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Diffstat (limited to 'papers/3.tex')
| -rw-r--r-- | papers/3.tex | 12 |
1 files changed, 6 insertions, 6 deletions
diff --git a/papers/3.tex b/papers/3.tex index a4ba95e..072a5d5 100644 --- a/papers/3.tex +++ b/papers/3.tex @@ -185,14 +185,14 @@ In this paper, I have examined three responses to Williamson's Explanation Chall \section{Appendix} \setlength{\tabcolsep}{0pt} \subsection{The proof for (Tracking)} -First, we can observe the following proof: \\ +First, we can observe the following proof: \begin{quote} -\begin{tabulary}{\textwidth}{Lr} +\begin{tabulary}{\textwidth}{p{0.457\textwidth}r} $\forall x (Xx \ba x=y), Haec(X)(z) \wedge z \neq y \vdash \bot \ $ & (Reductio, Cond. proof, Universal generalization) \\ \end{tabulary} -\begin{tabulary}{\textwidth}{Lr} +\begin{tabulary}{\textwidth}{p{0.805\textwidth}r} $\vdash \Box \forall x (Xx \ba x=y) \ra \Box \forall z \neg (Haec(X)(z) \wedge z \neq y)$ & (K) \\ -$\vdash \Box \forall x (Xx \ba x=y) \ra \neg \Diamond \exists z (Haec(X)(z) \wedge z \neq y)$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ & (Equivalence) \\ +$\vdash \Box \forall x (Xx \ba x=y) \ra \neg \Diamond \exists z (Haec(X)(z) \wedge z \neq y)$\ & (Equivalence) \\ $\vdash Haec(X)(y) \ra Tra(X)(y)$& \\ \end{tabulary} % terrible hack but it fixed the rendering, I am very sorry Mohit \end{quote} @@ -207,7 +207,7 @@ $\forall x (Xx \ba x=y), Haec(X)(z) \wedge z \neq y \vdash Xy \ba y=y \\ \end{quote} Then, from $\vdash Haec(X)(y) \ra Tra(X)(y)$, we can observe that: \\ \begin{quote} -\begin{tabulary}{\textwidth}{LR} +\begin{tabulary}{\textwidth}{p{0.7\textwidth}R} $\vdash \Box Haec(X)(y) \ra \Box Tra(X)(y)$ & (K) \\ $\vdash \Box \forall x (Xx \ba x=y) \ra \Box \Box \forall x (Xx \ba x=y)$ & (\textbf{4})\\ $\vdash Haec(X)(y) \ra \Box Haec(X)(y)$ & (Chaining conditionals) \\ @@ -216,7 +216,7 @@ $\vdash Haec(X)(y) \ra \Box Tra(X)(y)$ & \end{quote} \subsection{The proof for (o-Tracking)} \begin{quote} -\begin{tabulary}{\textwidth}{LR} +\begin{tabulary}{\textwidth}{p{0.705\textwidth}R} $\vdash Haec(X)(o) \ra Tra(X)(o)$ & (Proved above) \\ $\vdash \exists X Haec(X)(o) \ra \exists X Tra(X)(o)$ & (Derivable from $\forall$ rule) \\ $\vdash \Box \exists X Haec(X)(o) \ra \Box \exists X Tra(X)(o)$ & (K)\\ |