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

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\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}


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