GroundΒΆ
Let \(\mathcal{T} = (\mathcal{L}, \mathcal{M}, w_{r}, w_{e})\) where \(\mathcal{E} \neq \varnothing\). Given \(\mathcal{T}\), let \(\mathcal{T}' = (\mathcal{L}, \mathcal{M}', w_{r}, w_{e})\) where
and
Now let \(\mathcal{R}' := \{R'_{1}, \dots, R'_{\sigma(\mathcal{R}')}\}\) and \(\mathcal{Q}' := \{Q'_{1}, \dots, Q'_{\sigma(\mathcal{Q}')}\}\) be derived from \(\mathcal{T}'\). By definition, we have the following:
\(\forall i, \exists ! j\) such that \(R'_{i} \subseteq R_{j}.\)
\(\forall i, \exists ! j\) such that \(Q'_{i} \subseteq Q_{j}.\)
Similarly, let \(\mathcal{V}'\) and \(\mathcal{U}'\) to be derived from \(\mathcal{T}'\) respectively. Since \(\mathcal{V}' \cup \mathcal{U}' = \mathcal{R}'\), we get the following:
\(\forall i, \exists ! j\) such that \(V'_{i} \subseteq R_{j}.\)
\(\forall i, \exists ! j\) such that \(U'_{i} \subseteq R_{j}.\)
Furthermore, we also have
Note
This is due to \(\hat{Q}' \subseteq \hat{Q}\). More specifically, for any \(i\), \(V'_{i} \in \mathcal{R}'\) and \(V'_{i} \subseteq \hat{Q}'\). If we let \(V'_{i} = R'_{j}\), there is unique \(R_{k}\) where \(R_{j} \subseteq R_{k}\). By definition of \(V'_{i}\), we have \(R'_{j} \subseteq \hat{Q}'\). It follows
Hence, \(V_{l} = R_{k}\) for some \(l\) and \(V_{i} \subseteq V_{l}\).
However, it is worthwhile to note that we do not have
In particular, it is possible to have \(U'_{i} \subseteq V_{j}\) for some \(i\) and \(j\). We denote this by
where
and
Note
Note redacted.