Cleanup

Advanced/Definable Sets/parts/0 logic.tex

@@ -1,8 +1,9 @@
 \section{Logical Algebra}
 
 \definition{}
-Odds are, you are familiar with \textit{logical symbols}. \par
-In this handout, we'll use the following:
+\textit{Logical operators} operate on the values $\{\text{True}, \text{False}\}$, \par
+just like algebraic operators operate on numbers. \par
+In this handout, we'll use the following operators:
 \begin{itemize}
 	\item $\lnot$: not
 	\item $\land$: and
@@ -66,11 +67,15 @@ $\lnot A$ is the opposite of $A$, which is why it looks like a \say{negative} si
 \vspace{2mm}
 
 $A \rightarrow B$ is a bit harder to understand. Read aloud, this is \say{$A$ implies $B$.} \par
-The only time $\rightarrow$ is false is when $T \rightarrow F$. This may seem counterintuitive, but it makes sense. Think about it. \par
+The only time $\rightarrow$ is false is when $T \rightarrow F$. This may seem counterintuitive, but it will make more sense as we progress through this handout.
 
 \problem{}
 Evaluate the following.
 \begin{itemize}
+	\item $\lnot T$
+	\item $F \lor T$
+	\item $T \land T$
+	\item $(T \land F) \lor T$
 	\item $(T \land F) \lor T$
 	\item $(\lnot (F \lor \lnot T) ) \rightarrow T$
 	\item $(F \rightarrow T) \rightarrow (\lnot F \lor \lnot T)$
@@ -110,7 +115,7 @@ Evaluate the following.
 
 \problem{}
 Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par
-That is, show that these give the same result for the same $A$ and $B$.
+That is, show that these give the same result for the same $A$ and $B$. \par
 \hint{Use a truth table}
 
 \vfill

Advanced/Definable Sets/parts/1 structures.tex

@@ -86,7 +86,7 @@ For the sake of time, I will not provide a formal definition. It isn't particula
 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
 
-For example, $x$ is a free variable in the formula $\varphi(x) = x > 0$. \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.
 
 \definition{Definable Elements}
@@ -99,7 +99,7 @@ Can we define 2 in the structure $\Bigl( \mathbb{Z^+} ~\big|~ \{4, \times \} \Bi
 \hint{$\mathbb{Z}^+ = \{1, 2, 3, ...\}$. Also, $2 \times 2 = 4$.}
 
 \begin{solution}
-	$2$ is the only element in $\mathbb{Z}^+$ that satisfies $[x \text{ where } x \times x = 4]$.
+	$2$ is the only element in $\mathbb{Z}^+$ that satisfies $\varphi(x) = [x \times x = 4]$.
 \end{solution}
 
 \vfill
@@ -109,7 +109,7 @@ Can we define 2 in the structure $\Bigl( \mathbb{Z^+} ~\big|~ \{4, \times \} \Bi
 Try to define 2 in the structure $\Bigl( \mathbb{Z} ~\big|~ \{4, \times \} \Bigr)$.
 
 \begin{solution}
-	This isn't possible. We could try $[x \text{ where } x \times x = 4]$, but this is satisfied by both $2$ and $-2$. \\
+	This isn't possible. We could try $\varphi(x) = [x \times x = 4]$, but this is satisfied by both $2$ and $-2$. \par
 	We have no way to distinguish between negative and positive numbers.
 
 	\begin{instructornote}

Advanced/Definable Sets/parts/4 equivalence.tex

@@ -1,8 +1,9 @@
-\section{Equivalence (Bonus)}
+\section{Equivalence}
 
 \generic{Notation:}
 Let $S$ be a structure and $\varphi$ a formula. \par
-If $\varphi$ is true in $S$, we write $S \models \varphi$.
+If $\varphi$ is true in $S$, we write $S \models \varphi$. \par
+This is read \say{$S$ satisfies $\varphi$}
 
 \definition{}
 Let $S$ and $T$ be structures. \par