\documentclass[11pt]{article} \renewcommand{\thepage}{} \input{pset-defs} \begin{document} \begin{center}\large\bf CS 516---Software Foundations via Formal Languages---Spring 2022 \end{center} \begin{center}\Large\bf Problem Set 1 \end{center} \begin{center}\large\bf Due by 5pm on Friday, February 4 \\ Submission via Gradescope \end{center} \section*{Problem 1 (25 points)} Suppose $X$ is a set, and $x\in X$. What are the elements of $\emptyset\fun X$, $X\fun\emptyset$, $\{x\}\fun X$ and $X\fun\{x\}$? Prove that your answers are correct. \section*{Problem 2 (25 points)} Use mathematical induction (Theorem~1.2.1) to prove that, for all $n\in\nats$, if $n\geq 4$, then $2^n < n!$. Here $n!$ is the factorial of $n$, defined recursively by: \begin{enumerate}[\quad(1)] \item $0! = 1$; \item $(n + 1)! = (n + 1) * n!$, for all $n\in\nats$. \end{enumerate} \section*{Problem 3 (25 points)} Use strong induction (Theorem~1.2.4) to prove that, for all $n\in\nats$, if $n\geq 1$, then there are $i,j\in\nats$ such that $n=2^i(2j + 1)$. \section*{Problem 4 (25 points)} Define $f\in\ints\fun\ints$ by: \begin{displaymath} f\,n = \left\{ \begin{array}{ll} 0 & \eqtxtr{if} n = 0 , \\ 1 - n & \eqtxtr{if} n\geq 1, \\ -n - 1 & \eqtxtr{if} n\leq -1 . \end{array} \right. \end{displaymath} Given $n\in\ints$, define the $l$-th \emph{iteration of} $f$ \emph{on} $n$ ($f^l(n)\in\ints$), for $l\in\nats$, by recursion: \begin{align*} f^0(n) &= n , \\ f^{l + 1}(n) &= f(f^l(n)) . \end{align*} Then we have that: \begin{enumerate}[\quad(1)] \item For all $n\in\ints$, $f^1(n) = f\,n$. \item For all $n\in\ints$ and $l, m\in\nats$, $f^{l+m}(n)=f^m(f^l(n))$. \end{enumerate} \bigskip\noindent Let $R$ be the relation on $\ints$ defined by: for all $n,m\in\ints$, $n\mathrel{R}m$ iff $|n|<|m|$. Here $|n|$ is the absolute value of $n$: $n$, if $n\geq 0$; and $-n$, if $n\leq -1$. We have the following properties of absolute value, where $n,m\in\ints$: \begin{enumerate}[\quad(1)] \item if $n\leq m$, then $|n - m| = m - n$; \item if $m\leq n$, then $|n - m| = n - m$. \end{enumerate} Because $|\cdot|\in\ints\fun\nats$ and $<$ is well-founded on $\nats$, Proposition~1.2.10 tells us that $R$ is well-founded on $\ints$. \bigskip\noindent Use well-founded induction on $R$ (Theorem~1.2.8) to show that, for all $n\in\ints$, there is an $l\in\nats$ such that $f^l(n) = 0$. \end{document} %%% Local Variables: %%% mode: latex %%% TeX-master: t %%% End: