`restate` key with no `name` leads to `(,)` in restated title

[mbertucci47]

If restate is given to a theorem with no name key, then the restated theorem has (,) in the title. If a name is given, then there is no issue.

\documentclass{article}
\usepackage{amsthm,thmtools,kantlipsum}

\declaretheorem{theorem}

\begin{document}

\begin{theorem}[restate=foo]
\kant[1][2]
\end{theorem}

\foo*

\begin{theorem}[restate=foobar,name=Heading]
\kant[2][1]
\end{theorem}

\foobar*

\end{document}

[muzimuzhi]

Using name= as the default value of optional argument of \thmt@splitrestateargs seems to work. Need more tests.

\documentclass{article}
\usepackage{amsthm,thmtools,kantlipsum}

\declaretheorem{theorem}

\makeatletter
\renewcommand\thmt@splitrestateargs[1][name=]{%
  \g@addto@macro\thmt@storedoptargs{,#1}%
  \def\tmp@a##1\@{\def\thmt@storename{##1}}%
  \tmp@a
}
\makeatother

\begin{document}

% \begin{theorem}[restate={[name=]foo}]
\begin{theorem}[restate={foo}]
\kant[1][2]
\end{theorem}

\foo*

\begin{theorem}[restate=foobar,name=Heading]
\kant[2][1]
\end{theorem}

\foobar*

\end{document}

[mbertucci47]

I think the restate={[name=]foo} is a testing holdover, it should just say restate=foo, right?

[muzimuzhi]

Good catch, again! Example now corrected.

[mbertucci47]

Actually I think the comma in \thmt@storedoptargs{,#1} is unnecessary. Trying \ShowCommand{\thmt@storedoptargs} in a few places like

\documentclass{article}
\usepackage{amsthm,thmtools,kantlipsum}

\declaretheorem{theorem}
\makeatletter
\begin{document}

\begin{theorem}[restate=foo]
\kant[1][2]
\end{theorem}

\ShowCommand{\thmt@storedoptargs}

\foo*

\begin{theorem}[restate=foobar,name=Heading]
\kant[2][1]
\end{theorem}

\ShowCommand{\thmt@storedoptargs}

\foobar*

\begin{theorem}[restate=foobaraa,name=aaa,label=abc]
\ShowCommand{\thmt@storedoptargs}
\kant[2][1]
\end{theorem}

\end{document}

shows in the log things like

> \thmt@storedoptargs=macro:
->,.
<argument> \thmt@storedoptargs

and

> \thmt@storedoptargs=macro:
->,,name=Heading.
<argument> \thmt@storedoptargs 

and

> \thmt@storedoptargs=macro:
->,,name=aaa,label=abc.
<argument> \thmt@storedoptargs 

So it's putting an extra comma, which with no other keys is interpreted as a title. Just deleting the comma in the definition seems to fix it.

By the way, I noticed this syntax in only one other place, the definition of \thmt@define@thmuse@key in thm-kv.sty. Not sure if it matters there.

[muzimuzhi]

Partial findings: After

\begin{theorem}[restate={foo}]
\kant[1][2]
\end{theorem}

\foo* will finally call \begin{theorem}[{,}] ... \end{theorem}. And it's this intermediate form that caused the problem, see simplified example below.

To support old usage \begin{theorem}[<name>] as well as the extended \begin{theorem}[name=<name>], thmtools tries to distinguish valid key-value (like name=<name>) list between invalid (like New name). The latter is then treated as <name> as a whole. See \thmt@garbleoptarg. The problem here is that valid but meaningless key-value lists like , and , (or any combination of comma and space) are wrongly treated as invalid, hence used as theorem name.

\documentclass{article}
\usepackage{amsthm,thmtools,kantlipsum}

\declaretheorem{theorem}

\begin{document}
\begin{theorem}[{,}]
  content
\end{theorem}

\begin{theorem}[{, }]
  content
\end{theorem}

\begin{theorem}[{name=New name}]
  content
\end{theorem}

\begin{theorem}[{,name=New name,name=New name 2,,,,}]
  content
\end{theorem}
\end{document}

Not sure if kvsetkeys, the underneath key-value parser package has such a user interface. If not, then adding two guarding key-value pairs like \begin{theorem}[@kv=on,<normal kv-list>,@kv=off] can be a guaranteed, though slow solution.

[mbertucci47]

Yes I suppose this is a broader issue than the one I reported. However, in this specific case, doesn't removing the comma in \g@addto@macro\thmt@storedoptargs{,#1}% fix the issue without causing any harm?

[muzimuzhi]

Just not general enough. (We still have to add words to inform users why the innocent \begin{theorem}[,] will give unexpected output.) If possible I would avoid doing so. But as my previous comment started with "partial findings", right now I don't have enough info to answer the question "if it's possible to solve the general form, lightly?".