by Dennis
Last Updated January 11, 2018 10:23 AM

So I have two sets of equations that I'd like to be aligned, but they're broken up by two lines of text, the second of which should be indented.

**MWE with Comments**

```
\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}
```

The desired formatting (the one produced by the second attempt in the MWE, the one with the alignment character error) is as follows:

How can I get this accomplished?

`\\`

is defined for the alignment, but you can use `\newline`

```
\intertext{(The later-than relation on stages is given the obvious definition: $t > s =_{df} s < t$.)\newline
%
Then, the defined operators:}
```

(But you get a terrible line break with the text width in the MWE, just before the forced break)

An alternative solution (with desired indent and vertical space between lines in the inter text):

```
\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}
```

**Edit:** Before David Carlisle comment I was not aware of `\endgraf`

. By it can be rude solution:

```
\smallskip\newline
\hspace*{\parindent}
```

replaced by simple

```
\endgraf
```

I'd use `\linewidth`

for the width of the `\parbox`

. There is a small catch: the macro called by `align`

(or the *-variant thereof) doesn't accept `\par`

inside it, so you can't use `\par`

(or a blank line) for ending paragraphs in the `\longintertext`

macro I suggest. A workaround is to use `\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}
```

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