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I want to have a theorem-like environment, without the usual "Theorem x." labeling.

I tried to define an empty label text, which results in a dot and space being used as the label text - there has to be an easy way to do this?

Why do I want to do this? I am typesetting a proof by induction; for the induction base, I want to restate the theorem that is proved, restated for the base case. (i.e., replace i with 1). The theorem environment WITHOUT the label text is exactly what I want.

  • To others who know more than me: Is there a way to use \@gobble to remove the dot and the space? – Sergio Parreiras May 14 '14 at 12:51
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    If you use the ntheorem package to format your theorem-like structures, there is a \theoremstyle{empty} command. – Bernard May 14 '14 at 12:58
  • @Bernard: would you consider writing your comment as an answer? – Sergio Parreiras May 14 '14 at 13:00
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If you use the ntheorem package to format your theorem-like structures, there is a \theoremstyle{empty} command. Here is an example: I define a quotedthm with an empty style. It inherits all the design characteristics that are declared at the time it is defined (except the style of course). In particular, you can give it an optional argument such as the name of the quoted theorem, and it will be typed in smallcaps in my example. I add a theorem indent with respect to the surrounding environment (proof) — which is a specific particularity of ntheorem as is a correct automatic placement of the end-of-proof logo, even when the proof ends in a equation.

\documentclass[12pt,a4paper]{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fourier}
\usepackage{heuristica}
\usepackage{latexsym}
\usepackage{amsmath}
\usepackage[thmmarks, amsmath]{ntheorem}
\usepackage{latexsym}

\theoremstyle{plain}
\theoremheaderfont{\scshape\upshape}
\theorembodyfont{\itshape}
\theoremseparator{.}
\newtheorem{thm}{Theorem}
{
\theoremstyle{empty}
\theoremindent 1cm
\newtheorem{quotedthm}{}}

\theoremstyle{nonumberplain}
\theorembodyfont{\normalfont}
\theoremsymbol{\ensuremath{\Box}}
\newtheorem{proof}{Proof}
\begin{document}


\begin{thm}[Euclid’s lemma]
If a prime number $ p $ divides a product of integers, it divides at least one of them.
\end{thm}

\begin{thm}[Gauss’s lemma]
If an integer $ n $ divides the product of two integers, and it is coprime with one of them, it divides the other.
\end{thm}

\begin{proof}
Remember Euclid’s lemma :
\begin{quotedthm}
If a prime number $ p $ divides a product of integers, it divides at least one of them.
\end{quotedthm}
Or remember
\begin{quotedthm}[Euclid : ]
If a prime number $ p $ divides a product of integers, it divides at least one of them.
\end{quotedthm}
In other words :
 \[ p \in \mathbf{P} \wedge (p \mid ab) \Rightarrow (p\mid a) \vee (p\mid b). \]%
\end{proof}

\end{document}

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