answer space in xsim

Question

I am trying to use xsim to design a workbook for students, with a second version available for tutors. Therein will be exercises/questions/homework, however you want to call it, and their respective solutions.

From what I reviewed so far, xsim, seems to be a mostly well-suited package to achieve this. However, there is one piece of information I could not find in the documentation: How do I switch between an empty answer space/blank/box and the printed solutions?

Ideally, I want a way for the answer space to take up exactly the phantom space of the typeset solution.

Attached you find a MWE of how I want it to work, however I wish to have the procedure automated.

The following MWE produces (notice the solution/print = true)

\documentclass{article}
\usepackage{mwe}
\usepackage{amssymb, amsmath}
\usepackage{xsim}
\usepackage{tcolorbox}



\DeclareExerciseEnvironmentTemplate{tcolorbox}
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{\GetExerciseName~\GetExerciseProperty{counter}}%
    \GetExercisePropertyT{subtitle}{ \textit{\PropertyValue}}%
    ]%
}
{\endtcolorbox}

\DeclareExerciseType{problem}{
exercise-env = problem ,
solution-env = answer ,
exercise-name = Problem ,
solution-name = Answer ,
exercise-template = tcolorbox ,
solution-template = tcolorbox
}

\xsimsetup{
    answer/print=true
}


\begin{document}
    \begin{problem}
        Show that $(\mathbb{R}, +, \cdot)$ is a field.
    \end{problem}

    \begin{answer}
        Let $a, b, c \in \mathbb{R}$ arbitrary. Then, we show the following properties:
        \begin{description}
            \item[Commutativity of addition]  $a + b = b+ a$.
            \item[Associativity of addition] $a +(b + c) = (a+b) +c$.
            \item[Additive identity] $0 \in \mathbb{R}$, and $a +0 = a $.
            \item[Additive inverse] for each $a \in \mathbb{R}$, there exists $-a \in \mathbb{R}$, sch that $a + (-a) = 0$.
            \item[Commutativity of multiplication] $a \cdot b = b \cdot a$.
            \item[Associativity of multiplications] $a \cdot (b \dot c) = (a \cdot b) \cdot c$.
            \item[Multiplicative identity] $1 \in \mathbb{R}$, such that $1 \cdot a = a$.
            \item[Multiplicative inverse] for each $a \in \mathbb{R}$, there exists $a^{-1}\in \mathbb{R}$, such that $a \cdot a^{-1} = 1$.      
            \item[Distributivity] $a\cdot (b + c) = (a \cdot b) + (a \cdot c)$. 
        \end{description}
        These are the necessary and sufficient properties of a field, hence $(\mathbb{R}, +, \cdot)$ is a field.
    \end{answer}
\end{document}

but I want an empty box with the same dimensions when answers are disabled.

Update

Ultimately, I ended up ditching xsim, persued a different path, and came up with my own alternative solution, which is detailed in this other question and answer. However, it is still not an answer to this question.

Update

One step later than ultimately, I retuned to using xsim. See my answer with the solution below.

Answer 1

In the meantime I have come up with an answer, however this answer depends on the version of xsim in use.

xsim < v0.19b

See the updated MWE for example:

\documentclass{article}
\usepackage{mwe}
\usepackage{amssymb, amsmath}
\usepackage[]{xsim}
\usepackage{tcolorbox}

\xsimsetup{
    solution/print = true,
}

\SetExerciseParameters{exercise}{
    exercise-name = \XSIMtranslate{exercise},
    solution-name = \XSIMtranslate{solution},
    exercise-template = exercise,
    solution-template = solution,
    counter=section,
}



\DeclareExerciseEnvironmentTemplate{exercise}
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{\GetExerciseName~\GetExerciseProperty{counter}}%
    ]%
}%
{%
\endtcolorbox
% at this point, add invisible solution
\IfSolutionPrintTF{}{%
\tcolorbox[%
    upperbox=invisible,%
    beforeafter skip = .5\baselineskip ,%
    title =
    \textbf{\GetExerciseParameter{solution-name}~\GetExerciseProperty{counter}}%
    ]%
    \GetExerciseBody{solution}%
    \endtcolorbox%
}%
}%

\DeclareExerciseEnvironmentTemplate{solution}%
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{SOL: \GetExerciseName~\GetExerciseProperty{counter}}%
    ]%
}%
{\endtcolorbox}%


\begin{document}

\section{A section title}

\begin{exercise}
    Show that $(\mathbb{R}, +, \cdot)$ is a field.
\end{exercise}

\begin{solution}
    Let $a, b, c \in \mathbb{R}$ arbitrary. Then, we show the following properties:
    \begin{description}
        \item[Commutativity of addition]  $a + b = b+ a$.
        \item[Associativity of addition] $a +(b + c) = (a+b) +c$.
        \item[Additive identity] $0 \in \mathbb{R}$, and $a +0 = a $.
        \item[Additive inverse] for each $a \in \mathbb{R}$, there exists $-a \in \mathbb{R}$, sch that $a + (-a) = 0$.
        \item[Commutativity of multiplication] $a \cdot b = b \cdot a$.
        \item[Associativity of multiplications] $a \cdot (b \dot c) = (a \cdot b) \cdot c$.
        \item[Multiplicative identity] $1 \in \mathbb{R}$, such that $1 \cdot a = a$.
        \item[Multiplicative inverse] for each $a \in \mathbb{R}$, there exists $a^{-1}\in \mathbb{R}$, such that $a \cdot a^{-1} = 1$.
        \item[Distributivity] $a\cdot (b + c) = (a \cdot b) + (a \cdot c)$.
    \end{description}
    These are the necessary and sufficient properties of a field, hence $(\mathbb{R}, +, \cdot)$ is a field.
\end{solution}
\end{document}

produces the outputs (dependent on the printsolutions property):

xsim v0.20

Ideally, the same code should work. However, it does not until this bug is resolved. The workaround in the issue's comments, namely using

\xsimsetup{collect=true}

and later in the document body using

\printcollection[print=both]{all exercises}

unfortunately appends a filled-in copy of the solutions if the option solutions/print=false is set.

So the current workaround for this answer is: Move the solution body into the exercise for printing:

\documentclass{article}
\usepackage{mwe}
\usepackage{amssymb, amsmath}
\usepackage[]{xsim}
\usepackage{tcolorbox}

\xsimsetup{
    solution/print = true,  % toggle this line
    collect=true,
}

\SetExerciseParameters{exercise}{
    exercise-name = \XSIMtranslate{exercise},
    solution-name = \XSIMtranslate{solution},
    exercise-template = exercise,
    solution-template = solution,
    counter=section,
}



\DeclareExerciseEnvironmentTemplate{exercise}
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{\GetExerciseName~\GetExerciseProperty{counter}}%
    ]%
}%
{%
\endtcolorbox
% at this point, add invisible solution
\IfSolutionPrintTF{%
\tcolorbox[%
    beforeafter skip = .5\baselineskip ,%
    title =
    \textbf{\GetExerciseParameter{solution-name}~\GetExerciseProperty{counter}}%
    ]%
    \GetExerciseBody{solution}%
    \endtcolorbox%
}{%
\tcolorbox[%
    upperbox=invisible,%
    beforeafter skip = .5\baselineskip ,%
    title =
    \textbf{\GetExerciseParameter{solution-name}~\GetExerciseProperty{counter}}%
    ]%
    \GetExerciseBody{solution}%
    \endtcolorbox%
}%
}%

\DeclareExerciseEnvironmentTemplate{solution}%
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{\GetExerciseName~\GetExerciseProperty{counter}}%
    ]%
}%
{\endtcolorbox}%


\begin{document}
\section{A section title}


\begin{exercise}
    Show that $(\mathbb{R}, +, \cdot)$ is a field.
\end{exercise}

\begin{solution}
    Let $a, b, c \in \mathbb{R}$ arbitrary. Then, we show the following properties:
    \begin{description}
        \item[Commutativity of addition]  $a + b = b+ a$.
        \item[Associativity of addition] $a +(b + c) = (a+b) +c$.
        \item[Additive identity] $0 \in \mathbb{R}$, and $a +0 = a $.
        \item[Additive inverse] for each $a \in \mathbb{R}$, there exists $-a \in \mathbb{R}$, sch that $a + (-a) = 0$.
        \item[Commutativity of multiplication] $a \cdot b = b \cdot a$.
        \item[Associativity of multiplications] $a \cdot (b \dot c) = (a \cdot b) \cdot c$.
        \item[Multiplicative identity] $1 \in \mathbb{R}$, such that $1 \cdot a = a$.
        \item[Multiplicative inverse] for each $a \in \mathbb{R}$, there exists $a^{-1}\in \mathbb{R}$, such that $a \cdot a^{-1} = 1$.
        \item[Distributivity] $a\cdot (b + c) = (a \cdot b) + (a \cdot c)$.
    \end{description}
    These are the necessary and sufficient properties of a field, hence $(\mathbb{R}, +, \cdot)$ is a field.
\end{solution}



\printcollection[print=exercises]{all exercises}
\end{document}

This produces the following desired output (identical to the previous version, but with less straightforward code and possible additional unobserved drawbacks):