Fixed a typo (#44)

forallx-yyc-metatheory.tex

@@ -190,7 +190,7 @@ \chapter{Functional completeness}
 		$$(A \eand B \eand \enot C) \eor (A \eand \enot B \eand C) \eor (\enot A \eand B \eand \enot C)$$			
 It follows that the connectives of TFL are jointly functionally complete. We now prove each of the subsidiary results.
 
-\emph{Subsidiary Result \ref{expressive:eor}: functional completeness of `$\enot$' and `$\eor$'.} Observe that the scheme that we generate, using the truth table method of proving the DNF Theorem, will only contain the connectives `$\enot$', `$\eand$' and `$\eor$'. So it suffices to show that there is an equivalent scheme which contains only `$\enot$' and `$\eor$'. To show do this, we simply consider that
+\emph{Subsidiary Result \ref{expressive:eor}: functional completeness of `$\enot$' and `$\eor$'.} Observe that the scheme that we generate, using the truth table method of proving the DNF Theorem, will only contain the connectives `$\enot$', `$\eand$' and `$\eor$'. So it suffices to show that there is an equivalent scheme which contains only `$\enot$' and `$\eor$'. To demonstrate this, we simply consider that
 		\begin{align*}
 		(\metav{A} \eand \metav{B}) & \text{\quad and \quad} \enot(\enot \metav{A} \eor\enot \metav{B})
 		\end{align*}