\section{Bonus Problems} \definition{} The identity bird has sometimes been maligned, owing to the fact that whatever bird x you call to $I$, all $I$ does is to echo $x$ back to you. \vspace{2mm} Superficially, the bird $I$ appears to have no intelligence or imagination; all it can do is repeat what it hears. For this reason, in the past, thoughtless students of ornithology referred to it as the idiot bird. However, a more profound or- nithologist once studied the situation in great depth and dis- covered that the identity bird is in fact highly intelligent! The real reason for its apparently unimaginative behavior is that it has an unusually large heart and hence is fond of every bird! When you call $x$ to $I$, the reason it responds by calling back $x$ is not that it can't think of anything else; it's just that it wants you to know that it is fond of $x$! \vspace{2mm} Since an identity bird is fond of every bird, then it is also fond of itself, so every identity bird is egocentric. However, its egocentricity doesn't mean that it is any more fond of itself than of any other bird!. \problem{} The laws of the forest no longer apply. Suppose we are told that the forest contains an identity bird $I$ and that $I$ is agreeable. \ Does it follow that every bird must be fond of at least one bird? \vfill \problem{} Suppose we are told that there is an identity bird $I$ and that every bird is fond of at least one bird. \ Does it necessarily follow that $I$ is agreeable? \vfill \pagebreak \problem{} Suppose we are told that there is an identity bird $I$, but we are not told whether $I$ is agreeable or not. However, we are told that every pair of birds is compatible. \ Which of the following conclusiens can be validly drawn? \begin{itemize} \item Every bird is fond of at least one bird \item $I$ is agreeable. \end{itemize} \vfill \problem{} The identity bird $I$, though egocentric, is in general not hope- lessly egocentric. Indeed, if there were a hopelessly egocentric identity bird, the situation would be quite sad. Why? \vfill \definition{} A bird $L$ is called a lark if the following holds for any birds $x$ and $y$: \[ (Lx)y = x(yy) \] \problem{} Prove that if the forest contains a lark $L$ and an identity bird $I$, then it must also contain a mockingbird $M$. \vfill \pagebreak \problem{} Why is a hopelessly egocentric lark unusually attractive? \vfill \problem{} Assuming that no bird can be both a lark and a kestrel---as any ornithologist knows!---prove that it is impossible for a lark to be fond of a kestrel. \vfill \problem{} It might happen, however, that a kestrel is fond of a lark. \par Show that in this case, \textit{every} bird is fond of the lark. \vfill