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authorJacob Walchuk <jpw24@st-andrews.ac.uk>2025-07-07 13:27:49 +0100
committerJacob Walchuk <jpw24@st-andrews.ac.uk>2025-07-07 13:27:49 +0100
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@@ -205,7 +205,7 @@ $\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}{LR}
$\vdash \Box Haec(X)(y) \ra \Box Tra(X)(y)$ & (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) \\
@@ -214,7 +214,7 @@ $\vdash Haec(X)(y) \ra \Box Tra(X)(y)$ &
\end{quote}
\subsection{The proof for (o-Tracking)}
\begin{quote}
-\begin{tabulary}{\textwidth}{Lr}
+\begin{tabulary}{\textwidth}{LR}
$\vdash Haec(X)(o) \ra Tra(X)(o)$ & (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)\\