If you just want a linebreak, you can add that with \\ instead of \quad\allowbreak.
If you want to have two list items with the same number, you can set it using \item\setcounter{enumi}{1} where "1" should be a number one less than what you would like.
Here's an MWE with all both options:
\documentclass[]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\begin{enumerate}
\item $T(\phi_{1}(x)\barwedge\left(\phi_{2}(y)\veebar\phi_{3}(z)\right)),=,T(\left(\phi_{1}(x)\barwedge\phi_{2}(y) \right) \veebar\left(\phi_{1}(x)\barwedge\phi_{3}(z)\right))$\$T(\phi_{1}(x)\veebar \left(\phi_{2}(y)\barwedge\phi_{3}(z)\right)),=, T(\left(\phi_{1}(x)\veebar\phi_{2}(y)\right)\barwedge \left(\phi_{1}(x)\veebar\phi_{3}(z)\right))$.
\end{enumerate}
\hrule
\begin{enumerate}
\item\setcounter{enumi}{0} $T(\phi_{1}(x)\barwedge\left(\phi_{2}(y)\veebar\phi_{3}(z)\right)),=,T(\left(\phi_{1}(x)\barwedge\phi_{2}(y) \right) \veebar\left(\phi_{1}(x)\barwedge\phi_{3}(z)\right))$
\item $T(\phi_{1}(x)\veebar \left(\phi_{2}(y)\barwedge\phi_{3}(z)\right)),=, T(\left(\phi_{1}(x)\veebar\phi_{2}(y)\right)\barwedge \left(\phi_{1}(x)\veebar\phi_{3}(z)\right))$.
\item This is the third item
\item This is the fourth item
\end{enumerate}
\hrule
\begin{enumerate}
\item This is the first item
\item This is the second item
\item\setcounter{enumi}{2} $T(\phi_{1}(x)\barwedge\left(\phi_{2}(y)\veebar\phi_{3}(z)\right)),=,T(\left(\phi_{1}(x)\barwedge\phi_{2}(y) \right) \veebar\left(\phi_{1}(x)\barwedge\phi_{3}(z)\right))$
\item $T(\phi_{1}(x)\veebar \left(\phi_{2}(y)\barwedge\phi_{3}(z)\right)),=, T(\left(\phi_{1}(x)\veebar\phi_{2}(y)\right)\barwedge \left(\phi_{1}(x)\veebar\phi_{3}(z)\right))$.
\item This is the fourth item
\item This is the fifth item
\end{enumerate}
\end{document}
I've included two version of the \setcounter\enumi{} option to give you a better idea of how it works.

$ $and manually break the lines, or use display mode. – Peter Grill Jul 30 '23 at 21:05