new teachers, new work

4 files changed, 769 insertions, 0 deletions

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