z.$ If $x$ and $y$ are odd, then
+$xy$ is odd'' with direct proof.
+\answer
+Suppose $x,y\in\bb Z$ and that $x$ and $y$ are odd.
+Since $x$ is odd, there exists $j\in\bb Z$ such that $x = 2j+1.$
+Since $y$ is odd, there exists $k\in\bb Z$ such that $y = 2k+1.$
+$xy = (2j+1)(2k+1) = 4jk + 2j + 2k + 1 = 2(2jk + j + k) + 1.$
+Because $2jk + j + k$ is an integer, $xy$ is odd because it is one more
+than two times an integer.
+\endanswer
+
+\item{12.} Prove ``If $x\in\bb R$ and $0<x<4,$ then ${4\over x(4-x)}\geq
+1.$'' with direct proof.
+\answer
+Let $x\in\bb R$ and $0<x<4.$
+$$(x-2)^2 \geq 0 \to 4 - (x-2)^2\leq 4\to 4x-x^2 = x(4-x) \leq 4 \to
+{4\over x(4-x)}\geq {4\over 4} = 1.$$
+\endanswer
+
+\item{20.} Prove ``If $a$ is an integer, and $a^2|a,$ then
+$a\in\{-1,0,1\}.$'' with direct proof.
+
+\answer
+$a^2|a$ requires $a^2 \leq |a|.$ For $|a| > 1,$ $a^2 = |a|^2 > |a|,$ so
+we will check $a^2|a$ for the remaining cases $\{-1,0,1\}.$
+
+$n|m$ iff there is a $k\in\bb Z,$ $k\neq 0,$ $m = nk.$
+For $0,$  $0^2 = 1(0),$ so $0^2|0.$
+For $1,$  $1^2 = 1(1),$ so $1^2|1.$
+For $-1,$ $(-1)^2 = -1(-1),$ so $(-1)^2|1.$
+\endanswer
+
+\section{Problem not from the textbok}
+
+\item{1.} Prove that for all positive real numbers $x,$ the sum of $x$ and its
+reciprocal is greater than or equal to 2.
+
+\answer
+Let $x$ be a positive real number.
+For all real numbers $y,$ $y^2 \geq 0,$ so $(x-1)^2 \geq 0.$
+This is equal to
+$$x^2 - 2x + 1 \geq 0 \to x^2 + 1 \geq 2x \to x + 1/x \geq 2,$$
+since dividing by $x > 0$ is a valid algebraic operation.
+\endanswer
+
+\bye
diff --git a/gupta/hw4.tex b/gupta/hw4.tex
new file mode 100644
index 0000000..fe80e3f
--- /dev/null
+++ b/gupta/hw4.tex
@@ -0,0 +1,218 @@
+\newfam\bbold
+\def\bb#1{{\fam\bbold #1}}
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\font\bigbf=cmbx12 at 24pt
+
+\def\answer{\smallskip{\bf Answer.}\par}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+\let\endanswer\endproof
+\def\section#1{\medskip\goodbreak\noindent{\bf #1}}
+\let\impl\Rightarrow
+\def\nmid{\not\hskip2.5pt\mid}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-02-03 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 4 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 4: 26}
+
+\item{26.} Prove the following with direct proof: every odd integer is a
+difference of two squares.
+
+\answer
+Let $n$ be an odd integer.
+We will show that there are two perfect squares $a^2$ and $b^2$ (with
+$a,b\in\bb Z$) such that $n$ is their difference.
+
+Because it is odd, there exists an integer $k$ such that $n = 2k+1.$
+$$(k+1)^2-k^2 = k^2+2k+1-k^2 = 2k+1 = n,$$
+so any odd integer can be written as the difference of two squares.
+\endanswer
+
+\section{Hammack 5: 6, 12, 18, 20, 24, 28}
+
+\item{6.} Prove the following with contrapositive proof: suppose
+$x\in\bb R.$ If $x^3-x>0,$ then $x>-1.$
+
+\answer
+For contrapositive proof, let $x \leq -1.$
+We will prove that $x^3-x\leq 0.$
+
+We obtain $x+1 \leq 0$ and $x-1 \leq -2.$
+$$x(x-1)(x+1) \leq 0,$$
+because the product of three non-positive numbers is non-positive.
+% is this sufficient??
+\endanswer
+
+\item{12.} Prove the following with contrapositive proof: suppose
+$a\in\bb Z.$ If $a^2$ is not divisible by 4, then $a$ is odd.
+
+\answer
+For contrapositive proof, let $a$ be not odd (even). We will show that
+$a^2$ is divisible by 4.
+
+By the definition of even, there exists $k$ such that $a = 2k.$
+$a^2 = 4k^2,$ and $k^2\in\bb Z,$ so $a^2$ is divisible by 4.
+\endanswer
+
+\item{18.} Prove the following with either direct or contrapositive
+proof: for any $a,b\in\bb Z,$ it follows that $(a+b)^3\equiv a^3+b^3
+\pmod{3}$
+
+\answer
+Let $a,b\in\bb Z.$
+We will prove that $(a+b)^3\equiv a^3+b^3\pmod{3}.$
+
+$$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3(a^2b+ab^2),$$
+so $(a+b)^3\equiv a^3 + b^3 \pmod{3}$ by definition of modular
+equivalence.
+\endanswer
+
+\item{20.} Prove the following with either direct or contrapositive
+proof: if $a\in\bb Z$ and $a\equiv 1\pmod{5},$ then $a^2\equiv
+1\pmod{5}.$
+
+\answer
+Let $a\in\bb Z$ and $a\equiv 1\pmod{5}.$
+We will prove that $a^2\equiv 1\pmod{5}.$
+
+There exists $k\in\bb Z$ such that $a = 5k+1.$
+$$a^2 = (5k+1)^2 = 25k^2 + 10k + 1 = 5(5k^2+2k) + 1,$$
+so $a^2 \equiv 1\pmod{5}.$
+\endanswer
+
+\item{24.} Prove the following with either direct or contrapositive
+proof: if $a\equiv b \pmod{n}$ and $c\equiv d \pmod{n},$ then $ac\equiv
+bd \pmod{n}.$
+
+\answer
+Let $a,b,c,d\in\bb R.$
+Let $a\equiv b \pmod{n}$ and $c\equiv d \pmod{n}.$
+We will show that $ac\equiv bd\pmod{n}.$
+
+Therefore, there is $k\in\bb Z$ such that $a = b + nk,$ and there is
+$j\in\bb Z$ such that $c = d + nj.$
+$$ac = (b+nk)(d+nj) = bd + n(jb+dk+njk),$$
+so $ac \equiv bd \pmod{n},$ because $jb+dk+njk\in\bb Z.$
+\endanswer
+
+\item{28.} Prove the following with either direct or contrapositive
+proof: if $n\in\bb Z,$ then $4\nmid (n^2-3).$
+
+\smallskip
+{\bf Lemma.}
+Let $n\in\bb Z.$ We will show that, for some $k\in\bb Z,$ $n^2 = 4k$ or
+$n^2 = 4k+1.$
+$n$ can be written as exactly one of $4j,$ $4j+1,$ $4j+2,$ and $4j+3,$
+where $j\in\bb Z.$
+
+In the first case $n = 4j,$ $n^2 = 16j^2 = 4(4j^2),$ so with $k=4j^2,$
+$n^2 = 4k.$
+
+In the second case $n = 4j+1,$ $n^2 = 16j^2 + 8j + 1 = 4(4j^2+2j) + 1,$
+so with $k = 4j^2+2j,$ $n^2 = 4k + 1.$
+
+In the third case $n = 4j+2,$ $n^2 = 16j^2 + 16j + 4 = 4(4j^2+4j+1),$ so
+with $k = 4j^2+4j+1,$ $n^2 = 4k.$
+
+In the fourth case $n = 4j+3,$ $n^2 = 16j^2 + 24j + 9 = 4(4j^2+6j+2)+1,$
+so with $k = 4j^2+6j+2,$ $n^2 = 4k+1.$
+
+All of these values of $k$ are integers by integer closure.
+
+$n^2\neq 4k+2$ and $n^2\neq 4k+3$ for any integer $k$ because $n^2$ is
+an integer and it can be written as exactly one of $4k,$ $4k+1,$ $4k+2,$
+and $4k+3.$
+\endproof
+
+\answer
+Let $n\in\bb Z.$
+By the lemma, $n^2 = 4k$ or $n^2 = 4k+1.$
+In the case $n^2 = 4k,$ $n^2 - 3 = 4k - 3 = 4(k-1) + 1,$
+which is not divisible by $4.$
+In the case $n^2 = 4k+1,$ $n^2 - 3 = 4k - 2 = 4(k-1) + 2,$
+which is not divisible by $4.$
+\endanswer
+
+\section{Hammack 6: 4, 6, 8}
+
+\item{4.} Prove the following by contradiction: $\sqrt 6$ is irrational.
+
+\answer
+For the sake of contradiction, assume that $\sqrt 6$ is rational.
+Therefore, there exist coprime $p,q\in\bb Z$ such that $\sqrt 6 =
+{p\over q}.$
+
+$$p = q\sqrt 6 \to p^2 = 6q^2.$$
+$p^2$ is even only if $p$ is even ($2\mid p$), so $4\mid p^2,$ so $2\mid
+q^2,$ and thus $2\mid q.$
+Therefore, $(q,p) = 2\neq 1,$ meaning they're not coprime.
+\endanswer
+
+\item{6.} Prove the following by contradiction: if $a,b\in\bb Z,$ then
+$a^2-4b-2\neq 0.$
+
+\answer
+Let $a,b\in\bb Z,$ and for the sake of contradiction, let $a^2-4b-2 =
+0.$
+$$a^2-4b-2 \equiv 0 \equiv a^2-2 \to a^2 \equiv 2 \bmod{4}.$$
+Therefore, there exists $k\in\bb Z$ such that $a^2 = 4k+2.$
+By the lemma, $a^2 \neq 4k + 2.$
+\endanswer
+
+\item{8.} Prove the following by contradiction: suppose $a,b,c\in\bb Z.$
+If $a^2+b^2=c^2,$ then $a$ or $b$ is even.
+
+\answer
+Let $a,b,c\in\bb Z$ such that $a^2+b^2=c^2.$
+Suppose, for the sake of contradiction, $a$ and $b$ are odd.
+There exists $j$ and $k$ such that $a=2k+1$ and $b=2j+1.$
+
+Therefore, $a^2 + b^2 = (2k+1)^2 + (2j+1)^2 = 4k^2 + 4k + 1 + 4j^2 + 4j
++ 1 = 4(k^2+k+j^2+j) + 2 = c^2.$
+By the lemma, $c^2 \neq 4k + 2.$
+\endanswer
+
+\section{Problems not from the textbok}
+
+\item{1.} A perfect square is an integer $n$ for which there exists an
+integer $k$ such that $n = k^2.$ Prove that if $n$ is a positive integer
+such that $n\equiv 2\bmod 4$ or $n\equiv 3\bmod 4,$ then $n$ is not a
+perfect square.
+
+\answer
+This has already been proven in the above lemma.
+\endanswer
+
+\item{2.} After a grueling slog through the snow to reach Ponce City
+Market, you decide to reward yourself by buying three boxes of candy
+from Collier’s. One box contains mint candies, one chocolate candies,
+and the other is mixed. Unfortunately, all three boxes were incorrectly
+labeled! What is the smallest number of candies that you need to remove
+and sample to be able to correctly label all three boxes? Carefully
+justify your reasoning.
+
+\answer
+We only need to sample one candy.
+We sample the box labeled ``mixed,'' and without loss of generality we
+get a chocolate candy. This is the chocolate box.
+This box cannot be ``mixed'' because it is labeled incorrectly, and this
+box cannot be ``mint'' because the mint box doesn't have chocolate
+candy.
+Now, the box labeled ``mint'' must be the mixed box because it cannot be
+the chocolate box (we only have one of those) and it cannot be the mixed
+box because it is incorrectly labeled.
+By elimination, the last box is the mixed box.
+\endanswer
+
+\bye
diff --git a/gupta/hw5.tex b/gupta/hw5.tex
new file mode 100644
index 0000000..bb1bd43
--- /dev/null
+++ b/gupta/hw5.tex
@@ -0,0 +1,217 @@
+\newfam\bbold
+\def\bb#1{{\fam\bbold #1}}
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\font\bigbf=cmbx12 at 24pt
+
+\def\answer{\smallskip{\bf Answer.}\par}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+\let\endanswer\endproof
+\def\section#1{\medskip\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in%
+\noindent{\bf #1}}
+\let\impl\to
+\def\nmid{\hskip-3pt\not\hskip2.5pt\mid}
+\def\problem#1{\par\penalty-100\item{#1}}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-02-17 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 5 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Hammack 7: 6, 9, 12}
+
+\problem{6.} Suppose $x,y\in\bb R.$ Then $x^3+x^2y = y^2+xy$ if and only if
+$y = x^2$ or $y = -x.$
+
+$x^2(x+y) = y(x+y).$
+\answer
+$(\Rightarrow)$
+
+Let $x^3 + x^2y = y^2+xy.$
+We then have $x^2(x+y) = y(x+y).$
+If $y = -x,$ $x+y = 0 \impl x^2(x+y) = y(x+y).$
+Otherwise, we can divide by $x+y$ because it is nonzero, giving
+$y = x^2.$
+Therefore, $y=-x$ or $y=x^2.$
+
+$(\Leftarrow)$
+
+Let $y = -x$ or $y = x^2.$
+We will first consider the case $y = -x,$ then the case $y = x^2.$
+
+With $y = -x,$ $x^3 + x^2y = x^3 - x^3 = 0 = x^2 - x^2 = y^2 + xy.$
+
+If $y = x^2,$ $x^3 + x^2y = x^3 + x^4 = x^3 + x^4 = xy + y^2.$
+\endanswer
+
+\problem{9.} Suppose $a\in\bb Z.$ Prove that $14\mid a$ if and only if
+$7\mid a$ and $2\mid a.$
+
+\answer
+$(\Rightarrow)$
+
+Let $14\mid a.$
+This gives $a = 14k$ for some integer $k.$
+$a = 2(7k),$ so with $7k\in\bb Z,$ we get $2\mid a.$
+Similarly, $a = 7(2k),$ so with $2k\in\bb Z,$ we get $7\mid a.$
+
+$(\Leftarrow)$
+
+Let $7\mid a$ and $2\mid a.$
+These give $a = 7j$ for some $j$ and $a = 2k$ for some integers $j$ and
+$k.$
+A product of odd number $7$ and odd number $j$ cannot be even (and $a$
+is even because $2\mid a$), so $j$ must be even.
+Thus, there exists $l\in\bb Z$ such that $j = 2l.$
+This gives $a = 7(2l) = 14l \to 14\mid a.$
+\endanswer
+
+\problem{12.} There exist a positive real number $x$ for which $x^2 < \sqrt
+x.$
+
+\answer
+Observe that $x = 1/4$ gives $x^2 = 1/16$ and $\sqrt x = 1/2,$ so $1/16
+< 1/2.$
+\endanswer
+
+\section{Hammack 8: 12, 22, 28}
+
+\problem{12.} If $A,$ $B,$ and $C$ are sets, then $A-(B\cap C) =
+(A-B)\cup(A-C).$
+
+\answer
+$(\subseteq)$
+
+Let $x\in A - (B\cap C).$
+This gives us $x\in A$ and $x\not\in B\cap C.$
+We get $x\not\in B$ or $x\not\in C.$
+WLOG, let $x\not\in B.$
+$x\in A$ and $x\not\in B,$ so $x\in A-B,$ so $x\in (A-B)\cup(A-C).$
+
+$(\supseteq)$
+
+Let $x\in (A-B)\cup (A-C).$
+This gives $x\in A-B$ or $x\in A-C.$
+WLOG, let $x\in A-B.$
+Therefore, $x\in A$ and $x\not\in B.$
+This implies $x\not\in B\cap C,$ so $x\in A-(B\cap C).$
+
+Since we have $A-(B\cap C) \subseteq (A-B)\cup(A-C)$ and $(A-B)\cup(A-C)
+\subseteq A-(B\cap C),$
+we obtain $$A-(B\cap C) = (A-B)\cup(A-C).$$
+\endanswer
+
+\problem{22.} Let $A$ and $B$ be sets. Prove that $A\subseteq B$ if and
+only if $A\cap B = A.$
+
+\answer
+$(\Rightarrow)$
+
+Let $A\subseteq B.$
+Let $x\in A.$
+By subset, $x\in B.$
+And if and only if $x\in A$ and $x\in B,$ $x\in A\cap B,$ so $A = A\cap
+B.$
+
+$(\Leftarrow)$
+
+Let $A\cap B = A.$
+This implies $A\subseteq A\cap B,$ or $x\in A\impl x\in A\cap B\impl
+x\in B.$
+$x\in A\impl x\in B$ is the definition of $A\subseteq B.$
+\endanswer
+
+\problem{28.} Prove $\{12a+25b:a,b\in\bb Z\} = \bb Z.$
+
+\answer
+Let $A = \{12a+25b:a,b\in\bb Z\}.$ If $x\in A,$ $x\in\bb Z$ because the
+integers are closed under addition and mulitplication.
+
+If $x\in\bb Z,$ let $b = x$ and $a = -2x,$ giving us $12(-2x) + 25x =
+x\in A.$
+
+Therefore, since $A\subseteq\bb Z$ and $\bb Z\subseteq A,$ these two
+sets are equal.
+\endanswer
+
+\section{Hammack 9: 6, 28, 30, 34}
+Prove or disprove each of the following statements.
+
+\problem{6.} If $A,$ $B,$ $C,$ and $D$ are sets, then $(A\times
+B)\cup(C\times D) = (A\cup C)\times(B\cup D).$
+
+\answer
+{\bf Disproof.}
+
+Let $A = B = \{1\}$ and $C = D = \{2\}.$
+The set $A\times B = \{(1,1)\}.$
+The set $C\times D = \{(2,2)\}.$
+And the set $A\cup C = B\cup D = \{1,2\}.$
+
+$$(A\times B)\cup(C\times D) = \{(1,1),(2,2)\} \neq
+\{(1,1),(1,2),(2,1),(2,2)\} = (A\cup C)\times(B\cup D).$$
+\endanswer
+
+\problem{28.} Suppose $a,b\in\bb Z.$ If $a\mid b$ and $b\mid a,$ then $a =
+b.$
+
+\answer
+We will show this by contrapositive.
+Let $a\neq b.$ We will show that $a\nmid b$ or $b\nmid a.$
+
+WLOG, $a > b.$
+Immediately, $a\nmid b.$
+\endanswer
+
+\problem{30.} There exist integers $a$ and $b$ for which $42a + 7b = 1.$
+
+\answer
+{\bf Disproof.}
+
+Let $a,b\in\bb Z.$
+For the sake of contradiction, assume $42a + 7b = 1.$
+Dividing by 7, $6a + b = 1/7.$
+By closure of the integers, $6a+b\in\bb Z,$ and $1/7\not\in\bb Z,$
+giving a contradiction.
+\endanswer
+
+\problem{34.} If $X\subseteq A\cup B,$ then $X\subseteq A$ or $X\subseteq
+B.$
+
+\answer
+{\bf Disproof.}
+
+Let $A = \{1\},$ $B = \{2\},$ and $X = A\cup B.$
+Immediately, $X\subseteq A\cup B.$
+And then, $2\in X,$ but $2\not\in A,$ so $X\not\subseteq A.$
+Also, $1\in X,$ but $1\not\in B,$ so $X\not\in B.$
+\endanswer
+
+\section{Problem not from the textbok}
+
+\problem{1.} Let $A,$ $B,$ and $C$ be arbitrary sets. Prove that if
+$A-C\not\subseteq A-B,$ then $B\not\subseteq C.$
+
+\answer
+We will prove this by contrapositive.
+Let $B\subseteq C.$ We will show that $A-C\subseteq A-B.$
+
+If $x\in B,$ $x\in C,$ so by contrapositive, if $x\not\in C,$ $x\not\in
+B.$
+
+Let $y\in A-C.$
+By definition of setminus, $y\in A$ and $y\not\in C.$
+As established, this implies $y\not\in B.$
+Therefore, with $y\in A$ and $y\not\in B,$ $y\in A-B,$ so
+$A-C\subseteq A-B.$
+\endanswer
+
+\bye
diff --git a/howard/Makefile b/howard/Makefile
new file mode 100644
index 0000000..8e5dd3a
--- /dev/null
+++ b/howard/Makefile
@@ -0,0 +1,12 @@
+.POSIX:
+.SUFFIXES: .tex .pdf
+
+PDFTEX = pdftex
+
+.tex.pdf:
+	$(PDFTEX) $<
+
+all: hw1.pdf hw2.pdf hw3.pdf
+
+clean:
+	rm -f *.pdf *.log
diff --git a/howard/hw1.tex b/howard/hw1.tex
new file mode 100644
index 0000000..b9731b2
--- /dev/null
+++ b/howard/hw1.tex
@@ -0,0 +1,248 @@
+\newcount\indentlevel
+\newcount\itno
+\def\reset{\itno=1}\reset
+\def\afterstartlist{\advance\leftskip by .5in\par\advance\leftskip by -.5in}
+\def\startlist{\par\advance\indentlevel by 1\advance\leftskip by .5in\reset
+\aftergroup\afterstartlist}
+\def\alph#1{\ifcase #1\or a\or b\or c\or d\or e\or f\or g\or h\or
+    i\or j\or k\or l\or m\or n\or o\or p\or q\or r\or
+    s\or t\or u\or v\or w\or x\or y\or z\fi}
+\def\li#1\par{\medskip\item{\ifcase\indentlevel \number\itno.\or
+                                  \alph\itno)\else
+                                  (\number\itno)\fi
+                           }%
+         #1\smallskip\advance\itno by 1}
+\def\hline{\noalign{\hrule}}
+\let\impl\rightarrow
+\newskip\tableskip
+\tableskip=10pt plus 10pt
+
+\li Evaluate each of the conditional statements to true or false
+
+{\startlist
+    \li If $1+2=4,$ then $9+0= -9$
+
+        With $p$ $1+2=4$ and $q$ $9+0= -9,$ this statement is $p\impl
+        q.$
+        $1+2\neq 4,$ so this conditional is vacuously true.
+
+    \li $13 > 19$ only if 13 is prime
+
+        With $p$ 13 is prime and $q$ $13 > 19,$ this statement is
+        $p\impl q,$ and since $p$ is true and $q$ is false, this
+        statement is false.
+
+    \li Horses can fly whenever horses cannot fly
+
+        With $p$ ``horses cannot fly'' and $q$ ``horses can fly,'' this
+        is equivalent to $p\impl q.$ Since horses cannot fly, $p$ is
+        true and $q$ is false, so the statement is false.
+
+    \li $3\cdot3 = 9,$ if $9+9 = 18$
+
+        With $p$ $9+9 = 18$ and $q$ $3\cdot3=9,$ this statement is
+        equivalent to $p\impl q,$ and since both are true, this
+        statement is true.
+}
+
+
+\li Let $p$ and $q$ be propositions, where $p$ is the statement ``It is
+snowing outside,'' and $q$ is the statement ``It is June.'' Express each
+of the following propositions as an English sentence.
+
+{\startlist
+    \li $p\impl q.$
+
+    ``If it is snowing outside, it is June.''
+
+    \li $\lnot p \land q.$
+
+    ``It is not snowing outside, and it is June.''
+
+    \li $\lnot p \lor (p\land q).$
+
+    ``It is not snowing outside, or it is snowing outside in June.''
+}
+
+\li Consider the statement: ``If the TAs make the homework too hard,
+then the students will be sad.'' Write the converse, contrapositive, and
+inverse of the statement. Don't worry about the grammar/tense, we just
+want to see the correct idea.
+
+{\startlist
+    \li Converse
+
+    ``If the students will be sad, the TAs make the homework too hard.''
+
+    \li Contrapositive
+
+    ``If the students aren't sad, the TAs didn't make the homework too
+    hard.''
+
+    \li Inverse
+
+    ``If the TAs don't make the homework too hard, the students will be
+    happy.''
+}
+
+\li How many rows appear in a truth table for each of these compound
+propositions?
+
+{\startlist
+    \li $p\land q$
+
+    The number of rows is $2^v$ where v is the number of variables in
+    the expression. Therefore, this expression will have $2^2 = 4$ rows.
+
+    \li $\lnot p \impl (p\impl q)$
+
+    This still only has two variables, so it will have $2^2 = 4$ rows.
+
+    \li $(\lnot p\land q\land s)\lor(p\land\lnot q\land s)\lor(p\land
+    q\land\lnot s)$
+
+    This has three variables, so it will have $2^3 = 8$ rows.
+}
+
+\li Using the following propositions, translate the sentence ``You
+cannot see the movie if you are not over 18 years old and you do not
+have the permission of a parent'' to a compound proposition.
+
+$m := \hbox{``You can see the movie''}$
+
+$e := \hbox{``You are over 18 years old''}$
+
+$p := \hbox{``You have the permission of a parent''}$
+
+This is $(\lnot p\land\lnot e)\impl \lnot m.$
+
+\li Construct truth tables for the following propositions. Include all
+intermediate columns to receive full credit for each table.
+
+{\startlist
+    \li $p\lor q\land\lnot p$
+
+    \leavevmode
+    \halign{&\vrule\strut#&#\tabskip\tableskip&#&#\tabskip0pt\cr\hline
+
+        &&$p$&&&&$q$&&&&$\lnot p$&&&&$q\land\lnot p$&&&&
+        $p\lor q\land\lnot p$&&\cr\hline
+        &&T&&&&T&&&&F&&&&F&&&&T&&\cr\hline
+        &&T&&&&F&&&&F&&&&F&&&&T&&\cr\hline
+        &&F&&&&T&&&&T&&&&T&&&&T&&\cr\hline
+        &&F&&&&F&&&&T&&&&F&&&&F&&\cr\hline
+        }
+
+    \li $(p\lor\lnot q) \impl q$
+
+    \leavevmode
+    \halign{&\vrule\strut#&#\tabskip\tableskip&#&#\tabskip0pt\cr\hline
+
+        &&$p$&&&&$q$&&&&$\lnot q$&&&&$p\lor\lnot q$&&&&
+        $(p\lor\lnot q)\impl q$&&\cr\hline
+        &&T&&&&T&&&&F&&&&T&&&&T&&\cr\hline
+        &&T&&&&F&&&&T&&&&T&&&&F&&\cr\hline
+        &&F&&&&T&&&&F&&&&F&&&&T&&\cr\hline
+        &&F&&&&F&&&&T&&&&T&&&&F&&\cr\hline
+        }
+
+    \li $(\lnot q\impl\lnot q) \iff (\lnot p \impl \lnot q)$
+
+    \leavevmode
+    \halign{&\vrule\strut#&#\tabskip\tableskip&#&#\tabskip0pt\cr\hline
+
+        &&$p$&&&&$q$&&&&$\lnot p$&&&&$\lnot q$&&&&$\lnot q\impl\lnot q$&&&&
+        $\lnot p\impl\lnot q$&&&&
+        $(\lnot q\impl\lnot q)\iff(\lnot p\impl\lnot q)$&&\cr\hline
+        &&T&&&&T&&&&F&&&&F&&&&T&&&&T&&&&T&&\cr\hline
+        &&T&&&&F&&&&F&&&&T&&&&T&&&&T&&&&T&&\cr\hline
+        &&F&&&&T&&&&T&&&&F&&&&T&&&&F&&&&F&&\cr\hline
+        &&F&&&&F&&&&T&&&&T&&&&T&&&&T&&&&T&&\cr\hline
+        }
+}
+
+\li There is a spaceship where every passenger has exactly one role.
+Each passenger can either be a Crewmate or an Imposter. A Crewmate
+always tells the truth, and an Imposter always lies. For each question,
+determine the role of Person A and the role of Person B, or write
+``Cannot be determined'' for that person if there is not enough
+information. Explain your reasoning for full credit! (You can use a
+truth table or just plain English to explain.)
+
+{\startlist
+    \li Person A says ``I am a Crewmate, or B is a Crewmate,'' and
+    Person B says ``A is a Crewmate if I am an Imposter.''
+
+    If and only if A is a crewmate, their statement is true. Similarly,
+    if and only if B is a crewmate, their statement is true. We will use
+    this principle for all of the problems.
+
+    In this problem, if both are crewmates, both statements are true (B
+    is not an imposter, so their statement is vacuously true, and A is a
+    crewmate or B is a crewmate).
+
+    However, if both are imposters, both statements are false (neither
+    is a crewmate and A is not a crewmate even though B is an imposter).
+    This means we can't figure out any information.
+
+    \li Person A says ``I am an Imposter, and Person B is a Crewmate,''
+    and Person B says nothing.
+
+    Both must be imposters. If A is a crewmate, their statement ``I am
+    an imposter'' is a lie, so they are an imposter by contradiction.
+    Since A is an an imposter, this statement must be a lie, so ``Person
+    B is a crewmate'' is false, and B is an imposter.
+
+    \li Person A says ``Both Person B and I are Imposters,'' and Person
+    B says ``At least one of us is a Crewmate.''
+
+    By a similar logic as b, ``I am an imposter'' would be a lie if A
+    were a crewmate, so A is an imposter. And since that statement is
+    true, ``Person B is an imposter'' must be a lie, and Person B is a
+    crewmate. This makes B's statement true, validating our view.
+}
+
+\li Show that $(p\impl q)\lor\lnot p \equiv p\impl q$ using logical
+equivalences. Cite the laws of equivalences used to reach each step.
+
+\leavevmode
+\goodbreak
+\halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+&$(p\impl q)\lor\lnot p$&Given&&\cr
+&$(\lnot p\lor q)\lor\lnot p$&Conditional Identity&&\cr
+&$(q\lor\lnot p)\lor\lnot p$&Commutative Law&&\cr
+&$q\lor(\lnot p\lor\lnot p)$&Associative Law&&\cr
+&$q\lor\lnot p$&Idempotent Law&&\cr
+&$p\impl q$&Conditional Identity&&\cr
+\hline
+}
+
+\li Show that $\lnot ((p\land q)\lor p)\impl \lnot p$ is a tautology
+using:
+
+{\startlist
+    \li a truth table (include all intermediate columns)
+
+    \halign{&\vrule\strut#&#\tabskip\tableskip&#&#\tabskip0pt\cr\hline
+
+        &&$p$&&&&$q$&&&&$p\land q$&&&&$(p\land q)\lor p$&&&&
+        $\lnot((p\land q)\lor p)$&&&&
+        $\lnot p$&&&&$\lnot ((p\land q)\lor p)\impl \lnot p$&&\cr\hline
+        &&T&&&&T&&&&T&&&&T&&&&F&&&&F&&&&T&&\cr\hline
+        &&T&&&&F&&&&F&&&&T&&&&F&&&&F&&&&T&&\cr\hline
+        &&F&&&&T&&&&F&&&&F&&&&T&&&&T&&&&T&&\cr\hline
+        &&F&&&&F&&&&F&&&&F&&&&T&&&&T&&&&T&&\cr\hline
+    }
+
+    \li logical equivalences (cite the laws of equivalence used to reach
+    each step)
+
+    \halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+    &$\lnot ((p\land q)\lor p)\impl \lnot p$&Given&&\cr
+    &$\lnot p\impl \lnot p$&Absorption Law&&\cr
+    &$p\lor\lnot p$&Conditional Identity&&\cr
+    &$T$&Complement Law&&\cr
+    \hline
+    }
+}
+\bye
diff --git a/howard/hw2.tex b/howard/hw2.tex
new file mode 100644
index 0000000..f63b1ee
--- /dev/null
+++ b/howard/hw2.tex
@@ -0,0 +1,196 @@
+\newcount\indentlevel
+\newcount\itno
+\def\reset{\itno=1}\reset
+\def\afterstartlist{\advance\leftskip by .5in\par\advance\leftskip by -.5in}
+\def\startlist{\par\advance\indentlevel by 1\advance\leftskip by .5in\reset
+\aftergroup\afterstartlist}
+\def\alph#1{\ifcase #1\or a\or b\or c\or d\or e\or f\or g\or h\or
+    i\or j\or k\or l\or m\or n\or o\or p\or q\or r\or
+    s\or t\or u\or v\or w\or x\or y\or z\fi}
+\def\li#1\par{\medskip\penalty-100\item{\ifcase\indentlevel \number\itno.\or
+                                  \alph\itno)\else
+                                  (\number\itno)\fi
+                           }%
+         #1\smallskip\advance\itno by 1\relax}
+\def\hline{\noalign{\hrule}}
+\let\impl\rightarrow
+\newskip\tableskip
+\tableskip=10pt plus 10pt
+\def\\{\hfil\break}
+
+\li Express each of these statements using predicates and quanifiers
+
+{\startlist
+    \li ``Some pigs eat wheat.''
+
+    Let the domain be the set containing all animals. \\
+    Let $P(x):$ $x$ is a pig. \\
+    Let $W(x):$ $x$ eats wheat.
+
+    $$\exists x, P(x)\land W(x).$$
+
+    \li ``All board games are fun.''
+
+    Let the domain be the set of all board games. \\
+    Let $B(x):$ $x$ is a board game.
+    Let $F(x):$ $x$ is fun.
+
+    $$\forall x, B(x)\impl F(x).$$
+
+    \li ``Not all potatoes are sweet.''
+
+    Let the domain be the set of all plants. \\
+    Let $P(x):$ $x$ is a potato. \\
+    Let $S(x):$ $x$ is sweet.
+
+    $$\lnot\forall x, P(x)\impl S(x).$$
+}
+
+
+\li Determine the truth value of each of these statements if the domain
+of each variable consists of all integers.
+
+{\startlist
+    \li $\forall x\exists y (\root 5 \of x > y^2)$
+
+    False. $y^2 > 0$ for all $y,$ and $\root 5 \of {-1} = -1,$ disproving
+    this statement.
+
+    \li $\exists x\forall y((x = 2.5)\impl (y > x))$
+
+    False. $y = 0 < 2.5 = x.$
+
+    \li $\exists x\forall y(x^2 = y)$
+
+    False. $1 = x^2 = 0$ is not possible.
+
+    \li $\forall x\exists y(x^2 = y)$
+
+    True. The integers are closed under squaring.
+}
+
+\li Rewrite each of these statements so that no negation is to the left
+of a quantifier, so that every negation is to the left of a predicate
+(in other words, push the negation in past the quantifiers).
+
+{\startlist
+    \li $\lnot\forall y(\forall x R(x,y)\land\exists x S(x,y))$
+
+    $$\exists y(\exists x \lnot R(x,y)\lor\forall x \lnot S(x,y))$$
+
+    \li $\forall x\lnot\exists y\forall z(A(x,y)\impl B(x,z))$
+
+    $$\forall x\forall y\exists z\lnot(A(x,y)\impl B(x,z))$$
+
+    \li $\lnot\exists x\lnot\forall y\lnot\forall z(A(x,z)\land B(y,z))$
+
+    $$\forall x\forall y\exists z\lnot (A(x,z)\land B(y,z))$$
+
+    \li $\lnot\exists x\forall y\exists z(P(x,y)\iff Q(z))$
+
+    $$\forall x\exists y\forall z\lnot (P(x,y)\iff Q(z))$$
+
+}
+
+\li Let $S(x)$ be the statement ``$x$ has a sabre tooth tiger,'' let
+$D(x)$ be the statement ``x has a dhole,'' and let $H(x)$ be the
+statement ``x has a horse.'' Express each of these statements in terms
+of $S(x),$ $D(x),$ $H(x),$ quantifiers, and logical connectives. Let the
+domain consist of all people.
+
+{\startlist
+    \li Every person who owns a sabre tooth tiger also owns a dhole or a
+    horse.
+
+    $$\forall x S(x)\impl(D(x)\land H(x)).$$
+
+    \li No one owns a sabre tooth tiger, and at least one person owns a
+    horse.
+
+    $$\forall x\lnot S(x) \land \exists x H(x).$$
+
+    \li For each of the three animals---sabre tooth tigers, dholes, and
+    horses---there is a person who has this animal. Hint: All three
+    animals can, but do not have to be, owned by the same person.
+
+    $$\exists x S(x) \land \exists x D(x) \land \exists x H(x).$$
+}
+
+\li 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 (in other words, push the negation in past the quantifiers).
+Next, express the negation in English.
+
+{\startlist
+    \li Let the domain be all animals \\
+    Let $B(x):$ $x$ is a bird \\
+    Let $C(x):$ $x$ can chirp \\
+    ``There exists a bird that can chirp.''
+
+    $$\exists x B(x)\land C(x).$$
+
+    \li Let the domain be all animals \\
+    Let $C(x):$ $x$ is a cat \\
+    Let $E(x):$ $x$ eats fish \\
+    ``All cats eat fish.''
+
+    $$\forall x C(x)\impl E(x).$$
+
+    \li Let the domain be all animals
+    Let $D(x):$ $x$ is a dog \\
+    Let $M(x):$ $x$ can meow \\
+    ``There exists a dog, only if not all animals can meow.''
+
+    $$(\lnot\forall x M(x)) \impl (\exists x D(x)).$$
+}
+
+\li Determine the truth value of the statement $\exists x\forall
+y(x^3\leq y^4)$ if the domain for the variable consists of:
+
+{\startlist
+    \li the positive real numbers
+
+    This is not true because $0 < y < \root 4 \of x^3$ always exists for
+    any $x$ (the fourth root of a cube of a positive number is positive,
+    so this number is never zero)
+
+    \li the non-negative integers
+
+    This is true with $x=0$ because $y^4 \geq 0$ over this domain.
+
+    \li the nonzero real numbers
+
+    This is true with $x=-1$ because $y^4 \geq 0 > x^3 = -1.$
+}
+
+\li Show that $\lnot\forall x(P(x)\land\lnot Q(x))$ is logically
+equivalent to $\exists x(P(x)\impl Q(x)).$
+
+\halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+    &$\lnot \forall x(P(x)\land\lnot Q(x))$&Given&&\cr
+    &$\exists x \lnot(P(x)\land\lnot Q(x))$&DeMorgan's Law for Quantifiers&&\cr
+    &$\exists x (\lnot P(x)\lor\lnot\lnot Q(x))$&DeMorgan's Law&&\cr
+    &$\exists x (\lnot P(x)\lor Q(x))$&Double negation law&&\cr
+    &$\exists x (P(x)\impl Q(x))$&Conditional Identity&&\cr
+    \hline
+}
+
+\li Find a counterexample, if possible, to these universally quantified
+statements, where the domain for all variables consists of all real
+numbers. Otherwise, state that no counterexample exists.
+
+{\startlist
+    \li $\forall x\forall y(x^3\neq y^7)$
+
+    With $x=y=1,$ $x^3 = y^7.$
+
+    \li $\forall x\exists y(y = {x\over 2})$
+
+    No counterexample exists.
+
+    \li $\forall x\forall y((x\geq y)\impl(x^{100} > y)).$
+
+    With $x=y=0,$ $x\geq y,$ but $x^{100} \not> y.$
+}
+
+\bye
diff --git a/howard/hw3.tex b/howard/hw3.tex
new file mode 100644
index 0000000..ab61b0d
--- /dev/null
+++ b/howard/hw3.tex
@@ -0,0 +1,313 @@
+\newfam\bbold
+\def\bb#1{{\fam\bbold #1}}
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+
+\newcount\indentlevel
+\newcount\itno
+\def\reset{\itno=1}\reset
+\def\afterstartlist{\advance\leftskip by .5in\par\advance\leftskip by -.5in}
+\def\startlist{\par\advance\indentlevel by 1\advance\leftskip by .5in\reset
+\aftergroup\afterstartlist}
+\def\alph#1{\ifcase #1\or a\or b\or c\or d\or e\or f\or g\or h\or
+    i\or j\or k\or l\or m\or n\or o\or p\or q\or r\or
+    s\or t\or u\or v\or w\or x\or y\or z\fi}
+\def\li#1\par{\medskip\penalty-100\item{\ifcase\indentlevel \number\itno.\or
+                                  \alph\itno)\else
+                                  (\number\itno)\fi
+                           }%
+         #1\smallskip\advance\itno by 1\relax}
+\def\ul{\bgroup\def\li##1\par{\item{$\bullet$} ##1\par}}
+\let\endul\egroup
+\def\hline{\noalign{\hrule}}
+\let\impl\rightarrow
+\newskip\tableskip
+\tableskip=10pt plus 10pt
+\def\endproof{\leavevmode\quad\vrule height 6pt width 6pt
+depth 0pt\hskip\parfillskip\hbox{}{\parfillskip0pt\medskip}}
+
+\li Which rule of inference is used in each of these arguments?
+
+{\startlist
+    \li Richard is a Computer Science major. Richard is a Computer
+    Science major only if he attends Georgia Tech. Therefore, Richard
+    attends Georgia Tech.
+
+    This is modus ponens. With $p$ ``Richard is a Computer Science
+    Major,'' and $q$ ``he attends Georgia Tech,'' we have $p\impl q$ and
+    $p,$ concluding $q.$
+
+    \li Discrete Math exams are fun or Python is a boring language.
+    Discrete Math exams are not fun or Rohan is a TA for Discrete Math.
+    Therefore, Python is a boring language or Rohan is a TA for Discrete
+    Math.
+
+    This is resolution. With $p$ ``Discrete math exams are fun,'' $q$
+    ``Python is a boring language,'' and $r$ ``Rohan is a TA for
+    Discrete Math,'' we have $p\lor q$ and $\lnot p\lor r,$ concluding
+    $q\lor r.$
+
+    \li If it snows today, the school will close. If the schools are
+    closed, then you will not go to class. Therefore, you do
+    not go to class, if it snows today.
+
+    This is hypothetical syllogism.
+    Let $p$ ``it snows today,'' $q$ ``the schools will close,'' and $r$
+    ``you will not go to class.'' We have $p\impl q$ and $q\impl r,$
+    giving us $p\impl r.$
+
+    % hypothetical syllogism
+
+    \li You are a Discrete Math student whenever you are a Computer
+    Science major. You are not a Discrete Math student. Therefore, you
+    are not a Computer Science major.
+
+    This is modus tollens.
+    With $p$ ``you are a Computer Science major,'' and $q$ ``you are not
+    a Discrete Math student,'' we have $p\impl q$ and $\lnot q,$
+    concluding $\lnot p.$
+
+    % modus tollens
+}
+
+\li Using rules of inference, show that the premises below conclude with
+$d.$ Give the reason for each step as you show that $d$ is concluded.
+Each reason should be the name of a rule of inference and include which
+numbered steps are involved. For example, a reason for a step might be
+``Modus ponens $(2,3)$''
+
+\halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+    &Statement&Reason&&\cr\hline
+    &$a\lor b$&Premise&&\cr
+    &$b\lor c\impl e$&Premise&&\cr
+    &$d\lor\lnot e$&Premise&&\cr
+    &$\lnot a\lor c$&Premise&&\cr
+    &$b\lor c$&Resolution (1,4)&&\cr
+    &$e$&Modus Ponens (2,5)&&\cr
+    &$e$&Disjunctive Syllogism (3,6)&&\cr
+    \hline
+}
+
+\li Using the rules of inference, show that the following premises
+conclude with ``Yoshi wins the race, and Luigi rides a bike.'' Be sure
+to define all propositional variables for full credit (for example, you
+may define ``$t:$ Toad gets lost'' as one of your propositional
+variables). Remember, it is possible that you will use all premises, but
+it is also possible that some are not needed.
+
+\ul
+    \li Toad gets lost, and Luigi rides a bike.
+
+    \li If Luigi does not ride a bike, then Wario cheats.
+
+    \li Rosalina is the best princess in the race.
+
+    \li Wario does not cheat, or Toad does not get lost.
+
+    \li If Wario does not cheat and Toad gets lost, then Yoshi wins the
+    race.
+
+\endul
+
+\noindent Variable definitions:
+\ul
+    \li $t:$ Toad gets lost
+
+    \li $l:$ Luigi rides a bike.
+
+    \li $w:$ Wario cheats.
+
+    \li $y:$ Yoshi wins the race
+
+\endul
+
+\smallskip
+\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in
+\halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+    &Statement&Reason&&\cr\hline
+    &$t\land l$&Premise&&\cr
+    %&$\lnot l\impl w$&Premise&&\cr
+    &$\lnot w\lor \lnot t$&Premise&&\cr
+    &$(\lnot w\land t)\impl y$&Premise&&\cr
+    &$t$&Simplification (1)&&\cr
+    &$\lnot w$&Disjunctive Syllogism (2,4)&&\cr
+    &$\lnot w\land t$&Conjunction (4,5)&&\cr
+    &$y$&Modus ponens (3,6)&&\cr
+    &$l$&Simplification (1)&&\cr
+    &$y\land l$&Conjunction (7,8)&&\cr
+    \hline
+}
+\smallskip
+
+This is the conclusion ``Yoshi wins the race, and Luigi rides a bike.''
+\hfil
+\endproof
+
+\li For the following proof, there are blanks in many steps. Please fill
+out each blank with its correct statement or reason. Note that the
+domain for $x$ is all people, and Tashfia is a person.
+
+\def\ta{{\rm Tashfia}}
+\smallskip
+\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in
+\halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+    &Statement&Reason&&\cr\hline
+    &$\forall x(B(x)\impl C(x))$&Premise&&\cr
+    &$A(\ta)$&Premise&&\cr
+    &$\forall x(A(x)\impl\lnot C(x))$&Premise&&\cr
+    &$A(\ta)\impl\lnot C(\ta)$&Universal Instantiation (3)&&\cr
+    &$\lnot C(\ta)$&Modus ponens (2,4)&&\cr
+    &$B(\ta)\impl C(\ta)$&Universal Instantiation (1)&&\cr
+    &$\lnot B(\ta)$&Modus tollens (5,6)&&\cr
+    &$\exists x\lnot B(x)$&Existential generalization (7)&&\cr\hline
+}
+\smallskip
+\endproof
+
+\li The CS 2050 office hours cubicle is moving! The new cubicle has a
+width of $3x$ and a length of $y,$ where $x, y\in Z^+.$ Prove or
+disprove that the area of the cubicle is even whenever $x$ is even. Make
+sure to include the introduction, body, and conclusion. Clearly state
+your reasoning for all statements and use a two-column proof for the
+body whenever possible.
+
+Let $x,y\in Z^+$ represent arbitrary parameters of the desk with $x$
+even.
+Let the desk have width $w = 3x$ and length $l = y.$
+We will also say the desk has area $a = lw$ and prove that this area is
+even.
+
+\smallskip
+\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in
+\halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+    &Statement&Reason&&\cr\hline
+    &$a = lw$&Premise&&\cr
+    &$w = 3x$&Premise&&\cr
+    &$l = y$&Premise&&\cr
+    &$x$ is even&Premise&&\cr
+    &$x,y\in Z^+$&Premise&&\cr
+    &$a = 3xy$&Substitution (1,2,3)&&\cr
+    &$\exists j\in\bb Z, x = 2j$&Definition of even (4)&&\cr
+    &$x = 2k$&Existential Instantiation (7)&&\cr
+    &$a = 6ky$&Substitution (6,8)&&\cr
+    &$a = 2(3ky)$&Algebra (9)&&\cr
+    &$\exists j\in\bb Z, a = 2j$&Existential generalization (10)&&\cr
+    &$a$ is even&Definition of even (11)&&\cr
+    \hline
+}
+\smallskip
+\endproof
+
+\li Use a direct proof to show that if $n+9$ is odd, then $n^2-5n-14$ is
+even. Make sure to include the introduction, body, and conclusion.
+Clearly state your reasoning for all statements and use a two-column
+proof for the body whenever possible.
+
+Let $n$ be an integer such that $n+9$ is odd.
+We will show that $n^2-5n-14$ is even.
+
+\smallskip
+\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in
+\halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+    &Statement&Reason&&\cr\hline
+    &$n+9$ is odd&Premise&&\cr
+    &$\exists j\in\bb Z, n+9=2j+1$&Definition of Odd (1)&&\cr
+    &$n+9=2k+1$&Existential Instantiation (2)&&\cr
+    &$n = 2k-8$&Algebra (3)&&\cr
+    &$n^2-5n-14 = (4k^2-32k+64)-(10k+40)-14 = 2(2k^2-42k+5)$&Algebra (4)&&\cr
+    &$2k^2-42k+5\in\bb Z$&Closure of Integers (5)&&\cr
+    &$\exists j\in\bb Z, n^2-5n-14 = 2j$&Existential Generalization (5,6)&&\cr
+    &$n^2-5n-14$ is even&Definition of Even (7)&&\cr
+    \hline
+}
+\smallskip
+\endproof
+
+\li Let $n$ be an integer. Prove the statement ``If $3n^2+8$ is even,
+then $n$ is even.'' Make sure to include the introduction, body, and
+conclusion. Clearly state your reasoning for all statements and use a
+two-column proof for the body whenever possible.
+
+{\startlist
+    \li Prove the statement using a proof by contrapositive.
+
+    Assume for the sake of contrapositive that $n$ is odd.
+    We will show that $3n^2+8$ is odd.
+
+    \smallskip
+    \vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in
+    \halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+        &Statement&Reason&&\cr\hline
+        &$n$ is odd&Premise&&\cr
+        &$\exists j\in\bb Z, n=2j+1$&Definition of Odd (1)&&\cr
+        &$n=2k+1$&Existential Instantiation (2)&&\cr
+        &$3n^2+8=3(2k+1)^2+8$&Substitution (3)&&\cr
+        &$3n^2+8=12k^2+12k+11=2(6k^2+6k+5)+1$&Algebra (4)&&\cr
+        &$\exists j\in\bb Z, 3n^2+8=2j+1$&Existential Generalization (5)&&\cr
+        &$3n^2+8$ is odd&Definition of Odd (6)&&\cr
+        \hline
+    }
+    \smallskip
+    \endproof
+
+    \li Prove the statement using a proof by contradiction.
+
+    Let $3n^2+8$ be even and, for the sake of contradiction, let $n$ be
+    odd.
+
+    \smallskip
+    \vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in
+    \halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+        &Statement&Reason&&\cr\hline
+        &$n$ is odd&Premise&&\cr
+        &$3n^2+8$ is even&Premise&&\cr
+        &$\exists j\in\bb Z, n = 2j+1$&Definition of odd (1)&&\cr
+        &$n = 2k+1$&Existential instantiation (3)&&\cr
+        &$\exists j\in\bb Z, 3n^2 + 8 = 2j$&Definition of even (2)&&\cr
+        &$3n^2 + 8 = 2l$&Existential instantiation (5)&&\cr
+        &$3(2k+1)^2 + 8 = 2l$&Substitution (4,6)&&\cr
+        &$12k^2+12k+3 + 8 = 2l$&Algebra (7)&&\cr
+        &$l = (6k^2+6k+5) + 1/2$&Algebra (8)&&\cr
+        &$l\not\in\bb Z$&Integer Definition (9)&&\cr
+        \hline
+    }
+    \smallskip
+    We have reached a contradiction because $l$ is both an integer and
+    not an integer.
+    \endproof
+}
+
+\li Let $p$ be the product of 5 distinct integers, where each of the 5
+integers is between 1 and 63, inclusive. Prove or disprove that if $p$
+is odd, then at least one of these 5 integers in its product must be
+odd. Make sure to include the introduction, body, and conclusion.
+Clearly state your reasoning for all statements and use a two-column
+proof for the body whenever possible.
+
+Let $a_1,a_2,\ldots,a_5 \in\bb Z\cap[1,63],$ such that $i\neq j\impl
+a_i\neq a_j.$
+For the sake of contrapositive proof, let $a_i$ (for all $i$) be even.
+We will show that $p = a_1a_2a_3a_4a_5$ is even.
+
+\smallskip
+\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in
+\halign{\vrule\strut#\tabskip\tableskip&#\hfil&#\hfil&#\tabskip0pt&#\vrule\cr\hline
+    &Statement&Reason&&\cr\hline
+    &$a_1$ is even&Premise&&\cr
+    &$\exists j\in\bb Z, a_1=2j$&Definition of Even (1)&&\cr
+    &$a_1=2k$&Existential Instantiation (2)&&\cr
+    &$p = a_1a_2a_3a_4a_5$&Premise&&\cr
+    &$p = 2ka_2a_3a_4a_5$&Substitution (3,4)&&\cr
+    &$p = 2(ka_2a_3a_4a_5)$&Associative Property (5)&&\cr
+    &$\exists j\in\bb Z, p = 2j$&Existential Generalization (6)&&\cr
+    &$p$ is even&Definition of Even (7)&&\cr
+    \hline
+}
+\smallskip
+\endproof
+
+\bye
diff --git a/rosbruck/hostility.png b/rosbruck/hostility.png
new file mode 100644
index 0000000..01ab11b
--- /dev/null
+++ b/rosbruck/hostility.png
diff --git a/rosbruck/jung.png b/rosbruck/jung.png
new file mode 100644
index 0000000..90d44d0
--- /dev/null
+++ b/rosbruck/jung.png
diff --git a/rosbruck/surveys.tex b/rosbruck/surveys.tex
new file mode 100644
index 0000000..dd58cb4
--- /dev/null
+++ b/rosbruck/surveys.tex
@@ -0,0 +1,130 @@
+\input color
+\catcode`@=11
+\font\twelverm=ptmr7t at 12pt
+\def\rm{\fam\z@\twelverm}
+\font\twelvebf=ptmb7t at 12pt
+\def\bf{\fam\bffam\twelvebf}
+\catcode`@=12
+\rm
+\baselineskip=16pt
+
+\centerline{\bf Personality Surveys/Scales Results Sheet}
+
+{\bf Class Section:} B\par
+{\bf Name (Last, First):} Holden Rohrer\par
+{\bf GTID:} 903466338
+\bigskip
+
+{\bf\item{1.} Jung Typology Test (Myers-Briggs Indicator)}
+
+\smallskip
+\pdfximage width 4in{jung.png}\pdfrefximage\pdflastximage
+
+\iffalse
+\centerline{\color{red} ENFP}
+Extravert(41\%)  iNtuitive(38\%)  Feeling(41\%)  Perceiving(31\%)
+
+\item{$\bullet$} You have moderate preference of Extraversion over
+Introversion (41\%)
+\item{$\bullet$} You have moderate preference of Intuition over Sensing
+(38\%)
+\item{$\bullet$} You have moderate preference of Feeling over Thinking
+(41\%)
+\item{$\bullet$} You have moderate preference of Perceiving over Judging
+(31\%)
+\fi
+
+\bigskip
+{\bf\item{2.} Locus of Control (LOC) Survey (Julian Rotter)}
+
+18/23
+
+\bigskip
+{\bf\item{3.} Satisfaction with Life Survey}
+
+34/35
+
+\bigskip
+{\bf\item{4.} Hostility/Anger Scale}
+
+\smallskip
+\pdfximage width 4in{hostility.png}\pdfrefximage\pdflastximage
+
+\bigskip
+\vskip0pt plus 1in\goodbreak\vskip0pt plus -1in
+{\bf\item{5.} Summary} (1--2 pages single spaced, Times New Roman, size
+12, black font)
+\medskip\penalty200
+
+I was not very surprised with these results except by how clearly they
+confirmed my existing beliefs about myself.
+I consider myself very even-tempered and satisfied, my very low
+hostility/anger scale and high satisfaction with life confirmed.
+
+I've also taken the Myers-Briggs/Jung Typology test before, and last
+time I took it, I got ENFP, which is both who I believe I am and who I
+want to be, very personable and in-touch with my feelings.
+Interestingly, this is a change from results I've gotten in the past,
+which were always typically extraverted, but they used to be the more
+``analytical, considerate'' type of personality (sensing, thinking,
+judging) because I changed my mind about what sort of traits would make
+me happier.
+
+And the highly external locus of control makes a lot of sense with my
+socially-oriented politics (Marxism).
+I understand how strongly one's environments impact their choices, their
+disposition, and their ability to control their environment.
+It has also made me deeply sympathetic for other people's choices, which
+I think makes me a lot less hostile and more forgiving.
+
+My satisfaction with life correlates really well with my socio-economic
+status and the social desirability of my life plus
+non-competitiveness---meaning I can be content with my life and ignore
+anything ``negative'' that might happen.
+
+The main concern I have about these tests is that I'm rating myself more
+on my aspirations than accurate self-perception.
+Also, I've heard speculation around the inaccuracy and instability of
+the Meyers-Briggs test, so I'm personally a bigger fan of OCEAN
+(Openness, Conscientiousness, Extraversion, Agreeableness, and
+Neuroticism) traits because they were found to be stable over long
+periods of time and more closely respect the fact that personality is a
+spectrum instead of 16 tightly-aligned categories.
+
+These survey results don't give me any desire to improve myself because
+I'm already very satisfie with my life and my interpersonal relations.
+Probably my most surprising result would be that having an external
+locus of control doesn't demotivate me because I still believe that
+labor, social striving, and social changes do have power to influence
+change.
+I just reject the idea of individualist modes of change, and I believe
+that rewards aren't automatically going to be justly distributed.
+I'm glad that none of these results were surprising because that would
+mean I don't know myself very well.
+I also disagree with some of the external locus of control questions,
+like the one about war.
+That question makes more sense with a view of war as the result of a few
+people's choices, and doesn't really make sense with a social
+interpretation of politics.
+However, I still got an accurate approximation of my personality, so I
+can't complain.
+
+My Myers-Briggs results tell me that I like being around people and I
+get energized by their presence (extraversion), which isn't very
+surprising.
+Of course, I can still get drained from meeting too many new people or
+having a lot of small talk-y conversations, but I really do like and
+look forward to spending time around people I know even moderately well.
+I think that I fall into the category of moderately intuitive thinker,
+because I don't believe that my own beliefs are founded in any pure
+reasoning, and I really like working with theoretical ideas, which is
+why I'm considering work in a research or academic field.
+I also am very open and agreeable, so the ``perceiving type''
+(preferring freedom and flexibility) makes a lot of sense for my
+character, but I'm not opposed to settling down quickly on an idea if
+the situation demands it.
+I also try to prioritize people and emotions as much as I can over
+invalidating people with ``facts and logic,'' since that sort of thought
+process is simply rude.
+
+\bye
diff --git a/stanzione/Makefile b/stanzione/Makefile
new file mode 100644
index 0000000..4a3f928
--- /dev/null
+++ b/stanzione/Makefile
@@ -0,0 +1,25 @@
+.POSIX:
+.SUFFIXES: .tex .pdf
+
+PDFS = mm1.pdf
+
+PDFLATEX = pdflatex
+BIBER = biber
+
+all: $(PDFS)
+
+$(PDFS): sources.bib
+
+clean:
+	rm -f *.bbl *.blg *.log *.aux *.pdf *.run.xml *.bcf *.out
+
+.tex.pdf:
+	$(PDFLATEX) $<
+	$(BIBER) $*
+	$(PDFLATEX) $<
+	$(PDFLATEX) $<
+
+mm1.pdf: yang.jpg
+
+yang.jpg:
+	curl -o yang.jpg "https://upload.wikimedia.org/wikipedia/commons/7/7f/Yang-single_%28restoration%29.jpg"
diff --git a/stanzione/mm1.tex b/stanzione/mm1.tex
new file mode 100644
index 0000000..a03d49d
--- /dev/null
+++ b/stanzione/mm1.tex
@@ -0,0 +1,96 @@
+% Mastery Mailing 1
+\documentclass[12pt]{apa7}
+\usepackage[style=apa,backend=biber]{biblatex}
+\usepackage{graphicx}
+\setlength{\headheight}{15pt}
+
+% According to several sources, the following commands should be active
+% for an APA paper, but I just hate them.
+% \raggedright
+% \language255 % no hyphenation
+\parindent=.5in
+\linespread{1.85}
+
+\shorttitle{Mastery Mailing 1}
+
+\addbibresource{sources.bib}
+
+\leftheader{Rohrer}
+
+\begin{document}
+Hi Nana!
+\medskip
+
+I'm learning about meditation in psychology class, and I think you'd be
+interested in the subjective effects of ``open monitoring'' meditation.
+While some types of meditation focus on creating a ``trance'' state,
+many meditators report an increased sense of awareness of their own
+thoughts and surroundings \autocite[146]{textbook}.
+This really surprised me because I thought the medical effects of
+meditation---lowering heart rate, blood pressure, and stress
+levels---meant the conscious state of meditation was close to sleep.
+However, this may actually be true of some types of meditation.
+According to some psychologists, the term ``mindfulness'' has become
+conflated with so many different subjective experiences, news articles
+and new studies can't reliably mean 20-minute open-monitoring sessions
+or a global change in perception/awareness \autocite{doubt}.
+
+My textbook says that ``meditation involves using a mental or physical
+technique to induce a state of focused attention and heightened
+awereness'' \autocite[145]{textbook}.
+You're probably already familiar with some techniques of meditation,
+either from a religious (probably Buddhist) perspective or a secular
+perspective, but I'll examine a couple of techniques to see what
+psychologists are studying.
+There are two big categories: focused attention and open monitoring.
+Focused attention empties the mind of intrusive thoughts by thinking
+only about an object or your breathing or even movement (like in Tai
+Chi).
+Open monitoring is often transitioned to after focusing on breathing,
+but experts eventually can reach this state through ``effortless
+concentration'' \autocite{lutz}
+Open monitoring is acting as a passive observer of your own thoughts and
+external sensations.
+
+\begin{figure}[ht]
+    \begin{center}
+    \includegraphics[height=2.5in]{yang}
+    \par\emph{A Practioner of Tai Chi. Unknown author, Public domain, via Wikimedia Commons}
+    \end{center}
+\end{figure}
+
+This leads into the qualitative study that I found about meditation and
+depersonalization.
+Depersonalization is usually treated as a disease where the subject
+reports feeling disconnected from their own actions, as if they were
+watching someone else live their life.
+If this is brought on unexpectedly, it can be very distressing, and in
+certain cases, people have experienced it from meditation and
+immediately sought medical help \autocite{castillo}.
+This study also includes interviews with long-term meditation
+practitioners (not monks, just typical Western workers and
+businesspeople) who all experienced depersonalization.
+They report feeling mildly content and entirely lack strong emotions,
+and many are totally accustomed to work happening outside of their
+``self.''
+However, as one Mr. A said, ``I don't want to leave the thing sounding
+better than it is, because it's not bad, but it's not wonderful.''
+They describe what Yoga psychology calls enlightenment, especially the
+less persistent experience of derealization.
+Some meditators said they saw every object as being conscious, having
+auras, and vibrating to some degree.
+
+I thought of you when I learned about this because I know you are
+interested in enlightenment and mystic symbols, so people achieving
+these states after only a year or a few days of consistent meditation is
+interesting.
+I was really intrigued that these practitioners weren't just
+recommending the experience to everyone.
+Even though meditators who experience this accept their experience, I'm
+afraid of the loss of strong emotions because I like emotional highs
+even if it means I have to experience some disappointment and
+depression.
+
+\vfil\eject
+\printbibliography
+\end{document}
diff --git a/stanzione/sources.bib b/stanzione/sources.bib
new file mode 100644
index 0000000..fda2232
--- /dev/null
+++ b/stanzione/sources.bib
@@ -0,0 +1,43 @@
+@book{textbook,
+    title = "Discovering Psychology",
+    author = "Sandra Hockenbury and Susan Nolan",
+    edition = 8,
+    publisher = "Macmillan Learning",
+    year = 2018,
+}
+
+@article{doubt,
+    author = {Nicholas T. Van Dam and Marieke K. van Vugt and David R. Vago and Laura Schmalzl and Clifford D. Saron and Andrew Olendzki and Ted Meissner and Sara W. Lazar and Catherine E. Kerr and Jolie Gorchov and Kieran C. R. Fox and Brent A. Field and Willoughby B. Britton and Julie A. Brefczynski-Lewis and David E. Meyer},
+    title = {Mind the Hype: A Critical Evaluation and Prescriptive Agenda for Research on Mindfulness and Meditation},
+    journal = {Perspectives on Psychological Science},
+    volume = {13},
+    number = {1},
+    pages = {36-61},
+    year = {2018},
+    doi = {10.1177/1745691617709589},
+    note ={PMID: 29016274},
+    URL = {https://doi.org/10.1177/1745691617709589},
+}
+
+@article{lutz,
+    title = {Attention regulation and monitoring in meditation},
+    journal = {Trends in Cognitive Sciences},
+    volume = {12},
+    number = {4},
+    pages = {163-169},
+    year = {2008},
+    issn = {1364-6613},
+    doi = {10.1016/j.tics.2008.01.005},
+    url = {https://www.sciencedirect.com/science/article/pii/S1364661308000521},
+    author = {Antoine Lutz and Heleen A. Slagter and John D. Dunne and Richard J. Davidson},
+}
+
+@article{castillo,
+    title = "Depersonalization and Meditation",
+    author = "Richard J. Castillo",
+    pages = {158-168},
+    journal = "Psychiatry",
+    volume = 53,
+    year = {1990},
+    month = {May}
+}