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If an \item (in \begin{enumerate) contains a long line, I need to add a linebreak so the rest goes to another line (but with the same enumeration number). In the following code, I add \allowbreak in hoping to break the rest into a new line, but it does not work. How do I do it? Thanks.

\begin{enumerate}   
\item  $T(\phi_{1}(x)\barwedge\left(\phi_{2}(y)\veebar\phi_{3}(z)\right))\,=\,T(\left(\phi_{1}(x)\barwedge\phi_{2}(y) \right) \veebar\left(\phi_{1}(x)\barwedge\phi_{3}(z)\right))$, \quad\allowbreak $T(\phi_{1}(x)\veebar \left(\phi_{2}(y)\barwedge\phi_{3}(z)\right))\,=\, T(\left(\phi_{1}(x)\veebar\phi_{2}(y)\right)\barwedge \left(\phi_{1}(x)\veebar\phi_{3}(z)\right))$.
\end{enumerate}
CarLaTeX
  • 62,716
  • Adding a blank line will start a new paragraph with the same enumeration in text mode? In math mode, you need to start a new $ $ and manually break the lines, or use display mode. – Peter Grill Jul 30 '23 at 21:05
  • Note that the text that follow an \item in not an argument of \item. \item simply starts a new paragrpah and adds a label (inside the new left margin). You can have multiple paragraphs following one \item. – John Kormylo Jul 30 '23 at 22:58

1 Answers1

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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.

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