Ground
======

Let :math:`\mathcal{T} = (\mathcal{L}, \mathcal{M}, w_{r}, w_{e})`
where :math:`\mathcal{E} \neq \varnothing`. Given :math:`\mathcal{T}`, let
:math:`\mathcal{T}' = (\mathcal{L}, \mathcal{M}', w_{r}, w_{e})` where

.. math::

   \mathcal{M}' := \{M'_{1}, \dots, M'_{\sigma(\mathcal{M}')}\}

and

.. math::

   \tilde{M} \subseteq \tilde{M}'.

Now let :math:`\mathcal{R}' := \{R'_{1}, \dots, R'_{\sigma(\mathcal{R}')}\}` and
:math:`\mathcal{Q}' := \{Q'_{1}, \dots, Q'_{\sigma(\mathcal{Q}')}\}`
be derived from :math:`\mathcal{T}'`. By definition, we have the
following:

* :math:`\forall i, \exists ! j` such that :math:`R'_{i} \subseteq R_{j}.`
* :math:`\forall i, \exists ! j` such that :math:`Q'_{i} \subseteq Q_{j}.`

Similarly, let :math:`\mathcal{V}'` and :math:`\mathcal{U}'` to be derived
from :math:`\mathcal{T}'` respectively.
Since :math:`\mathcal{V}' \cup \mathcal{U}' = \mathcal{R}'`, we get the
following:

* :math:`\forall i, \exists ! j` such that :math:`V'_{i} \subseteq R_{j}.`
* :math:`\forall i, \exists ! j` such that :math:`U'_{i} \subseteq R_{j}.`

Furthermore, we also have

.. math::

   \forall i, \exists ! j ~\text{such that}~ V'_{i} \subseteq V_{j}.

.. note::

   This is due to :math:`\hat{Q}' \subseteq \hat{Q}`. More specifically,
   for any :math:`i`, :math:`V'_{i} \in \mathcal{R}'` and
   :math:`V'_{i} \subseteq \hat{Q}'`. If we let :math:`V'_{i} = R'_{j}`,
   there is unique :math:`R_{k}` where :math:`R_{j} \subseteq R_{k}`.
   By definition of :math:`V'_{i}`, we have
   :math:`R'_{j} \subseteq \hat{Q}'`. It follows

   .. math::

      R_{k} \cap \hat{Q}' \neq \varnothing \implies R_{k} \cap \hat{Q} \neq \varnothing \implies R_{k} \subseteq \hat{Q}.

   Hence, :math:`V_{l} = R_{k}` for some :math:`l`
   and :math:`V_{i} \subseteq V_{l}`.

However, it is worthwhile to note that we do *not* have

.. math::

   \forall i, \exists ! j ~\text{such that}~ U'_{i} \subseteq U_{j}.

In particular, it is possible to have :math:`U'_{i} \subseteq V_{j}` for some
:math:`i` and :math:`j`. We denote this by

.. math::

   \begin{align}
     \mathcal{U}' := &~ \{U'_{1}, \dots, U'_{\sigma(\mathcal{U}')}\} \\
     = &~ \mathcal{U}'_{\mathcal{V}} \cup \mathcal{U}'_{\mathcal{U}}
   \end{align}

where

.. math::

   \begin{align}
     \mathcal{U}'_{\mathcal{V}} := &~ \{U'_{\mathcal{V}, 1}, \dots, U'_{\mathcal{V}, \sigma(\mathcal{U}'_{\mathcal{V}})}\} \\
     \mathcal{U}'_{\mathcal{U}} := &~ \{U'_{\mathcal{U}, 1}, \dots, U'_{\mathcal{U}, \sigma(\mathcal{U}'_{\mathcal{V}})}\}
   \end{align}

and

.. math::

   \begin{align}
     \mathcal{U}'_{\mathcal{V}} \subseteq &~ \mathcal{V} \\
     \mathcal{U}'_{\mathcal{U}} \subseteq &~ \mathcal{U}.
   \end{align}

.. note::

   **Note redacted.**