Fixed Kestrel problems

Advanced/Lambda Calculus/parts/01 combinators.tex

@@ -2,43 +2,41 @@
 
 \definition{}
 A \textit{free variable} in a $\lm$-expression is a variable that isn't bound to any input. \par
-For example, $b$ is a free variable in $\lm a. b$. The same is true of $\star$ in any of the previous pages.
+For example, $b$ is a free variable in $(\lm a.a)~b$.
 
-A \textit{combinator} is a function with no free variables.
+\definition{Combinators}
+A \textit{combinator} is a lambda expression with no free variables.
 
-
+\vspace{1mm}
-
-\definition{The Kestrel}
 
 Notable combinators are often named after birds.\hspace{-0.5ex}\footnotemark{} We've already met a few: \par
 The \textit{Idiot}, $I = \lm a.a$ \par
 The \textit{Mockingbird}, $M = \lm f.ff$ \par
 The \textit{Cardinal}, $C = \lm fgx.(~ f(g(x)) ~)$
-
+The \textit{Kestrel}, $K = \lm ab . a$
-\footnotetext{Raymond Smullyan's \textit{To Mock a Mockingbird} is responsible for this.}
-
-\vspace{2ex}
-
-Another notable combinator is $K$, the \textit{Kestrel}:
-$$
-	K = \lm ab . a
-$$
 
 
 
 \problem{}
-What does the Kestrel do? Explain in plain English. \par
+If we give the Kestrel two arguments, it does something interesting: \par
-\hint{What is $(K~\heartsuit~\star)$?}
+It selects the first and rejects the second. \par
+Convince yourself of this fact by evaluating $(K~\heartsuit~\star)$.
 
-\vspace{2cm}
+\vfill
 
-\problem{}
+\problem{}<kitedef>
-Reduce $(K~I)$ to derive the \textit{Kite}. How does the Kite compare to the Kestrel? \par
+Modify the Kestrel so that it selects its \textbf{second} argument and rejects the first. \par
-We'll call the Kite KI.
 
 \begin{solution}
-	$\text{KI} = \lm ab . b$.
+	$\lm ab . b$.
 \end{solution}
 
+\vfill
+
+\problem{}
+We'll call the combinator from \ref{kitedef} the \textit{Kite}, $KI$. \par
+Show that we can also obtain the kite by evaluating $(K~I)$.
+
+
 \vfill
 \pagebreak