Sectioning paragraphs, such as for different cases inside a proof

Question

I'd like to have offset paragraphs inside a proof, say to delineate one case from another. I've been using

    \newtheoremstyle{component}{}{}{}{}{\itshape}{.}{.5em}{\thmnote{#3}#1}
    \theoremstyle{component}
    \newtheorem*{component}{}

and implementing it as

    \begin{component}[Case 1]
    ...
    \end{component}

which works nicely, and has the look I want.

Unfortunately, if this happens to fall towards the end of a proof, I need to manually add a \qedhere at the end, as otherwise the tombstone appears one line below.

Is there some simpler way to implement something similar, that would also solve the tombstone problem? I know I could just go with

    \medskip
    \noindent \emph{Case 1.}
    ...

    \medskip

but that's even more tedious, and is the whole reason I used the component theoremstyle in the first. Thanks for any suggestions you may have.

Answer 1

(Originally posted as a comment by Ulrich Schwarz)

IMO, just cope with the \qedhere. amsthm's behaviour is really quite simple once you get it: you need \qedhere if your environment ends with a list (in the TeXnical sense), which is, for common intents and purposes, displayed equations, subtheorems -- as you found out -- and itemize/enumerate. NTheorem's alternative approach is fragile and requires multiple passes, so yes, endmarks are tricky.

Answer 2

To avoid unneccesary code, it would probably be preferable to treat the cases like list items. By using the package enumitem, you can define a customized list environment with the appropriate behaviour

\newlist{pcases}{enumerate}{1}
\setlist[pcases]{
  label=\underline{Case~\arabic*:}\protect\thiscase.~,
  ref=\arabic*,
  align=left,
  labelsep=0pt,
  leftmargin=0pt,
  labelwidth=0pt,
  parsep=0pt
}
\newcommand{\case}[1][]{%
  \if\relax\detokenize{#1}\relax
    \def\thiscase{}%
  \else
    \def\thiscase{~#1}%
  \fi
  \item
}

You can use it like this:

\begin{proof}
some text
    \begin{pcases}
      \case[$k = 0$] This case is obvious.
      \case[$k > 0$] Then there exists a...
    \end{pcases}
\end{proof}

You would still have to write the \qedhere.