\documentclass[12pt]{article}

\usepackage{times,mathptm}
\pagestyle{myheadings}



\parindent 0in
\setlength{\parindent}{0in}
\setlength{\parskip}{1ex}

\topmargin -0.1in \headheight\baselineskip
\textheight 8.5in         % 1in top and bottom margin
\textwidth 6.5in        % 1in left and right margin
\oddsidemargin 0in
\evensidemargin 0in




\usepackage{amsmath}
\usepackage{amssymb}


% \setlength{\itemsep}{0pt}

\markright{\footnotesize CS 451 Programming Paradigms, Spring 2001}

\begin{document}

\section*{Homework set 9: $\lambda$-calculus --- due Monday 5 March}

Total number of points available on this homework is 100. Full credit
is equivalent to 100 points.

\begin{enumerate}


\item (30 pts.)      
Show that the following terms have a normal form:
\begin{enumerate}
\item $(\lambda y. yyy) ((\lambda ab.a) \textsf{\textbf{I}} (\textsf{\textbf{S}}\textsf{\textbf{S}}))$
\item $(\lambda yz. zy) ((\lambda x. xxx)(\lambda x. xxx))(\lambda w. \textsf{\textbf{I}})$
\item $\textsf{\textbf{S}}\textsf{\textbf{S}}\textsf{\textbf{S}}\textsf{\textbf{S}}\textsf{\textbf{S}}\textsf{\textbf{S}}\textsf{\textbf{S}}$
\end{enumerate}




\item (40 pts.)      
For each of the following $\lambda$-expressions either find its normal form
or show that it has no normal form:
\begin{enumerate}
\item $(\lambda x. x \: x) (\lambda x. x)$
\item $(\lambda x. x \: x) (\lambda x. x \: x)$
\item $\textsf{\textbf{Y}}$ \ \emph{(see below)}
\item $\textsf{\textbf{Y}} (\lambda y. y)$
\end{enumerate}



\item (30 pts.)     
A $\lambda$-expression
$\textsf{\textbf{Fix}}$ 
with the property that 
$\textsf{\textbf{Fix}} \: E = E (\textsf{\textbf{Fix}} \: E)$ for any $E$ 
is called a fixed-point operator (or combinator). 
One well known fixed-point operator is $\textsf{\textbf{Y}}$, defined as
$\textsf{\textbf{Y}} \triangleq \lambda f. (\lambda x. f (x \: x))(\lambda x. f (x \: x))$. 
Show that $\textsf{\textbf{Y}}$ is a fixed-point operator.





\end{enumerate}


\end{document}