diff --git a/Advanced/Definable Sets/parts/1 structures.tex b/Advanced/Definable Sets/parts/1 structures.tex
index 87ec66d..e76492e 100644
--- a/Advanced/Definable Sets/parts/1 structures.tex	
+++ b/Advanced/Definable Sets/parts/1 structures.tex	
@@ -55,18 +55,12 @@ If you look at our set of constant symbols, you'll see that the only integers we
 
 \vspace{1mm}
 
-Say we want the number 2. We could use the function $+$ to define it: $2 \coloneqq [x \text{ where } 1 + 1 = x]$ \par
-We would write this as $2 \coloneqq [x \text{ where } +(1, 1) = x]$ in proper \say{functional} notation.
+To \say{define} an element of a set, we need to write a sentence that is only true for that element. For example, if we want to define 2 in the structure above, we could use the sentence $\varphi(x) = [1 + 1 = x]$. \par
+Clearly, this is only true when $x = 2$.
 
 
 \problem{}
-Can we define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, 1, +, -, <\} \Bigr)$? If so, how?
-
-\vfill
-
-\problem{}
-Can we define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, +, -, <\} \Bigr)$? \par
-\hint{In this problem, $1$ has been removed from the set of constant symbols.}
+Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, 1, +, -, <\} \Bigr)$.
 
 \vfill
 \pagebreak
@@ -87,7 +81,13 @@ A formula can contain one or more \textit{free variables.} These are denoted $\v
 Formulas with free variables let us define \say{properties} that certain objects have. \par
 
 For example, $x$ is a free variable in the formula $\varphi(x) = [x > 0]$. \par
-$\varphi(3)$ is true and $\varphi(-3)$ is false.
+$\varphi(3)$ is true and $\varphi(-3)$ is false. \par
+
+\vspace{2mm}
+
+This \say{free variable} notation is much like the function notation you are used to: \par
+$\varphi(x) = [x > 0]$ is similar to $f(x) = x + 1$, since the values of $\varphi(x)$ and $f(x)$ depend on $x$.
+
 
 \definition{Definable Elements}
 Say $S$ is a structure with a universe $U$. \par
@@ -95,7 +95,7 @@ We say an element $e \in U$ is \textit{definable in $S$} if we can write a formu
 
 
 \problem{}
-Can we define 2 in the structure $\Bigl( \mathbb{Z^+} ~\big|~ \{4, \times \} \Bigr)$? \par
+Define 2 in the structure $\Bigl( \mathbb{Z^+} ~\big|~ \{4, \times \} \Bigr)$. \par
 \hint{$\mathbb{Z}^+ = \{1, 2, 3, ...\}$. Also, $2 \times 2 = 4$.}
 
 \begin{solution}
@@ -106,7 +106,8 @@ Can we define 2 in the structure $\Bigl( \mathbb{Z^+} ~\big|~ \{4, \times \} \Bi
 
 
 \problem{}
-Try to define 2 in the structure $\Bigl( \mathbb{Z} ~\big|~ \{4, \times \} \Bigr)$.
+Try to define 2 in the structure $\Bigl( \mathbb{Z} ~\big|~ \{4, \times \} \Bigr)$. \par
+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
diff --git a/Advanced/Definable Sets/parts/2 quantifiers.tex b/Advanced/Definable Sets/parts/2 quantifiers.tex
index a717741..1006a94 100644
--- a/Advanced/Definable Sets/parts/2 quantifiers.tex	
+++ b/Advanced/Definable Sets/parts/2 quantifiers.tex	
@@ -60,9 +60,15 @@ Which are true in $\mathbb{R}^+_0$? \par
 
 \problem{}
 Does the order of $\forall$ and $\exists$ in a formula matter? \par
-What's the difference between $\exists x ~ \forall y ~ (x < y)$ and $\forall y ~ \exists x ~ (x < y)$? \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.}
 
+\begin{solution}
+	If $\exists x$ is inside $\forall y$, $x$ depends on $y$. We can have a different value of $x$ for every $y$. \par
+	If $\exists x$ is outside, $x$ is fixed \textit{before} we check all $y$.
+\end{solution}
+
+
 \vfill
 
 \problem{}
diff --git a/Advanced/Definable Sets/parts/3 sets.tex b/Advanced/Definable Sets/parts/3 sets.tex
index 7a5b354..bea0e09 100644
--- a/Advanced/Definable Sets/parts/3 sets.tex	
+++ b/Advanced/Definable Sets/parts/3 sets.tex	
@@ -102,9 +102,6 @@ You can repharase this question as \say{given $a,b \in \mathbb{Z}$, can you writ
 Consider the structure $S = ( \mathbb{R} ~|~ \{0, \diamond \} )$ \par
 The relation $a \diamond b$ holds if $| a - b | = 1$
 
-\problempart{}
-Define 0 in $S$.
-
 \problempart{}
 Define $\{-1, 1\}$ in $S$.