diff --git a/Advanced/Mock a Mockingbird/main.tex b/Advanced/Mock a Mockingbird/main.tex index 4e81ab3..ceba1af 100755 --- a/Advanced/Mock a Mockingbird/main.tex +++ b/Advanced/Mock a Mockingbird/main.tex @@ -57,225 +57,7 @@ Based on a book of the same name. } - \section{Introduction} - - A certain enchanted forest is inhabited by talking birds. Each of these birds has a name, and will respond whenever it hears the name of another. Suppose you are exploring this forest and come across the bird $A$. You call the name of bird $B$. $A$ hears you and responds with the name of some other bird, which we will designate $AB$. - - Bird $AB$ is, by definition, $A$'s response to $B$. - - \vspace{2mm} - - As you wander around this forest, you quickly discover two interesting facts: - \begin{enumerate}[itemsep = 1mm] - \item $A$'s responds to $B$ mustn't be the same as $B$'s response to $A$. - \item Given three birds $A$, $B$, and $C$, $(AB)C$ and $A(BC)$ are not necessarily the same bird. \\ - Bird $A(BC)$ is $A$'s response to bird $BC$, while $(AB)C$ is $AB$'s response to $C$. \\ - Thus, $ABC$ is ambiguous. Parenthesis are mandatory. - \end{enumerate} - - \vspace{2mm} - - You also find that this forest has two laws: - \begin{enumerate}[itemsep = 1mm] - \item $L_1$, \textit{The Law of Composition}: \\ - For any two birds $A$ and $B$, there must be a bird $C$ so that $Cx = A(Bx)$ - \item $L_2$, \textit{The Law of the Mockingbird}: \\ - The forest must contain the Mockingbird $M$, which always satisfies $Mx = xx$. \\ - In other words, the Mockingbird's response to any bird $x$ is the same as $x$'s response to itself. - \end{enumerate} - - \vfill - - \definition{} - We say a bird $A$ is fond of a bird $B$ if $A$ responds to $B$ with $B$. \\ - In other words, $A$ is fond of $B$ if $AB = B$. - - \vfill - - \definition{} - We say a bird $C$ \textit{composes} $A$ with $B$ if for any bird $x$, - $$ - Cx = A(Bx) - $$ - In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$. - - \vfill - \pagebreak - - \section{To Mock a Mockingbird} - - \problem{} - The bear, a lifelong resident of the forest, tells you that any bird $A$ is fond of at least one other bird. \\ - Complete his proof. - \begin{alltt} - let A \cmnt{Let A be any any bird.} - let Cx := A(Mx) \cmnt{Define C as the composition of A and M} - - \cmnt{The rest is up to you.} - CC = ?? - \end{alltt} - - - \begin{helpbox} - \texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\ - \texttt{Def:} $A$ is fond of $B$ if $AB = B$ - \end{helpbox} - - - \begin{solution} - \begin{alltt} - let A \cmnt{Let A be any any bird.} - let Cx := A(Mx) \cmnt{Define C as the composition of A and M} - CC = A(MC) - = A(CC) \qed{} - \end{alltt} - \end{solution} - - \vfill - \problem{} - We say a bird $A$ is \textit{egocentric} if it is fond if itself. - - Show that the laws of the forest guarantee that at least one bird is egocentric. - - - \begin{helpbox} - \texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\ - \texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\ - \texttt{Lem:} Any bird is fond of at least one bird. - \end{helpbox} - - \begin{solution} - \begin{alltt} - \cmnt{We know M is fond of at least one bird.} - let E so that ME = E - - ME = E \cmnt{By definition of fondness} - ME = EE \cmnt{By definition of M} - \thus{} EE = E \qed{} - \end{alltt} - \end{solution} - - \vfill - \pagebreak - - - \problem{} - We say a bird $A$ is \textit{agreeable} if for all birds $B$, there is at least one bird $x$ on which $A$ and $B$ agree. \\ - This means that $Ax = Bx$. - - \begin{helpbox} - \texttt{Def:} $Mx := xx$ - \end{helpbox} - - \begin{solution} - We know that $Mx = xx$. \\ - - From this definition, we see that $M$ agrees with any $x$ on $x$ itself. - \end{solution} - - \vfill - - \problem{} - Take two birds $A$ and $B$. Let $C$ be their composition. \\ - Show that $A$ must be agreeable if $C$ is agreeable. \\ - The bear has again given you a hint. - \begin{alltt} - \cmnt{Given information} - let A, B - let Cx := A(Bx) - - let D \cmnt{Arbitrary bird} - let Ex := D(Bx) \cmnt{Define E as the composition of D and B} - Cy = ?? - \end{alltt} - - \begin{helpbox}[0.65] - \texttt{Def:} $A$ is agreeable if $Ax = Bx$ for all $B$ with some $x$. \\ - \texttt{Law:} For any $A, B$, there is C defined by Cx = A(Bx) - \end{helpbox} - - - \begin{solution} - \begin{alltt} - \cmnt{Given information} - let A, B - let Cx := A(Bx) - - let D \cmnt{Arbitrary bird} - let Ex := D(Bx) \cmnt{Define E as the composition of D and B} - Cy = Ey \cmnt{For some y, because C is agreeable} - \thus{} A(By) = Ey - \thus{} A(By) = D(By) \qed{} - \end{alltt} - \end{solution} - - \vfill - \pagebreak - - \problem{} - Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ defined by $Dx = A(B(Cx))$ - - \begin{solution} - \begin{alltt} - let A, B, C - - \cmnt{Invoke the Law of Composition:} - let Q := BC - let D := AQ - - D = AQ - = A(BC) \qed{} - \end{alltt} - \end{solution} - - \vfill - - - \problem{} - We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\ - Note that $x$ and $y$ may be the same bird. \\ - Show that any two birds in this forest are compatible. \\ - \begin{alltt} - let A, B - let Cx := A(Bx) - \end{alltt} - - \begin{helpbox} - \texttt{Law:} Law of composition \\ - \texttt{Lem:} Any bird is fond of at least one bird. - \end{helpbox} - - \begin{solution} - \begin{alltt} - let A, B - - let Cx := A(Bx) \cmnt{Composition} - let y := Cy \cmnt{Let C be fond of y} - - Cy = y - \thus{} A(By) = y - - let x := By \cmnt{Rename By to x} - Ax = y \qed{} - \end{alltt} - \end{solution} - - \vfill - - \problem{} - Show that any bird that is fond of at least one bird is compatible with itself. - - \begin{solution} - \begin{alltt} - let A - let x so that Ax := x - Ax = x \qed{} - \end{alltt} - That's it. - \end{solution} - - - \vfill - \pagebreak + \input{parts/00 intro} + \input{parts/01 tmam} \end{document} \ No newline at end of file diff --git a/Advanced/Mock a Mockingbird/parts/00 intro.tex b/Advanced/Mock a Mockingbird/parts/00 intro.tex new file mode 100644 index 0000000..eb070f3 --- /dev/null +++ b/Advanced/Mock a Mockingbird/parts/00 intro.tex @@ -0,0 +1,44 @@ +\section{Introduction} + +A certain enchanted forest is inhabited by talking birds. Each of these birds has a name, and will respond whenever it hears the name of another. Suppose you are exploring this forest and come across the bird $A$. You call the name of bird $B$. $A$ hears you and responds with the name of some other bird, which we will designate $AB$. + +Bird $AB$ is, by definition, $A$'s response to $B$. + +\vspace{2mm} + +As you wander around this forest, you quickly discover two interesting facts: +\begin{enumerate}[itemsep = 1mm] + \item $A$'s responds to $B$ mustn't be the same as $B$'s response to $A$. + \item Given three birds $A$, $B$, and $C$, $(AB)C$ and $A(BC)$ are not necessarily the same bird. \\ + Bird $A(BC)$ is $A$'s response to bird $BC$, while $(AB)C$ is $AB$'s response to $C$. \\ + Thus, $ABC$ is ambiguous. Parenthesis are mandatory. +\end{enumerate} + +\vspace{2mm} + +You also find that this forest has two laws: +\begin{enumerate}[itemsep = 1mm] + \item $L_1$, \textit{The Law of Composition}: \\ + For any two birds $A$ and $B$, there must be a bird $C$ so that $Cx = A(Bx)$ + \item $L_2$, \textit{The Law of the Mockingbird}: \\ + The forest must contain the Mockingbird $M$, which always satisfies $Mx = xx$. \\ + In other words, the Mockingbird's response to any bird $x$ is the same as $x$'s response to itself. +\end{enumerate} + +\vfill + +\definition{} +We say a bird $A$ is fond of a bird $B$ if $A$ responds to $B$ with $B$. \\ +In other words, $A$ is fond of $B$ if $AB = B$. + +\vfill + +\definition{} +We say a bird $C$ \textit{composes} $A$ with $B$ if for any bird $x$, +$$ + Cx = A(Bx) +$$ +In other words, this means that $C$'s response to $x$ is the same as $A$'s response to $B$'s response to $x$. + +\vfill +\pagebreak \ No newline at end of file diff --git a/Advanced/Mock a Mockingbird/parts/01 tmam.tex b/Advanced/Mock a Mockingbird/parts/01 tmam.tex new file mode 100644 index 0000000..12429a3 --- /dev/null +++ b/Advanced/Mock a Mockingbird/parts/01 tmam.tex @@ -0,0 +1,175 @@ +\section{To Mock a Mockingbird} + +\problem{} +The bear, a lifelong resident of the forest, tells you that any bird $A$ is fond of at least one other bird. \\ +Complete his proof. +\begin{alltt} + let A \cmnt{Let A be any any bird.} + let Cx := A(Mx) \cmnt{Define C as the composition of A and M} + + \cmnt{The rest is up to you.} + CC = ?? +\end{alltt} + + +\begin{helpbox} + \texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\ + \texttt{Def:} $A$ is fond of $B$ if $AB = B$ +\end{helpbox} + + +\begin{solution} + \begin{alltt} + let A \cmnt{Let A be any any bird.} + let Cx := A(Mx) \cmnt{Define C as the composition of A and M} + CC = A(MC) + = A(CC) \qed{} + \end{alltt} +\end{solution} + +\vfill +\problem{} +We say a bird $A$ is \textit{egocentric} if it is fond if itself. + +Show that the laws of the forest guarantee that at least one bird is egocentric. + + +\begin{helpbox} + \texttt{Law:} There exists a Mockingbird, $Mx := xx$ \\ + \texttt{Def:} $A$ is fond of $B$ if $AB = B$ \\ + \texttt{Lem:} Any bird is fond of at least one bird. +\end{helpbox} + +\begin{solution} + \begin{alltt} + \cmnt{We know M is fond of at least one bird.} + let E so that ME = E + + ME = E \cmnt{By definition of fondness} + ME = EE \cmnt{By definition of M} + \thus{} EE = E \qed{} + \end{alltt} +\end{solution} + +\vfill +\pagebreak + + +\problem{} +We say a bird $A$ is \textit{agreeable} if for all birds $B$, there is at least one bird $x$ on which $A$ and $B$ agree. \\ +This means that $Ax = Bx$. + +\begin{helpbox} + \texttt{Def:} $Mx := xx$ +\end{helpbox} + +\begin{solution} + We know that $Mx = xx$. \\ + + From this definition, we see that $M$ agrees with any $x$ on $x$ itself. +\end{solution} + +\vfill + +\problem{} +Take two birds $A$ and $B$. Let $C$ be their composition. \\ +Show that $A$ must be agreeable if $C$ is agreeable. \\ +The bear has again given you a hint. +\begin{alltt} + \cmnt{Given information} + let A, B + let Cx := A(Bx) + + let D \cmnt{Arbitrary bird} + let Ex := D(Bx) \cmnt{Define E as the composition of D and B} + Cy = ?? +\end{alltt} + +\begin{helpbox}[0.65] + \texttt{Def:} $A$ is agreeable if $Ax = Bx$ for all $B$ with some $x$. \\ + \texttt{Law:} For any $A, B$, there is C defined by Cx = A(Bx) +\end{helpbox} + + +\begin{solution} + \begin{alltt} + \cmnt{Given information} + let A, B + let Cx := A(Bx) + + let D \cmnt{Arbitrary bird} + let Ex := D(Bx) \cmnt{Define E as the composition of D and B} + Cy = Ey \cmnt{For some y, because C is agreeable} + \thus{} A(By) = Ey + \thus{} A(By) = D(By) \qed{} + \end{alltt} +\end{solution} + +\vfill +\pagebreak + +\problem{} +Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$ defined by $Dx = A(B(Cx))$ + +\begin{solution} + \begin{alltt} + let A, B, C + + \cmnt{Invoke the Law of Composition:} + let Q := BC + let D := AQ + + D = AQ + = A(BC) \qed{} + \end{alltt} +\end{solution} + +\vfill + + +\problem{} +We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\ +Note that $x$ and $y$ may be the same bird. \\ +Show that any two birds in this forest are compatible. \\ +\begin{alltt} + let A, B + let Cx := A(Bx) +\end{alltt} + +\begin{helpbox} + \texttt{Law:} Law of composition \\ + \texttt{Lem:} Any bird is fond of at least one bird. +\end{helpbox} + +\begin{solution} + \begin{alltt} + let A, B + + let Cx := A(Bx) \cmnt{Composition} + let y := Cy \cmnt{Let C be fond of y} + + Cy = y + \thus{} A(By) = y + + let x := By \cmnt{Rename By to x} + Ax = y \qed{} + \end{alltt} +\end{solution} + +\vfill + +\problem{} +Show that any bird that is fond of at least one bird is compatible with itself. + +\begin{solution} + \begin{alltt} + let A + let x so that Ax := x + Ax = x \qed{} + \end{alltt} + That's it. +\end{solution} + + +\vfill +\pagebreak \ No newline at end of file