# Displaying Counter Value from Equation Tag

I am trying to create the following, but haven't been able to find anything here or elsewhere online...

I want to create an equation with a custom counter and then call the value of the counter when it was used in that equation. For example consider this.

\newcounter{opt}
$ax + b = c \refstepcounter{opt} \label{eq:1}$
\ref{eq:1}


This final \ref call produces the value 1, as I desire. However, I'd like to tag the equation with a non-standard tag, eg \tag*{$(P_1)$}. I can do this as follows.

\newcounter{opt}
$ax + b = c \refstepcounter{opt} \label{eq:1} \tag*{(P_\theopt)}$
\ref{eq:1}


This gives the correct tag, however now the \ref call produces (P_1), instead of just 1.

How can I adapts this so that I get the desired tag and I can reference the equation and get the value of the counter?

This, "Referring to label and value of counter at that point", is a highly related post. It's doing something a bit more in-depth than I am trying to do, and I can't understand what's going on well enough to adapt it...

If it helps, the context is the following. I have an optimisation problem, and I want to label two equations (P_1) and (C_1) using the same counter opt. I then want to call this as (P,C)_1; I was going to do this by writing (P,C)_\ref{eq:1}.

I suppose, all I actually want to be able to do is to write a different thing in the brackets () on the equation line to what I \ref produces.

If there's a better way of doing this, I'm all ears!

• If the equation is labeled (P<sub>1</sub>), why should the reference be just 1? – egreg Sep 29 '17 at 8:42
• Sorry, if I wasn't clear. I'm not questioning why the above does what it does -- indeed, it's labelled (P_1), as you say. However, I want to be able to call from the label just the value 1. This is what I can't work out how to do. \\ Hopefully the context I've added makes it clear! :) – Sam T Sep 29 '17 at 8:45
• Do you want, for certain equations, a counter independent from the general equation counter, or only change the way the general counter is displayed? – Bernard Sep 29 '17 at 9:30
• I'd like this whole part to be separate from the general equation counter – Sam T Sep 29 '17 at 9:33
• But if the two equations are different, you can't refer them with the same value of the counter! – Bernard Sep 29 '17 at 9:38

Use a protected command, so it will be written as such in the .aux file and you can define it to act differently in different contexts.

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

\newcounter{xtag}
\NewDocumentCommand{\xtag}{sm}{%
\IfBooleanF{#1}{\refstepcounter{xtag}}%
\tag{\choosetag{#2}{\thextag}}%
}

\ExplSyntaxOn
\NewDocumentCommand{\choosetag}{mm}
{
\bool_if:NTF \l_samt_choosetag_ref_bool
{
#2
}
{
$\mathrm{#1}\sb{#2}$
}
}
\NewDocumentCommand{\xref}{m}
{
\group_begin:
\bool_set_true:N \l_samt_choosetag_ref_bool
\ref{#1}
\group_end:
}
\NewDocumentCommand{\xeqref}{om}
{
\IfNoValueTF{#1}
{
\eqref{#2}
}
{
\textnormal{(#1)}$\sb{\xref{#2}}$
}
}

\bool_new:N \l_samt_choosetag_ref_bool
\ExplSyntaxOff

\begin{document}

\begin{gather}
1=1 \xtag{P}  \label{a} \\
2=2 \xtag*{C} \label{b} \\
3=3 \xtag{P}  \label{c}
\end{gather}

\xeqref{a} or \xeqref[P,C]{a}

\xeqref{c}

\end{document}


The command \xtag advances the counter, but the variant \xtag* doesn't. • Grazie mille! That's exactly what I wanted :) -- I have no idea what the group and bool stuff means, but I understand the rest... mostly! There are a couple of other situations I wanted a similar thing for, and I should be able to adapt your answer to those, so thanks very much! :) – Sam T Sep 29 '17 at 11:42
• The \choosetag command changes its behavior according to truth or falsehood of the boolean; in \xref the boolean is set to true, so only the second argument is used. I fixed a slip. – egreg Sep 29 '17 at 11:51
• Ok, thanks. And yeah, I noticed the change -- what difference did it make? – Sam T Sep 29 '17 at 11:57
• @SamT You'd realize it if typesetting \xeqref in a theorem statement (or, in general, in italic context). – egreg Sep 29 '17 at 11:59
• Ah, because it was forcing it to be upright? – Sam T Sep 29 '17 at 12:02