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选择命题、假设、推论等的编号

tag-icon • tag-icon numbering theorems
选择命题、假设、推论等的编号

我有以下输出

在此处输入图片描述

但我希望他们能说出来假设 1和命题 1相反。我在序言中使用了以下内容：

\documentclass[12pt,letterpaper]{article}
\usepackage[margin=1in]{geometry}
\usepackage[english]{babel}
\usepackage{amsmath}
\setcounter{secnumdepth}{3} %the depth of section numbering 0 means no numbering of any section, sub-section and on.
%used for begining theorems and propostions numbered according to the section they are in
\newtheorem{thm}{Theorem}[subsection]% theorem counter resets every \subsection
\renewcommand{\thethm}{\arabic{thm}}% Remove subsection from theorem counter representation
\newtheorem{proposition}[thm]{Proposition}
\newtheorem{assumption}[thm]{Assumption}}

这基本上是我的测试文档的一部分：

\begin{assumption}
    \label{Assumption 1}
    \begin{minipage}[t]{.7\linewidth}
        \begin{enumerate}
            \item[i)] 
            \item[ii)] 
        \end{enumerate}
    \end{minipage}
\end{assumption}

\begin{proposition}
    \label{P1}
    Proposition...
\end{proposition}
\noindent {\it Proof}. See Appendix \ref{proofs}. $\blacksquare$

我希望的是，假设按假设编号。例如假设 2紧随其后假设 1. 以及按命题编号的命题：命题 1，命题 2等等。推论和引理等也有类似的想法。我该如何实现这一点？

答案1

从上面的评论来看，序言中的以下内容解决了该问题：

\documentclass[12pt,letterpaper]{article}
\usepackage[margin=1in]{geometry}
\usepackage[english]{babel}
\usepackage{amsmath}
\setcounter{secnumdepth}{3} %the depth of section numbering 0 means no numbering of any section, sub-section and on.
%used for begining theorems and propostions numbered according to the section they are in
\newtheorem{thm}{Theorem}[subsection]% theorem counter resets every \subsection
\renewcommand{\thethm}{\arabic{thm}}% Remove subsection from theorem counter representation 
%Numbers the propositions, lemmas, corollaries, etc within themselves
\newtheorem{proposition}{Proposition}[subsection] \renewcommand{\theproposition}{\arabic{proposition}}
\newtheorem{assumption}{Assumption}[subsection]  \renewcommand{\theassumption}{\arabic{assumption}}
\newtheorem{lemma}{Lemma}[subsection] 
\renewcommand{\thelemma}{\arabic{lemma}}
\newtheorem{corollary}{Corollary}[subsection] \renewcommand{\thecorollary}{\arabic{corollary}}

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