# Align does not increment the equation counter

The align environment is not advancing the equation counter. Here's the code that produced the output below:

\documentclass{report}

\usepackage{mathtools,amsthm}

\begin{document}
\numberwithin{section}{chapter}
\numberwithin{equation}{section}
\let\realequation\equation
\def\equation{\setcounter{equation}{\arabic{subsection}}%
\refstepcounter{subsection}%
\realequation}

\theoremstyle{definition}
\newtheorem{thm}[subsection]{Theorem}
\newtheorem{lemma}[subsection]{Lemma}
\newtheorem{corol}[subsection]{Corollary}
\newtheorem{defn}[subsection]{Definition} % definition numbers are dependent on theorem numbers
\newtheorem{defns}[subsection]{Definitions} % definition numbers are dependent on theorem numbers
\newtheorem{prop}[subsection]{Proposition} % proposition numbers are dependent on theorem numbers
\newtheorem{exmp}[subsection]{Example} % same for example numbers
\newtheorem*{remark}{Remark}
\newtheorem{enum_remark}[subsection]{Remark}
\newtheorem{exercises}[subsection]{Exercises}
\newtheorem*{opex}{Opening Exercises}

\begin{align} \label{nchoosed}
\begin{split}
\prescript{}{n}C_d = \binom{n}{d}
&=  \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
&= \frac{n!}{d! (n-d)!}
\end{split}
\end{align}

some text
$$a = b$$
\end{document}


The document is set up so that a single counter runs through each chapter and gets advanced by subsections, theorems, equations, etc.

• the numbering is determined by code you are not showing so it is impossible to guess what is wrong. Please edit your example so that it is a complete small document that shows the problem. – David Carlisle Jun 13 at 15:11
• @DavidCarlisle thanks I've updated the code – jtbrasel Jun 13 at 15:40
• well it wasn't runnable as posted but I guessed some needed packages, it doesn't make the posted image but i suppose now shows the issue – David Carlisle Jun 13 at 15:45
• align advances the equation counter but you have redefined equation to ignore that and use the subsection counter, so naturally they both use 1 – David Carlisle Jun 13 at 15:48
• ahhh, ok, so I need to redefine align so that it also uses the subsection counter? – jtbrasel Jun 13 at 15:50

align advances the equation counter but you have redefined equation to ignore that and use the subsection counter, so naturally they both use 1

This makes subsection and equation counter the same, so if any construct increases one, the value of the other will have changed.

\documentclass{report}

\usepackage{mathtools,amsthm}

\numberwithin{section}{chapter}
\numberwithin{equation}{section}

% make equation counter an alias for subsection
\makeatletter
\let\c@equation\c@subsection
\makeatother

\theoremstyle{definition}
\newtheorem{thm}[subsection]{Theorem}
\newtheorem{lemma}[subsection]{Lemma}
\newtheorem{corol}[subsection]{Corollary}
\newtheorem{defn}[subsection]{Definition} % definition numbers are dependent on theorem numbers
\newtheorem{defns}[subsection]{Definitions} % definition numbers are dependent on theorem numbers
\newtheorem{prop}[subsection]{Proposition} % proposition numbers are dependent on theorem numbers
\newtheorem{exmp}[subsection]{Example} % same for example numbers
\newtheorem*{remark}{Remark}
\newtheorem{enum_remark}[subsection]{Remark}
\newtheorem{exercises}[subsection]{Exercises}
\newtheorem*{opex}{Opening Exercises}

\begin{document}

first align
\begin{align} \label{nchoosed}
\begin{split}
\prescript{}{n}C_d = \binom{n}{d}
&=  \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
&= \frac{n!}{d! (n-d)!}
\end{split}
\end{align}

some text
$$a = b$$

second align
\begin{align} \label{nchoosedz}
\begin{split}
\prescript{}{n}C_d = \binom{n}{d}
&=  \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
&= \frac{n!}{d! (n-d)!}
\end{split}
\end{align}

some text
$$a = b$$
\end{document}

• Thanks so much! that's the fix. – jtbrasel Jun 13 at 16:02