[Tex/LaTex] Why won’t the boolean work correctly

Question

Sorry for the very open question title, but I can't really figure out what is wrong exactly. I wanted to save a little time when writing semantic types, which are usually subscript but are often recursive, leading to types within types, like this:

D_<e,<s,t>>

The easiest solution seemed to me to use a boolean which checks whether I am within Type Brackets (<>) already; if so, \type would not use \textsubscript, if not, \type would set the boolean to true and use \textsubscript; the last command in \type would set the boolean to false if it had been false when the command was called. This is a minimal example:

\documentclass{article}
\usepackage{etoolbox}

\newbool{intype}
\newcommand{\type}[1]{\ifbool{intype}{$\langle$#1$\rangle$}{\setbool{intype}{true}\textsubscript{$\langle$#1$\rangle$}}\setbool{intype}{false}}

\begin{document}
D\type{\type{$\tau$,\type{$s,t$}},\type{$\tau$,\type{$s$,$t$}}}
\end{document}

The expected output is D_{<<τ,<s,t>>,<τ,<s,t>>}.

The output I get is D_{<<τ,<s,t>>,<τ,<s,t>>.}

This is really confusing – If my solution doesn't work, why isn't the <s,t> bit in the first pair of brackets lower than the τ?

_{Sorry for that long line of code btw, as far as I can tell, adding spaces or line breaks produces unwanted spaces in the output.}

Answer 1

You are setting it false too soon, but you can just use the grouping to set it back automatically:

\documentclass{article}
\usepackage{etoolbox}

\newbool{intype}
\newcommand{\type}[1]{%
  \ifbool{intype}%
  {$\langle$#1$\rangle$}%
  {\textsubscript{\setbool{intype}{true}$\langle$#1$\rangle$}}%
  }

\begin{document}
D\type{\type{$\tau$,\type{$s,t$}},\type{$\tau$,\type{$s$,$t$}}}
\end{document}