# Animate subitemize in beamer

I know there is a way to 'animate' the enumerate in beamer using the following:

\begin{enumerate}
\item<1-> Item 1;
\item<2-> Item 2;
\item<3-> Item 3;
\item<4-> Item 4.
\end{enumerate}


However I wanted to 'animate' a subitem and display it's number, e.g., 1.x - 1.2. Is this possible? I want something like this:

\begin{enumerate}
\item<1-> Item 1;
\begin{enumerate}
\item<1.1-> Item 1.1;
\item<1.2-> Item 1.2.
\end{enumerate}
\item<2-> Item 2;
\item<3-> Item 3;
\item<4-> Item 4.
\end{enumerate}

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## 2 Answers

If you just want to sequentially list all items in the list (including subitems), you don't have to manually specify each item. You can use the default overlay specification for the environment:

\begin{enumerate}[<default overlay specification>]
%...
\end{enumerate}


Here is an example:

\documentclass{beamer}% http://ctan.org/pkg/beamer
\begin{document}
\begin{frame}
\begin{enumerate}[<+->]
\item Item 1;
\begin{enumerate}
\item Item 1.1;
\item Item 1.2.
\end{enumerate}
\item Item 2;
\item Item 3;
\item Item 4.
\end{enumerate}
\end{frame}
\end{document}


The <default overlay specification> is inherited by subenvironments. See the beamer documentation (section 12.1 Itemizations, Enumerations, and Description, p 111).

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I think you just need to shift all your numbers after 1 so that things work properly. E.g. swap 1.1- to 2- and 1.2- to 3- if you wan them to stay on the screen thereafter. If you want them to go away when you are "done" with "Item 1", then you need to put an upper limit, or something like:

\begin{enumerate}
\item<1-> Item 1;
\begin{enumerate}
\item<2-3> Item 1.1;
\item<3> Item 1.2.
\end{enumerate}
\item<4-> Item 2;
\item<5-> Item 3;
\item<6-> Item 4.
\end{enumerate}

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No way to make 1.x, i.e., 1.1, 1.2? –  Marcos Roriz Junior Feb 15 '12 at 19:23
Not that I know of. Werner's answer shows how to sequentially list the items. From your question, I interpreted it as you wanted to have the subitems disappear when you were done with them. If that is not the case, then Werner's solution is optimal. –  cm2 Feb 16 '12 at 16:58
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