因此，我有两组想要对齐的方程式，但它们被两行文本分开，其中第二行应该缩进。
MWE 与评论
\documentclass{article}
\usepackage{amsmath,mathtools,amssymb}
\DeclarePairedDelimiter{\sbar}{\lvert}{\rvert}
\newcommand{\tsbar}[1]{\sbar{#1}^\sigma_s}
\begin{document}
For this first one, I can't figure out an appropriate width for the parbox and I can't figure out how to indent the second line of text (neither the indent command nor an hspace works).
\begin{align*}
\tsbar{\phi x_1,\dots,v_k} &= T \text{ iff } \langle \sigma(v_1),\dots,\sigma(v_k) \rangle \in \sbar{\phi}_s;\\
\tsbar{\lnot \phi} &= T \text{ iff } \tsbar{\phi} \neq T;\\
\tsbar{\phi \rightarrow \psi} &= T \text{ iff } \tsbar{\phi}\neq T \text{ or } \tsbar{\psi}=T;\\
\tsbar{\forall x \phi} &= T \text{ iff, for every } a\in D_s,\ \sbar{\phi}^{[v/a]} = T;\\
\tsbar{\Box_> \phi} &= T \text{ iff, for every } t > s, \tsbar{\phi} = T;\\
\tsbar{\Box_< \phi} &= T \text{ iff, for every } t < s, \tsbar{\phi} = T.\\
\parbox{15em}{(Later-than is given the obvious definition: $t > s =_{df} s < t$.)\\[1ex]
%
\indent Then, the defined operators:}\\
\tsbar{\Box_\geq \phi} &= T \text{ iff, for every } t \geq s, \tsbar{\phi} = T;\\
\tsbar{\Box_\leq \phi} &= T \text{ iff, for every } t \leq s, \tsbar{\phi} = T;\\
\tsbar{\Box \phi} &= T \text{ iff, for every } t \leq s \vee s \leq t, \tsbar{\phi} = T.
\end{align*}
This one gives me everything I want, but yields a ``misplaced alignment character'' error that only disappears when I remove the double backslash before the second line in the intertext (and I haven't figured out how to produce the requisite spacing without it).
\begin{align*}
\tsbar{\phi x_1,\dots,v_k} &= T \text{ iff } \langle \sigma(v_1),\dots,\sigma(v_k) \rangle \in \sbar{\phi}_s;\\
\tsbar{\lnot \phi} &= T \text{ iff } \tsbar{\phi} \neq T;\\
\tsbar{\phi \rightarrow \psi} &= T \text{ iff } \tsbar{\phi}\neq T \text{ or } \tsbar{\psi}=T;\\
\tsbar{\forall x \phi} &= T \text{ iff, for every } a\in D_s,\ \sbar{\phi}^{[v/a]} = T;\\
\tsbar{\Box_> \phi} &= T \text{ iff, for every } t > s, \tsbar{\phi} = T;\\
\tsbar{\Box_< \phi} &= T \text{ iff, for every } t < s, \tsbar{\phi} = T.\\
\intertext{(The later-than relation on stages is given the obvious definition: $t > s =_{df} s < t$.)\\
%
Then, the defined operators:}
\tsbar{\Box_\geq \phi} &= T \text{ iff, for every } t \geq s, \tsbar{\phi} = T;\\
\tsbar{\Box_\leq \phi} &= T \text{ iff, for every } t \leq s, \tsbar{\phi} = T;\\
\tsbar{\Box \phi} &= T \text{ iff, for every } t \leq s \vee s \leq t, \tsbar{\phi} = T.
\end{align*}
\end{document}
所需的格式（MWE 中第二次尝试产生的格式，具有对齐字符错误的格式）如下：
我怎样才能实现这个目标？
答案1
我会使用\linewidth来表示 的宽度\parbox。有一个小问题：由 调用的宏align（或其 * 变体）不接受\par它内部的内容，因此您不能使用\par（或空白行）来结束\longintertext我建议的宏中的段落。一种解决方法是使用\endgraf。
\documentclass{article}
\usepackage{amsmath,mathtools,amssymb}
\DeclarePairedDelimiter{\sbar}{\lvert}{\rvert}
\newcommand{\tsbar}[1]{\sbar{#1}^\sigma_s}
\newlength{\normalparindent}
\AtBeginDocument{\setlength{\normalparindent}{\parindent}}
\newcommand{\longintertext}[1]{%
\intertext{%
\parbox{\linewidth}{%
\setlength{\parindent}{\normalparindent}
\noindent#1%
}%
}%
}
\begin{document}
For this first one, I can't figure out an appropriate width for the parbox and I can't figure out how
to indent the second line of text (neither the indent command nor an hspace works).
\begin{align*}
\tsbar{\phi x_1,\dots,v_k} &= T \text{ iff } \langle \sigma(v_1),\dots,\sigma(v_k) \rangle \in \sbar{\phi}_s;\\
\tsbar{\lnot \phi} &= T \text{ iff } \tsbar{\phi} \neq T;\\
\tsbar{\phi \rightarrow \psi} &= T \text{ iff } \tsbar{\phi}\neq T \text{ or } \tsbar{\psi}=T;\\
\tsbar{\forall x \phi} &= T \text{ iff, for every } a\in D_s,\ \sbar{\phi}^{[v/a]} = T;\\
\tsbar{\Box_> \phi} &= T \text{ iff, for every } t > s, \tsbar{\phi} = T;\\
\tsbar{\Box_< \phi} &= T \text{ iff, for every } t < s, \tsbar{\phi} = T.\\
\longintertext{%
(Later-than is given the obvious definition: $t > s =_{df} s < t$.)\endgraf
Then, the defined operators:
}
\tsbar{\Box_\geq \phi} &= T \text{ iff, for every } t \geq s, \tsbar{\phi} = T;\\
\tsbar{\Box_\leq \phi} &= T \text{ iff, for every } t \leq s, \tsbar{\phi} = T;\\
\tsbar{\Box \phi} &= T \text{ iff, for every } t \leq s \vee s \leq t, \tsbar{\phi} = T.
\end{align*}
\end{document}
答案2
您可以使用两种可能性：
\intertext{text... \endgraf text...}（受到推崇的）\intertext{text... \smallskip\newline\hspace*{\parindent} text...}
接受的答案已经显示了第一种可能性。因此，我仅展示第二种可能性：
\documentclass{article}
\usepackage{amsmath,mathtools,amssymb}
\DeclarePairedDelimiter{\sbar}{\lvert}{\rvert}
\newcommand{\tsbar}[1]{\sbar{#1}^\sigma_s}
\begin{document}
For this first one, I can't figure out an appropriate width for the parbox and I can't figure out how to indent the second line of text (neither the indent command nor an hspace works).
\begin{align*}
\tsbar{\phi x_1,\dots,v_k}
&= T \text{ iff } \langle \sigma(v_1),\dots,\sigma(v_k) \rangle \in \sbar{\phi}_s;\\
\tsbar{\lnot \phi}
&= T \text{ iff } \tsbar{\phi} \neq T;\\
\tsbar{\phi \rightarrow \psi}
&= T \text{ iff } \tsbar{\phi}\neq T \text{ or } \tsbar{\psi}=T;\\
\tsbar{\forall x \phi}
&= T \text{ iff, for every } a\in D_s,\ \sbar{\phi}^{[v/a]} = T;\\
\tsbar{\Box_> \phi}
&= T \text{ iff, for every } t > s, \tsbar{\phi} = T;\\
\tsbar{\Box_< \phi}
&= T \text{ iff, for every } t < s, \tsbar{\phi} = T.\\
\intertext{(Later-than is given the obvious definition: $t > s =_{df} s < t$.)
\smallskip\newline
\hspace*{\parindent}
Then, the defined operators:}
\tsbar{\Box_\geq \phi}
& = T \text{ iff, for every } t \geq s, \tsbar{\phi} = T;\\
\tsbar{\Box_\leq \phi}
& = T \text{ iff, for every } t \leq s, \tsbar{\phi} = T;\\
\tsbar{\Box \phi}
& = T \text{ iff, for every } t \leq s \vee s \leq t, \tsbar{\phi} = T.
\end{align*}
\end{document}
答案3
\\是为对齐定义的，但您可以使用\newline
\intertext{(The later-than relation on stages is given the obvious definition: $t > s =_{df} s < t$.)\newline
%
Then, the defined operators:}
（但是在强制换行之前，MWE 中的文本宽度会出现一个可怕的换行符）
