修改对齐环境改变方程编号的位置

Question

我怎样才能改变align环境（或基于创建新的环境align），以便方程式列中的对齐不是rlrlrl…但rrcll同时保持方程式编号在行末的位置？

我想出了下面的代码，它按照我想要的方式执行对齐部分。但是方程编号将变量空间移向方程。该空间取决于方程的长度。

关于如何定义新环境，我从第二个答案中得到了这个问题。

\documentclass{article}
\usepackage{amsmath,environ}

\makeatletter
\NewEnviron{gleichungssystem}
  {\def\align@preamble{%
     &\hfil
      \strut@
      \setboxz@h{\@lign$\m@th\displaystyle{####}$}%
      \ifmeasuring@\savefieldlength@\fi
      \set@field
      \tabskip2\tabcolsep
     &\hfil
      \strut@
      \setboxz@h{\@lign$\m@th\displaystyle{####}$}%
      \ifmeasuring@\savefieldlength@\fi
      \set@field
      \tabskip\tabcolsep
     &\hfil
      \strut@
      \setboxz@h{\@lign$\m@th\displaystyle{####}$}%
      \ifmeasuring@\savefieldlength@\fi
      \set@field
      \hfil
      \tabskip\tabcolsep
     &\setboxz@h{\@lign$\m@th\displaystyle{####}$}%
      \ifmeasuring@\savefieldlength@\fi
      \set@field
      \hfil
      \tabskip5\tabcolsep
     &\setboxz@h{\@lign$\m@th\displaystyle{####}$}%
      \ifmeasuring@\savefieldlength@\fi
      \set@field
      \hfil
      \tabskip\alignsep@
  }%
  \begin{align}\BODY\end{align}}
\makeatother

\begin{document}
\begin{gleichungssystem}
  &0 &= &\dot{x}(0) & \\
  \stackrel{xxx}{\Leftrightarrow} &0 & = & c_1 i \omega_0 \cdot e^{i \omega_0 0} - c_2 i \omega_0 \cdot e^{-i \omega_0 0} &\left| e^{0} = 1 \right. \\
  \Leftrightarrow &0 &= &i \omega_0 c_1 - i \omega_0 c_2 &\left| \cdot \frac{1}{i\omega_0} \right. \\
  \Leftrightarrow &0 &= &c_1 - c_2 &\left| + c_2 \right. \\
  \Leftrightarrow &c_2 &= &c_1 & \label{umformung_randbed-frei-unged1_bsp1-2}
\end{gleichungssystem}
\begin{gleichungssystem}
  &0 &= &\dot{x}(0) & \\
  \stackrel{xxx}{\Leftrightarrow} &0 & = & c_1 i \omega_0 \cdot e^{i \omega_0 0} - c_2 i \omega_0 \cdot e^{-i \omega_0 0} &
\end{gleichungssystem}

\begin{equation}
  a + b = c
\end{equation}

\end{document} 

Answer 1

对于这种复杂的对齐，您可以使用IEEEeqnarray。您可以在以下位置找到指南https://moser-isi.ethz.ch/docs/typeset_equations.pdf

\documentclass{article}
\usepackage{amsmath,IEEEtrantools}

\newenvironment{gleichungssystem}
 {\begin{IEEEeqnarray}{c'lCLL}}
 {
}

\begin{document}
\begin{IEEEeqnarray}{c'lCLL}
  & 0 &=& \dot{x}(0)
\\
\overset{xxx}{\Leftrightarrow}
  & 0 &=& c_1 i \omega_0 \cdot e^{i \omega_0 0} - 
          c_2 i \omega_0 \cdot e^{-i \omega_0 0} 
  & \left| e^{0} = 1 \right.
\\
\Leftrightarrow
  & 0 &=& i \omega_0 c_1 - i \omega_0 c_2
  & \left| \cdot \frac{1}{i\omega_0} \right.
\\
\Leftrightarrow
  & 0 &= &c_1 - c_2
  & \left| + c_2 \right.
\\
  \Leftrightarrow &c_2 &= &c_1 & \label{umformung_randbed-frei-unged1_bsp1-2}
\end{IEEEeqnarray}

\begin{IEEEeqnarray}{c'RCL}
  & 0 &=& \dot{x}(0)
\\
\overset{xxx}{\Leftrightarrow}
  & 0 &=& c_1 i \omega_0 \cdot e^{i \omega_0 0} - c_2 i \omega_0 \cdot e^{-i \omega_0 0}
\\
\IEEEeqnarraymulticol{4}{c}{a + b = c}
\end{IEEEeqnarray}

\end{document} 

Answer 2

也许你应该尝试一下这个包witharrows：

\documentclass{article}
\usepackage{witharrows}
\begin{document}
\begin{DispWithArrows}[format=rrcll]
L_1     \quad & a+b  &{} = {}& c + d & \quad L_1     \\
L_1+L_2 \quad &  x   &{} = {}& y     & \quad L_1+L_2 
\end{DispWithArrows}
\end{document}

Answer 3

既然使用环境如此简单，为何要重新发明轮子alignat？

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{alignat}{3}
  & & \quad 0 &= \dot{x}(0) \\
  & \overset{xxx}{\Leftrightarrow} &0 & = c_1 i \omega_0 \cdot e^{i \omega_0 0} - c_2 i \omega_0 \cdot e^{-i \omega_0 0} &\quad & \left|\; e^{0} = 1 \right. \\
   & \Leftrightarrow &0 &= i \omega_0 c_1 - i \omega_0 c_2 & & \left|{} \cdot \frac{1}{i\omega_0} \right. \\
  & \Leftrightarrow &0 &= c_1 - c_2 & & \left|{} + c_2 \right. \\
  & \Leftrightarrow &c_2 & = c_1 \label{umformung_randbed-frei-unged1_bsp1-2}
\end{alignat}

\begin{alignat}{2}
  & &\quad 0 &= \dot{x}(0) \\
  & \stackrel{xxx}{\Leftrightarrow} &0 & = c_1 i \omega_0 \cdot e^{i \omega_0 0} - c_2 i \omega_0 \cdot e^{-i \omega_0 0} 
\end{alignat}

\begin{equation}
  a + b = c
\end{equation}

\end{document}