Updates from Overleaf

1 files changed, 35 insertions, 29 deletions

diff --git a/papers/2.tex b/papers/2.tex
index d480e1c..146fa6a 100644
--- a/papers/2.tex
+++ b/papers/2.tex
@@ -571,35 +571,39 @@ could achieve this through an example. Consider a 5-element Sorites with
 objects \textbf{a}, \textbf{b}, \textbf{c}, \textbf{d}, and \textbf{e}.
 Suppose that Alfred's initial categorizations are:
 
-\[D_{1}F = \{ a,b\}\]
-
-\[{\sim}D_{1}F\; \& \; {\sim}D_{1}{\sim}F = \{ c\}\]
-
-\[D_{1}{\sim}F = \{ d,e\}\]
-
+\begin{center}
+\begin{tabulary}{\textwidth}{RCL}
+$D_1 F$ & = & $\{ \mathbf{a}, \mathbf{b} \}$ \\
+${\sim}D_1 F \ \& \ {\sim}D_1{\sim}F$ & = & $\{ \mathbf{c} \}$ \\
+$D_{1}{\sim}F$ & = & $\{ \mathbf{d}, \mathbf{e}\}$ \\
+\end{tabulary}
+\end{center}
 Alfred considers the pair \textbf{b} and \textbf{c} again. He realizes
 that he cannot tell the difference, concluding that \textbf{b} is also
 borderline. He adjusts his p-set accordingly, forming a new \(t_{2}\)
 p-set.
 
-\[D_{2}F = \{ a\}\]
-
-\[{\sim}D_{2}F\ \&\ {\sim}D_{2}{\sim}F = \{ b,c\}\]
-
-\[D_{2}{\sim}F = \{ d,e\}\]
+\begin{center}
+\begin{tabulary}{\textwidth}{RCL}
+$D_2 F$ & = & $\{ \mathbf{a} \}$ \\
+${\sim} D_2 F \ \& \ {\sim}D_2 {\sim}F$ & = & $\{ \mathbf{b}, \mathbf{c} \}$ \\
+$D_2 {\sim}F$ & = & $\{ \mathbf{d}, \mathbf{e} \}$
+\end{tabulary}
+\end{center}
 
 \begin{center}
 \includegraphics[width=4.50937in,height=2.12793in]{papers/figures/2-3.pdf}
  \end{center}
 The \(t_{1}\) division, from the perspective of \(t_{2}\) becomes:
 
-\emph{\hfill\break
-}\[{D_{2}D}_{1}F = \{ a\}\]
-
-\[{\sim}D_{2}D_{1}F\ \; \& \; {\sim}D_{2}{{\sim}D}_{1}F = \{ b\}\]
+\begin{center}
+\begin{tabulary}{\textwidth}{RCL}
+$D_2 D_1 F$ & = & $\{ \mathbf{a} \}$ \\
+${\sim} D_2 D_1 F \ \& \ {\sim}D_{2}{{\sim}D}_{1}F$ & = & $\{ \mathbf{b} \}$ \\
+$D_{2}{\sim}D_{1}F\ \&\ D_{2}{\sim}D_{1}{\sim}F $ & = & $\{ \mathbf{c} \}$
+\end{tabulary}
+\end{center}
 
-\[D_{2}{\sim}D_{1}F\ \&\ D_{2}{\sim}D_{1}{\sim}F = \{ c\}\]
-\\
 \begin{center}
 \includegraphics[width=4.79722in,height=2.19101in]{papers/figures/2-4.pdf}
 \end{center}
@@ -610,11 +614,13 @@ Now suppose that at time \(t_{3}\), he looks at the pair $\mathbf{a}$ and
 $\mathbf{b}$. Since he cannot tell the difference, he decides that b is
 also a definite case, adjusting the p-set again.
 
-\[D_{3}F = \{ a,b\}\]
-
-\[{\sim}D_{3}F \; \& \; {\sim}D_{3}{\sim}F = \{ c\}\]
-
-\[D_{3}{\sim} F = \{ d,e\}\]
+\begin{center}
+\begin{tabulary}{\textwidth}{RCL}
+$D_3 F$ & = & $\{ \mathbf{a}, \mathbf{b} \}$ \\
+${\sim} D_3F \ \& \ {\sim}D_{3}{\sim}F$ & = & $\{ \mathbf{c} \}$ \\
+$D_{3}{\sim} F $ & = & $\{ \mathbf{d}, \mathbf{e} \}$
+\end{tabulary}
+\end{center}
 
 \begin{center}
     \includegraphics[width=4.46286in,height=2.2071in]{papers/figures/2-5.pdf}
@@ -624,13 +630,13 @@ of \(t_{3}\), $\mathbf{b}$ was not a definite borderline case at
 \(t_{2}\). Thus, the \(t_{2}\) division, from the \(t_{3}\) perspective,
 is:
 
-\[{D_{3}D}_{2}F = \{ a\}\]
-
-\[\sim D_{3}D_{2}F\ \&\ \sim D_{3}{\sim D}_{2}F = \{ b\}\]
-
-\[D_{3}\sim D_{2}F\ \&\ D_{3}\sim D_{2}\sim F = \{ c\}\]
-
-\\
+\begin{center}
+\begin{tabulary}{\textwidth}{RCL}
+$D_3 D_2 F$ & = & $\{ \mathbf{a} \}$ \\
+${\sim}D_3 D_3 F \ \& \ {\sim}D_2 {\sim}D_2 F$ & = & $\{ \mathbf{b} \}$ \\
+$D_3 {\sim}D_2 F \ \& \ D_3 {\sim}D_2 {\sim}F$ & = & $\{ \mathbf{c} \}$
+\end{tabulary}
+\end{center}
 
 \begin{center}
     \includegraphics[width=4.875in,height=2.1236in]{papers/figures/2-6.pdf}