Edits

Advanced/Definable Sets/main.tex

@@ -9,6 +9,7 @@
 
 % for \coloneqq, a centered :=
 \usepackage{mathtools}
+\usepackage{units}
 
 \uptitlel{Advanced 2}
 \uptitler{\smallurl{}}

Advanced/Definable Sets/parts/0 logic.tex

@@ -81,13 +81,33 @@ Evaluate the following.
 	\item $\texttt{F} \lor \texttt{T}$
 	\item $\texttt{T} \land \texttt{T}$
 	\item $(\texttt{T} \land \texttt{F}) \lor \texttt{T}$
-	\item $(\texttt{T} \land \texttt{F}) \lor \texttt{T}$
+	\item $(\lnot (\texttt{F} \lor \lnot \texttt{T}) ) \rightarrow \lnot \texttt{T}$
-	\item $(\lnot (\texttt{F} \lor \lnot \texttt{T}) ) \rightarrow \texttt{T}$
 	\item $(\texttt{F} \rightarrow \texttt{T}) \rightarrow (\lnot \texttt{F} \lor \lnot \texttt{T})$
 \end{itemize}
 
+\begin{solution}
+	\texttt{F}
+	\texttt{T}
+	\texttt{T}
+	\texttt{T}
+	\texttt{F}
+	\texttt{T}
+\end{solution}
+
 \vfill
 \pagebreak
+
+
+
+
+
+
+
+
+
+
+
+
 \begin{instructornote}
 	We can also think of $[x \geq 0] \rightarrow b$ as follows:
 	if $x$ isn't the kind of object we care about, we evaluate true and
@@ -117,8 +137,18 @@ Evaluate the following.
 	\item $(A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)$ for any $A, B$
 \end{itemize}
 
+\begin{instructornote}
+	Note that the last formula is the contrapositive of $A \rightarrow B$.
+\end{instructornote}
+
+\begin{solution}
+	All are true.
+\end{solution}
+
 \vfill
 
+% Show that A -> B ^ B -> A = T iff A = B
+
 \problem{}
 Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par
 That is, show that these expressions always evaluate to the same value given

Advanced/Definable Sets/parts/1 structures.tex

@@ -22,7 +22,7 @@ Symbols come in three types:
 
 	\item \textit{Function symbols}, which let us navigate between elements of our universe. \par
 		Examples: $+, \times, \sin{x}, \sqrt{x}$ \par
-		\note{In this handout, symbols we usually call \say{operators} are also called functions. \par
+		\note{Note that symbols we usually call \say{operators} are functions under this definition. \par
 		The only difference between $a + b$ and $+(a, b)$ is notation.}
 		\vspace{2mm}
 
@@ -31,7 +31,7 @@ Symbols come in three types:
 		\vspace{2mm}
 \end{itemize}
 
-The equality check $=$ is \textbf{not} a relation symbol. It is included in every structure by default. \par
+The equality check $=$ is \textit{not} a relation symbol. It is included in every structure by default. \par
 By definition, $a = b$ is true if and only if $a$ and $b$ are the same element of our universe.
 
 
@@ -61,10 +61,15 @@ others, we must define them using the tools this structure offers.
 
 \vspace{2mm}
 
+% NOTE: this is a great example for typesetting.
+% The line breaks here are ugly without a centered sentence.
 To \say{define} an element of a set, we need to write a sentence that is only true for that element. \par
-For example, if we want to define 2 in the structure above, \par
+If we want to define 2 in the structure above,
-we could use the sentence \say{$2$ is the $x$ that satisfies $[1 + 1 = x]$.} \par
+we could use the following sentence:
-This is a valid definition because $2$ is the \textbf{only} element of $\mathbb{Z}$ for which $[1 + 1 = x]$
+\begin{center}
+	\say{$2$ is the $x$ that satisfies $[1 + 1 = x]$.} \par
+\end{center}
+This is a valid definition because $2$ is the \textit{only} element of $\mathbb{Z}$ for which $[1 + 1 = x]$
 evaluates to \texttt{true}.
 
 
@@ -81,6 +86,9 @@ Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, 1, +, -, <\} \Bigr)$.
 
 
 
+
+
+
 Let us formalize what we found in the previous two problems. \par
 
 \definition{Formulas}
@@ -89,31 +97,51 @@ of constants, functions, relations, \par and logical operators.
 
 \vspace{2mm}
 
-You already know what a \say{well-formed} string is: $1 + 1$ is fine, $\sqrt{+}$ is nonsense. \par
+You already know what a \say{well-formed string} is: $1 + 1$ is fine, $\sqrt{+}$ is nonsense. \par
-For the sake of time, I will not provide a formal definition. It isn't particularly interesting.
+For the sake of time, I will not provide a formal definition --- it isn't particularly interesting.
 
 \vspace{2mm}
 
-As a quick example, the formula $\phi(x) = [1 + 1 = x]$ evaluates to \texttt{true} when $x$ is 2 \par
+As a quick example, the formula $\psi \coloneqq [\lnot (1 = 1)]$ is always false, \par
-and to \texttt{false} otherwise.
+and $\varphi(x) \coloneqq [1 + 1 = x]$ evaluates to \texttt{true} only when $x$ is 2.
+
+
+
 
 
 \definition{Free Variables}
 A formula can contain one or more \textit{free variables.} These are denoted $\varphi{(a, b, ...)}$. \par
-Formulas with free variables let us define \say{properties} that certain objects have. \par
+Formulas with free variables let us define \say{properties} that certain objects have.
-
-For example, $x$ is a free variable in the formula above. \par
-$\varphi(2)$ is \texttt{true} and $\varphi(-3)$ is \texttt{false}. \par
 
 \vspace{2mm}
 
-This \say{free variable} notation is much like the function notation we are used to: \par
+For example, consider the two formulas from the previous definition, $\psi$ and $\varphi$:
-$\varphi(x) = [x > 0]$ is similar to $f(x) = x + 1$, since the values of $\varphi(x)$ and $f(x)$ depend on $x$.
+\begin{itemize}
+	\item $\psi \coloneqq [\lnot (1 = 1)]$ \par
+	There are no free variables in this formula. \par
+	In any structure, $\psi$ is always either \texttt{true} or \texttt{false}.
+
+	\vspace{2mm}
+
+	\item $\varphi(x) \coloneqq [1 + 1 = x]$ \par
+	This formula has one free variable, labeled $x$. \par
+	The value of $\varphi(x)$ depends on the $x$ we're talking about: \par
+	$\varphi(72)$ is false, and $\varphi(2)$ is true.
+\end{itemize}
+
+\vspace{2mm}
+
+\note{
+	This \say{free variable} notation is very similar to the function notation we are used to: \par
+	The values of both $\varphi(x) \coloneqq [x > 0]$ and $f(x) = x + 1$ depend on $x$.
+}
+
+
 
 
 \definition{Definable Elements}
-Say $S$ is a structure with a universe $U$. \par
+Let $S$ be a structure over a universe $U$. \par
-We say an element $e \in U$ is \textit{definable in $S$} if we can write a formula that only $e$ satisfies.
+We say an element $x \in U$ is \textit{definable in $S$} if we can write a formula $\varphi(x)$ that only $x$ satisfies.
 
 
 \problem{}
@@ -121,10 +149,16 @@ Define 2 in the structure $\Bigl( \mathbb{Z^+} ~\big|~ \{4, \times \} \Bigr)$. \
 \hint{$\mathbb{Z}^+ = \{1, 2, 3, ...\}$. Also, $2 \times 2 = 4$.}
 
 \begin{solution}
-	$2$ is the only element in $\mathbb{Z}^+$ that satisfies $\varphi(x) = [x \times x = 4]$.
+	$2$ is the only element in $\mathbb{Z}^+$ that satisfies $\varphi(x) \coloneqq [x \times x = 4]$.
 \end{solution}
 
+
 \vfill
+\pagebreak
+
+
+
+
 
 
 \problem{}
@@ -132,8 +166,10 @@ Try to define 2 in the structure $\Bigl( \mathbb{Z} ~\big|~ \{4, \times \} \Bigr
 Why can't you do it?
 
 \begin{solution}
-	This isn't possible. We could try $\varphi(x) = [x \times x = 4]$, but this is satisfied by both $2$ and $-2$. \par
+	We could try $\varphi(x) \coloneqq [x \times x = 4]$, but this is satisfied by both $2$ and $-2$. \par
-	We have no way to distinguish between negative and positive numbers.
+	We have no way to distinguish between negative and positive numbers. \par
+
+	\note{This problem is intentionally hand-wavy. We don't have the tools to write a proper proof.}
 
 	\begin{instructornote}
 		Actually, it is. Bonus problem: how? \par
@@ -145,12 +181,21 @@ Why can't you do it?
 
 
 \problem{}
-What numbers are definable in the structure $\Bigl( \mathbb{R}^+_0 ~\big|~ \{1, 2, \div \} \Bigr)$?
+Consider the structure $\Bigl( \mathbb{R}^+_0 ~\big|~ \{1, 2, \div \} \Bigr)$
+
+\begin{itemize}
+	\item Define $2^2$
+	\item Define $2^n$ for all positive integers $n$
+	\item Define $2^{-n}$ for all positive integers $n$
+
+	\item What other numbers can we define in this structure? \par
+	\hint{There is at least one more \say{class} of numbers we can define.}
+\end{itemize}
+
 
 \begin{solution}
-	We can define powers of two, positive and negative.
+	As far as I've seen, we can define any $2^{\nicefrac{a}{b}}$ for $a, b \in \mathbb{Z}$. \par
-
+	For example, $\phi(x) \coloneqq [2 = x \div (1 \div x)]$ defines $\sqrt{2}$.
-	If you're clever, you can define many more: $\sqrt{2}, \sqrt[3]{2}, ...$.
 \end{solution}
 
 \vfill

Advanced/Definable Sets/parts/2 quantifiers.tex

@@ -5,7 +5,7 @@ We will now add two more: $\forall$ (for all) and $\exists$ (exists).
 
 \definition{}
 $\forall$ and $\exists$ are \textit{quantifiers}. They allow us to make statements about arbitrary symbols. \par
-\hint{Quantifiers are aptly named: they tell us \textit{how many} symbols satisfy a certain sentence.}
+\note{Quantifiers are aptly named: they tell us \textit{how many} symbols satisfy a certain sentence.}
 
 
 \vspace{2mm}
@@ -26,22 +26,28 @@ For example, $\exists ~ (0 < x)$ means \say{there is a positive number in our se
 \vspace{4mm}
 
 \problem{}
-Which of the following are true in $\mathbb{Z}$? \par
+Which of the following are true in $\mathbb{Z}$? Which are true in $\mathbb{R}^+_0$? \par
-Which are true in $\mathbb{R}^+_0$? \par
+\note{$\mathbb{R}^+_0$ is the set of positive real numbers and zero.}
-\hint{$\mathbb{R}^+_0$ is the set of positive real numbers and zero.} \par
 
 \begin{itemize}[itemsep = 1mm]
-
 	\item $\forall x ~ (x \geq 0)$
 	\item $\lnot (\exists x ~ (x = 0))$
-
 	\item $\forall x ~ [\exists y ~ (y \times y = x)]$
-
+	\item $\forall xy ~ \exists z ~ (x < z < y)$ \tab
-	\item $\forall xy ~ \exists z ~ (x < z < y)$ \tab \note{This is a compact way to write $\forall x ~ (\forall y ~ (\exists z ~ (x < z < y)))$}
+	\note{This is a compact way to write $\forall x ~ (\forall y ~ (\exists z ~ (x < z < y)))$}
-
+	\item $\lnot \exists x ~ ( \forall y ~ (x < y) )$
-	\item $\lnot \exists x ~ ( \forall y ~ (x < y) )$ %\tab~\tab \note{Solution is below.}
 \end{itemize}
 
+\begin{solution}
+	\begin{itemize}
+		\item \say{all $x$ are positive} \tab $\mathbb{R}^+_0$
+		\item \say{zero doesn't exist} \tab neither
+		\item \say{square roots exist} \tab $\mathbb{R}^+_0$
+		\item \say{this set is dense} \tab\null\tab $\mathbb{R}^+_0$
+		\item \say{there is no minimum} \tab $\mathbb{Z}$
+	\end{itemize}
+\end{solution}
+
 %\begin{examplesolution}
 %	Here is a solution to the last part: $\lnot \exists x ~ ( \forall y ~ (x < y) )$ \par
 %
@@ -60,10 +66,22 @@ Which are true in $\mathbb{R}^+_0$? \par
 \vfill
 \pagebreak
 
+
+
+
+
+
+
+
+
+
 \problem{}
 Does the order of $\forall$ and $\exists$ in a formula matter? \par
 What's the difference between $\exists x ~ \forall y ~ (x \leq y)$ and $\forall y ~ \exists x ~ (x \leq y)$? \par
-\hint{In $\mathbb{R}^+$, the first is false and the second is true. $\mathbb{R}^+$ does not contain zero.}
+\hint{
+	Consider $\mathbb{R}^+$\hspace{-1.3ex},\hspace{0.8ex} the set of positive reals. Zero is not positive. \par
+	Which of the above formulas is true in $\mathbb{R}^+$\hspace{-1.3ex},\hspace{0.8ex} and which is false?
+}
 
 \begin{solution}
 	If $\exists x$ is inside $\forall y$, $x$ depends on $y$. We may pick a different value of $x$ for every $y$. \par
@@ -76,6 +94,9 @@ What's the difference between $\exists x ~ \forall y ~ (x \leq y)$ and $\forall
 \problem{}
 Define 0 in $\Bigl( \mathbb{Z} ~\big|~ \{\times\} \Bigr)$
 
+\begin{solution}
+	$\varphi(x) \coloneqq \bigl[~ \forall y ~ x \times y = x ~\bigr]$
+\end{solution}
 
 \vfill
 
@@ -83,6 +104,10 @@ Define 0 in $\Bigl( \mathbb{Z} ~\big|~ \{\times\} \Bigr)$
 \problem{}
 Define 1 in $\Bigl( \mathbb{Z} ~\big|~ \{\times\} \Bigr)$
 
+\begin{solution}
+	$\varphi(x) \coloneqq \bigl[~ \forall y ~ x \times y = y ~\bigr]$
+\end{solution}
+
 
 \vfill
 \pagebreak
@@ -91,6 +116,10 @@ Define 1 in $\Bigl( \mathbb{Z} ~\big|~ \{\times\} \Bigr)$
 \problem{}
 Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, <\} \Bigr)$
 
+\begin{solution}
+	$\varphi(x) \coloneqq \bigl[~ (x<0) \land \lnot \exists y ~ (x < y < 0) ~\bigr]$
+\end{solution}
+
 \vfill
 
 %\problem{}
@@ -100,10 +129,10 @@ Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, <\} \Bigr)$
 
 \problem{}
 Let $\varphi(x)$ be a formula. \par
-Write a formula equivalent to $[~ \forall x ~ \varphi(x) ~]$ using only logical symbols and $\exists$.
+Write a formula equivalent to $\forall x ~ \varphi(x)$ using only logical symbols and $\exists$.
 
 \begin{solution}
-	$[~ \forall x ~ \varphi(x) ~]$ is true if and only if $[~ \lnot \exists x ~ \lnot \varphi(x) ~]$ is true.
+	$\forall x ~ \varphi(x)$ is true if and only if $\lnot \exists x ~ \lnot \varphi(x)$ is true.
 \end{solution}
 
 \vfill

Advanced/Definable Sets/parts/3 sets.tex

@@ -1,10 +1,10 @@
 \section{Definable Sets}
 
-Armed with $(), \land, \lor, \lnot, \rightarrow, \forall,$ and $\exists$, we have enough tools to define sets.
+Armed with $(), \land, \lor, \lnot, \rightarrow, \forall,$ and $\exists$, we have the tools to define sets.
 
 \definition{Set-Builder Notation}
 Say we have a sentence $\varphi(x)$. \par
-The set of all elements that satisfy that sentence can be written as follows:
+The set of all elements that satisfy that sentence may be written as follows:
 \begin{equation*}
 	\{ x ~|~ \varphi(x) \}
 \end{equation*}
@@ -20,20 +20,18 @@ $$
 
 \definition{Definable Sets}
 Let $S$ be a structure with a universe $U$. \par
-We say a subset $M$ of $U$ is \textit{definable} if we
+We say a subset $M$ of $U$ is \textit{definable} if we can write a formula \par
-can write a formula that is true for some $x$ iff $x \in M$.
+that is true for some $x$ if and only if $M$ contains $x$.
 
 \vspace{4mm}
 
-For example, consider the structure $\big\langle~ \mathbb{Z} ~\big|~ \{+\} ~\big\rangle$ \par
+For example, consider the structure $\bigl( \mathbb{Z} ~\big|~ \{+\} \bigr)$. \par
-
+Only even numbers satisfy the formula $\varphi(x) \coloneqq \bigl[\exists y ~ (y + y = x)\bigr]$, \par
-Only even numbers satisfy the formula $\varphi(x) = \exists y ~ (y + y = x)$, \par
 so we can define \say{the set of even numbers} as $\{ x ~|~ \exists y ~ (y + y = x) \}$. \par
 Remember---we can only use symbols that are available in our structure!
 
 \problem{}
-When is the empty set definable?
+The empty set is definable in any structure. How?
-
 \begin{solution}
 	Always: $\{ x ~|~ \lnot (x = x) \}$
 \end{solution}
@@ -43,18 +41,31 @@ When is the empty set definable?
 
 \problem{}
 Define $\{0, 1\}$ in $\Bigl( \mathbb{Z}^+_0 ~\big|~ \{<\} \Bigr)$
+\hint{Define 0 and 1 as elements first, and remember that we can use logical symbols.}
 
-\begin{instructornote}
+\begin{solution}
-	Here's an interesting fact:
+	$\varphi_0(x) \coloneqq \bigl[~ \lnot \exists y ~ y < x ~\bigr]$ \par
+	$\varphi_1(x) \coloneqq \bigl[~ (0 < x) ~\land~ \lnot \exists y ~ (x < y < 0) ~\bigr]$
+
+	\vspace{2mm}
+
+	Our final solution is $\{ x ~|~ \varphi_0(x) \lor \varphi_1(x) \}$.
+
+	\note{A finite set of definable elements is always definable. \par
+	An infinite set of definable elements might not be definable.}
+\end{solution}
 
-	A finite set of definable elements is always definable. \note{(Why?)} \par
-	An infinite set of definable elements might not be definable.
-\end{instructornote}
 
 \vfill
 
 \problem{}
-Define the set of prime numbers in $\Bigl( \mathbb{Z} ~\big|~ \{\times, \div, <\} \Bigr)$
+Define the set of prime numbers in $\Bigl( \mathbb{Z} ~\big|~ \{\times, \div, <\} \Bigr)$. \par
+\hint{A prime number is an integer that is positive and is only divisible by 1 and itself.}
+
+\begin{solution}
+	$\psi(x) \coloneqq \bigl[~ \exists y ~ (0<y<x) ~\bigr]$ \tab \note{\say{$x$ is positive and isn't 0 or 1}} \par
+	$\varphi(x) \coloneqq \bigl[~ (x<0) \land \lnot \exists ab ~ (\psi(a) \land \psi(b) \land a \times b = x) \bigr]$
+\end{solution}
 
 
 \vfill
@@ -65,33 +76,83 @@ Define the set of prime numbers in $\Bigl( \mathbb{Z} ~\big|~ \{\times, \div, <\
 
 
 
+
+
+
+
 \problem{}
 Define $\mathbb{R}^+_0$ in $\Bigl( \mathbb{R} ~\big|~ \{\times\} \Bigr)$ \par
 
+\begin{solution}
+	$\varphi(x) \coloneqq \bigl[ \exists y ~ y \times y = x \bigr]$
+\end{solution}
+
 \vfill
 
 \problem{}
-Let $\bigtriangleup$ be a relational symbol. $a \bigtriangleup b$ is true if and only if $a$ divides $b$. \par
+Let $\bigtriangleup$ be a relational symbol. $a \bigtriangleup b$ is only true if $a$ divides $b$. \par
 Define the set of prime numbers in $\Bigl( \mathbb{Z}^+ ~\big|~ \{ \bigtriangleup \} \Bigr)$ \par
 
+\begin{solution}
+	$
+		\varphi(x) \coloneqq
+		\bigl[~ \lnot \exists abc ~ \bigl(
+			(a \bigtriangleup x) \land
+			(b \bigtriangleup x) \land
+			(c \bigtriangleup x) \land
+			\lnot (a = b) \land
+			\lnot (a = c) \land
+			\lnot (b = c)
+		\bigr) ~\bigr]
+	$
+\end{solution}
+
 \vfill
 
+
+
 \theorem{Lagrange's Four Square Theorem}
 Every natural number may be written as a sum of four integer squares.
 
 \problem{}
 Define $\mathbb{Z}^+_0$ in $\Bigl( \mathbb{Z} ~\big|~ \{\times, +\} \Bigr)$
 
+\begin{solution}
+	$\varphi(x) \coloneqq \bigl[~ \exists abcd ~ (a^2 + b^2 + c^2 + d^2 = x) ~\bigr]$,
+	where $a^2 \coloneqq a \times a$.
+\end{solution}
+
 \vfill
 
+
+
+
 \problem{}
 Define $<$ in $\Bigl( \mathbb{Z} ~\big|~ \{\times, +\} \Bigr)$ \par
 \hint{We can't formally define a relation yet. Don't worry about that for now. \\
-You can repharase this question as \say{given $a,b \in \mathbb{Z}$,\\*/ write a sentence that is only true if $a < b$?}}
+You can repharase this question as \say{given $x,y \in \mathbb{Z}$, write a formula $\varphi(x, y)$ that is only true if $x < y$}}
+
+\begin{solution}
+	Let $\psi(x)$ be the formula from the previous problem.
+
+	\vspace{2mm}
+
+	$\varphi(x, y) \coloneqq \bigl[~ \lnot (x=y) \land \exists d ~ \bigl(\psi(d) \land (x + d = y)\bigr) ~\bigr]$
+\end{solution}
+
 
 \vfill
 \pagebreak
 
+
+
+
+
+
+
+
+
+
 \problem{}
 Consider the structure $S = ( \mathbb{R} ~|~ \{0, \diamond \} )$ \par
 The relation $a \diamond b$ holds if $| a - b | = 1$
@@ -99,9 +160,18 @@ The relation $a \diamond b$ holds if $| a - b | = 1$
 \problempart{}
 Define $\{-1, 1\}$ in $S$.
 
+\begin{solution}
+	$\varphi(x) \coloneqq \bigl[ 0 \diamond x \bigr]$
+\end{solution}
+
 \problempart{}
 Define $\{-2, 2\}$ in $S$.
 
+\begin{solution}
+	$\varphi(x) \coloneqq \bigl[~ \forall a ~ (0 \diamond x \rightarrow a \diamond x) \land \lnot (x = 0) ~\bigr]$
+\end{solution}
+
+
 \vfill
 
 \problem{}
@@ -110,20 +180,34 @@ Let $S$ be the structure $( \mathcal{P} ~|~ \{\subseteq\})$ \par
 
 \problempart{}
 Show that the empty set is definable in $S$. \par
-\hint{Defining $\{\}$ with $\{x ~|~ x \neq x\}$ is \textbf{not} what we need here. \\
+\hint{Defining $\{\}$ with $\{x ~|~ \lnot x = x\}$ is \textbf{not} what we need here. \\
 We need $\varnothing \in \mathcal{P}$, the \say{empty set} element in the power set of $\mathbb{Z}^+_0$.}
 
+\begin{solution}
+	$\varphi(x) \coloneqq \bigl[~ \forall y ~ x \subseteq y ~\bigr]$ \par
+	Note that we can use the same property to define 0 in $( \mathbb{Z} ~|~ \{\leq\})$
+\end{solution}
+
 \vfill
 
 \problempart{}
 Let $x \Bumpeq y$ be a relation on $\mathcal{P}$. $x \Bumpeq y$ holds if $x \cap y \neq \{\}$. \par
 Show that $\Bumpeq$ is definable in $S$.
 
+\begin{solution}
+	Let $\psi(x)$ be the formula from the previous problem.
+
+	\vspace{2mm}
+
+	$\varphi(x, y) \coloneqq \bigl[~ \exists x ~ (a \subseteq x) \land (a \subseteq y) \land \lnot \psi(a) ~\bigr]$
+\end{solution}
+
 \vfill
 
 \problempart{}
 Let $f$ be the function on $\mathcal{P}$ defined by $f(x) = \mathbb{Z}^+_0 - x$. This is called the \textit{complement} of $x$. \par
-Show that $f$ is definable in $S$.
+Show that $f$ is definable in $S$. \par
+\hint{You can define a function by writing a formula $\varphi(x, y)$ that is only true when $y = f(x)$.}
 
 \vfill
 \pagebreak