Adding optional arguments to environment

Question

I'm trying to understand and modify this code:

\newtheorem*{nonamethm}{\nonamethmname}
\newcommand{\nonamethmname}{}

\newenvironment{genthm}[1]
 {\renewcommand{\nonamethmname}{#1}\nonamethm}
 {\endnonamethm}
\newenvironment{genthm*}[1]
 {\renewcommand{\nonamethmname}{#1}\nonamethmcheck}
 {\endnonamethm}

\newcommand\nonamethmcheck[1][]{%
  \if\relax\detokenize{#1}\relax
    \nonamethm\relax
  \else
    \nonamethm[#1]%
  \fi
  \mbox{}%
}

It was proposed by @egreg in one of my previous questions. It works perfectly, but I don't understand why and now I'd like to extend its functionality.

With this code I can just write

\begin{genthm}
Theorem statement
\end{genthm}

instead of

\newtheorem{arbitraryname}
\begin{arbitraryname}
Theorem statement
\end{arbitraryname}

I'd like to understand how this code works. As far as I know \relax effectively does nothing, but I'm having troubles understanding the other pieces. For example, does the % symbol do anything, or is it just an "empty" comment? What about the empty command \newcommand{\nonamethmname}{}?

Finally I'd like to know what's the best way to extend the above commands in such a way that I can optionally specify the theorem style. Ideally, I could do \begin{genthm}{definition} Definition statement. \end{genthm} to avoid having to write \theoremstyle{definition}. If the second argument is omitted, it should be defaulted to \theoremstyle{plain}, which is the default.

Is this possible?

Answer 1

The trick is quite simple: we define a generic unnumbered theorem, where the theorem label is not explicitly given as a word, but given as a macro that we can redefine at will.

The redefinition happens in the environment around this newly defined theorem environment: when you say

\begin{genthm}{Name}

the environment redefines \nonamethmname to the argument, in this case Name and starts the nonamethm environment that will typeset the statement. At the end, we close nonamethm. In genthm* we delay the opening of the nonamethm environment looking first for the “attribution” optional argument, because we want to issue \mbox{} so an enumerate will start after a line break. The test \if\relax\detokenize{#1}\relax returns true if the optional argument is missing. The \relax after \begin{nonamethm} in the definition of genthm* is just to stop its lookup for an optional argument (we already know it's not there), so gaining some nanosecond.

In order to do what you want, you need to define a generic unnumbered theorem for each style you plan to use, because the style is chosen at definition time.

It's easy to extend the definitions to allow for your request, but it's handier to use xparse.

\documentclass{article}
\usepackage{amsthm}
\usepackage{xparse}

\newtheorem*{nonamethmplain}{\nonamethmname}
\theoremstyle{definition}
\newtheorem*{nonamethmdefinition}{\nonamethmname}
\theoremstyle{remark}
\newtheorem*{nonamethmremark}{\nonamethmname}
\newcommand{\nonamethmname}{}

\NewDocumentEnvironment{genthm}{O{plain}m}
 {\renewcommand{\nonamethmname}{#2}\begin{nonamethm#1}}
 {\end{nonamethm#1}}
\NewDocumentEnvironment{genthm*}{O{plain}mo}
 {\renewcommand{\nonamethmname}{#2}%
  \IfNoValueTF{#3}
    {\begin{nonamethm#1}\relax}%
    {\begin{nonamethm#1}[#3]}%
  \mbox{}}
 {\end{nonamethm#1}}

\begin{document}

\begin{genthm}{Hairy ball theorem}
Whatever it says
\end{genthm}

\begin{genthm}[definition]{Transcendental number}
A \emph{transcendental} number is a number that is not algebraic.
Easy, no?
\end{genthm}

\begin{genthm}[remark]{Easy remark}[J. de La Palice]
Everything is easy when it is.
\end{genthm}

\begin{genthm*}[definition]{Bunch of definitions}
\begin{enumerate}
\item A
\item B
\item C
\end{enumerate}
\end{genthm*}

\end{document}

The genthm environment now looks for an optional argument (default plain) and appends it to nonamethm in order to choose the correct style.

Note that the final attribution optional argument is still honored, thanks to the features in xparse.

Please, be friendly to your readers and provide them theorem numbers.

Answer 2

I just stole the code from egreg's answer to the linked question Make `enumerate` start from a new line

and redefined the genthm environment, wrapping * and unstarred version together (with some minor caveat, the position of the star) and added the optional argument.

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[italian]{babel}

\usepackage{xparse}
\usepackage{amsthm}

\newtheorem*{nonamethm}{\nonamethmname}
\newcommand{\nonamethmname}{}


\newcommand\nonamethmcheck[1][]{%
  \if\relax\detokenize{#1}\relax
    \nonamethm\relax
  \else
    \nonamethm[#1]%
  \fi
  \mbox{}%
}


\NewDocumentEnvironment{genthm}{s+om}{%
  \IfBooleanTF{#1}{% Check for starred version 
    \IfValueTF{#2}{%  Check for optional 2nd argument
      \theoremstyle{#2}%
    }{}%
    \renewcommand{\nonamethmname}{#3}\nonamethmcheck%
  }{% No, it's the unstarred versions
    \IfValueTF{#2}{%
      \theoremstyle{#2}%
    }{}
    \renewcommand{\nonamethmname}{#3}\nonamethm%
  }%
}{%
  \endnonamethm%
}



\begin{document}

%\begin{aenv}*
%Something
%\end{aenv}

\begin{genthm}*{Teoremi del quoziente}
\begin{enumerate}
\item Se $\lim\limits_{x \to \alpha} |f(x)| = +\infty$,
  allora $\lim\limits_{x \to \alpha} \frac{1}{f(x)} = 0$.

\item Se $\lim\limits_{x \to \alpha} f(x) = 0$,
  allora $\lim\limits_{x \to \alpha} \frac1{|f(x)|} = +\infty$.

\item Se $\lim\limits_{x \to \alpha} f(x) = l \neq 0$,
  allora $\lim\limits_{x \to \alpha} \frac1{f(x)} = \frac{1}{l}$.

\item Se $\lim\limits_{x \to \alpha} f(x) = l$ e $\lim\limits_{x \to \alpha} g(x) = m \neq 0$,
  allora $\lim\limits_{x \to \alpha} \frac{f(x)}{g(x)} = \frac{l}{m}$.
\end{enumerate}
\end{genthm}

\begin{genthm}*{Teoremi del quoziente}
Siano $f$ e $g$ funzioni definite in un intorno bucato di~$\alpha$.
\begin{enumerate}
\item Se $\lim\limits_{x \to \alpha} |f(x)| = +\infty$,
  allora $\lim\limits_{x \to \alpha} \frac{1}{f(x)} = 0$.

\item Se $\lim\limits_{x \to \alpha} f(x) = 0$,
  allora $\lim\limits_{x \to \alpha} \frac1{|f(x)|} = +\infty$.

\item Se $\lim\limits_{x \to \alpha} f(x) = l \neq 0$,
  allora $\lim\limits_{x \to \alpha} \frac1{f(x)} = \frac{1}{l}$.

\item Se $\lim\limits_{x \to \alpha} f(x) = l$ e $\lim\limits_{x \to \alpha} g(x) = m \neq 0$,
  allora $\lim\limits_{x \to \alpha} \frac{f(x)}{g(x)} = \frac{l}{m}$.
\end{enumerate}
\end{genthm}

\end{document}