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Diffstat (limited to 'papers')
| -rw-r--r-- | papers/2.tex | 64 |
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} |