Approximation operators via TD-matroids on two sets

Abstract

Rough set theory is an extension of set theory with two additional unary set-theoretic operators known as approximation in order to extract information and knowledge. It needs the basic, or say definable, knowledge to approximate the undefinable knowledge in a knowledge space using the pair of approximation operators. Many existed approximation operators are expressed with unary form. How to mine the knowledge which is asked by binary form with rough set has received less research attention, though there are strong needs to reveal the answer for this challenging problem. There exist many information with matroid constraints since matroid provides a platform for combinatorial algorithms especially greedy algorithm. Hence, it is necessary to consider a matroidal structure on two sets no matter the two sets are the same or not. In this paper, we investigate the construction of approximation operators expressed by binary form with matroid theory, and the constructions of matroidal structure aided by a pair of approximation operators expressed by binary form.

• First, we provide a kind of matroidal structure—TD-matroid defined on two sets as a generalization of Whitney classical matroid.

• Second, we introduce this new matroidal construction to rough set and construct a pair of approximation operators expressed with binary form.

• Third, using the existed pair of approximation operators expressed with binary form, we build up two concrete TD-matroids.

• Fourth, for TD-matroid and the approximation operators expressed by binary form on two sets, we seek out their properties with aspect of posets, respectively.

• Through the paper, we use some biological examples to explain and test the correct of obtained results. In summary, this paper provides a new approach to research rough set theory and matroid theory on two sets, and to study on their applications each other.

Appendices

Appendix

1. Proof of Lemma 7

Proof

Let $low(A,B)=\{(X_j,Y_j), j\in \mathcal {J}\}$ and $upr(A,B)=\{(X_p, Y_p), p\in \mathcal {P}\}$. It follows $|\mathcal {J}|, |\mathcal {P}|<\infty$ since $|S|,|T|<\infty$.

Next, we prove item (1).

$X_j\subseteq A$ and $B\subseteq Y_j$ hold since $(X_j,Y_j)\sqsubseteq (A,B) \; (j\in \mathcal {J})$. So, we obtain $\cup _{j\in \mathcal {J}}X_j\subseteq A$ and $B\subseteq \cap _{j\in \mathcal {J}}Y_j$. And further, we get $\underline{MR}(A,B)=(\cup _{j\in \mathcal {J}}X_j, \cap _{j\in \mathcal {J}} Y_j)\sqsubseteq (A,B)$.

If $|\mathcal {P}|=0$, then $upr(A,B)=\emptyset$. Under this case, we know $\overline{MR}(A,B)=(S,\emptyset )$ by Definition 8. So, $(A,B)\sqsubseteq \overline{MR}(A,B)$ holds since $A\subseteq S$ and $B\supseteq \emptyset$.

If $|\mathcal {P}|\ne \emptyset$, then $A\subseteq X_p$ and $B\supseteq Y_p$ hold since $(A,B)\sqsubseteq (X_p,Y_p)\;(p\in \mathcal {P})$. So, we obtain $A\subseteq \cap _{p\in \mathcal {P}}X_p$ and $B\supseteq \cup _{p\in \mathcal {P}}Y_p$. And further, we get $(A,B)\sqsubseteq (\cap _{p\in \mathcal {P}}X_p, \cup _{p\in \mathcal {P}} Y_p)=\overline{MR}(A,B)$.

Next, we prove item (2).

Let $(A,B)\in \mathcal {T}$. Then, $A\subseteq A$ and $B\subseteq B$ together mean $(A,B)\in low(A,B)$ and $(A,B)\in upr(A,B)$.

On the one hand, there is one and only one $j_0\in \mathcal {J}$ satisfying $X_{j_0}=A$ and $Y_{j_0}=B$ since $(A,B)\in low(A,B)$. So, we get $A=X_{j_0}\subseteq \cup _{j\in \mathcal {J}}X_j\subseteq A$ and $B=Y_{j_0}\supseteq \cap _{j\in \mathcal {J}}Y_j\supseteq B$ according to the definition of $\underline{MR}(A,B)$ and item (1). Thus, we obtain $\underline{MR}(A,B)=(\cup _{j\in \mathcal {J}}X_j,\cap _{j\in \mathcal {J}}Y_j)=(A,B)$.

On the other hand, there is one and only one $p_0\in \mathcal {P}$ satisfying $X_{p_0}=A$ and $Y_{p_0}=B$ since $(A,B)\in upr(A,B)$. So, we get $A\subseteq \cap _{p\in \mathcal {P}}X_p\subseteq X_{p_0}=A$ and $B\supseteq \cup _{p\in \mathcal {P}}Y_p\supseteq Y_{p_0}=B$ according to the definition of $\overline{MR}(A,B)$ and item (1). Thus, we obtain $\overline{MR}(A, B)=(\cap _{p\in \mathcal {P}}X_p,\cup _{p\in \mathcal {P}}Y_p)=(A,B)$.

Next, we prove item (3).

Suppose $\underline{MR}(A,B)=(A,B)=\overline{MR}(A,B)$. Then, it infers the following formulas

$$\begin{aligned}&X_j\subseteq \cup _{j\in \mathcal {J}}X_j=A=\cap _{p\in \mathcal {P}}X_p\subseteq X_p\, (j\in \mathcal {J}, p\in \mathcal {P})\, \textcircled {1} .\\&Y_p\subseteq \cup _{p\in \mathcal {P}}Y_p=B=\cap _{j\in \mathcal {J}}Y_j\subseteq Y_j\, (j\in \mathcal {J},p\in \mathcal {P})\,\, \textcircled {2} . \end{aligned}$$

It induces $(A,Y_p)\sqsubseteq (X_p,Y_p)\in \mathcal {T}$ according to $Y_p\subseteq Y_p$ and the formula $\textcircled {1}\; (p\in \mathcal {P})$. Combining formula $\textcircled {2}$ with (T2) and $(X_p,Y_p)\in \mathcal {T}\,(p\in \mathcal {P})$, we get $(A,Y_p)\in \mathcal {T}$ satisfying $(A,B)\sqsubseteq (A,Y_p)$ for any $p\in \mathcal {P}$. Thus, $(A,Y_p)\in upr(A,B)$ holds for any $p\in \mathcal {P}$. That is to say, there is $p_1\in \mathcal {P}$ such that $X_{p_1}=A$ satisfying $(X_{p_1}, Y_p)\in \mathcal {T}\, (p\in \mathcal {P})$. Additionally, $(X_p,B)\sqsubseteq (X_p,Y_p)\in \mathcal {T}$ holds since $B\supseteq Y_p$ holds by formula $\textcircled {2}\;(p\in \mathcal {P})$. Taking this result with (T2), it induces $(X_p,B)\in \mathcal {T}\,(p\in \mathcal {P})$. Combining formula $\textcircled {1}$ with $(A,B)\sqsubseteq (X_p,B)\,(p\in \mathcal {P})$, we get $(X_p,B)\in upr(A,B)$ for any $p\in \mathcal {P}$. That is to say, there is $p_2\in \mathcal {P}$ such that $Y_{p_2}=B$ and $(X_p, Y_{p_2})\in \mathcal {T}\, ( p\in \mathcal {P})$. Especially, $(X_{p_1}, Y_{p_2})\in \mathcal {T}$ holds since $p_1\in \mathcal {P}$. In other words, $(A,B)\in \mathcal {T}$ is followed since $X_{p_1}=A$ and $Y_{p_2}=B$.

2. Proof of Lemma 8

Proof

Let $low(A_i,B_i) =\{(X_{ij}, Y_{ij}), j\in \mathcal {J}_i\}$ and $upr(A_i,B_i) =\{(X_{ip}, Y_{ip}), p\in \mathcal {P}_i\}\, (i=1,2)$.

To prove item (1).

$(X_{1j}, Y_{1j})\sqsubseteq (A_1,B_1)\sqsubseteq (A_2,B_2)$ follows $(X_{1j}, Y_{1j})\in low (A_2,B_2)\, (j\in \mathcal {J}_1)$. So, we get $\cup _{j\in \mathcal {J}_1}X_{1j}\subseteq \cup _{j\in \mathcal {J}_2}X_{2j}$ and $\cap _{j\in \mathcal {J}_1}Y_{1j}\supseteq \cap _{j\in \mathcal {J}_2}Y_{2j}$. Furthermore, we obtain $\underline{MR}(A_1,B_1)=(\cup _{j\in \mathcal {J}_1}X_{1j}, \cap _{j\in \mathcal {J}_1}Y_{1j})\sqsubseteq (\cup _{j\in \mathcal {J}_2}X_{2j},\cap _{j\in \mathcal {J}_2}Y_{2j})=\underline{MR}(A_2,B_2)$.

To prove item (2).

$(A_1,B_1)\sqsubseteq (A_2,B_2)\sqsubseteq (X_{2p},Y_{2p})$ follows $(X_{2p},Y_{2p})\in upr(A_1,B_1)\,(p\in \mathcal {P}_2)$. So, $upr(A_2,B_2)\subseteq upr(A_1,B_1)$ holds. This demonstrates that the case of $upr(A_1,B_1)=\emptyset$ and $upr(A_2,B_2)\ne \emptyset$ does not exist. In other words, it exists and only exists the following cases:

Case 1. $upr(A_i,B_i)\ne \emptyset \,\, (i=1,2)$;

Case 2. $upr(A_i,B_i)=\emptyset \,\, (i=1,2)$;

Case 3. $upr(A_1,B_1)\ne \emptyset$ and $upr(A_2,B_2)=\emptyset$.

Under Case 1, we get $\cap _{p\in \mathcal {P}_1}X_{1p}\subseteq \cap _{p\in \mathcal {P}_2}X_{2p}$ and $\cup _{p\in \mathcal {P}_1}Y_{1p}\supseteq \cup _{p\in \mathcal {P}_2}Y_{2p}$ according to $upr(A_2,B_2)\subseteq upr(A_1,B_1)$. Furthermore, we obtain $\overline{MR}(A_1,B_1)=(\cap _{p\in \mathcal {P}_1}X_{1p},\cup _{p\in \mathcal {P}_1}Y_{1p}) \sqsubseteq (\cap _{p\in \mathcal {P}_2}X_{2p},\cup _{p\in \mathcal {P}_2}Y_{2p}) = \overline{MR}(A_2,B_2)$.

Under Case 2, we obtain $\overline{MR}(A_i,B_i)=(S,\emptyset )\, (i=1,2)$ in view of Definition 8. That is, $\overline{MR}(A_1,B_1)\sqsubseteq \overline{MR}(A_2,B_2)$ holds.

Under Case 3, we obtain $\overline{MR}(A_2,B_2)=(S,\emptyset )$ and the existence of $\overline{MR}(A_1,B_1)$ with $(\overline{MR}(A_1,B_1)\ne (S,\emptyset )$ in light of Definition 8. It is easy to see $\overline{MR}(A_1,B_1)\sqsubseteq (S,\emptyset )$. Namely, it has $\overline{MR}(A_1,B_1)\sqsubseteq \overline{MR}(A_2,B_2)$.

3. Proof of Theorem 3(2)

Proof

Since $\mathbb {K}$ is a formal context, it follows $O\ne \emptyset$ and $P\ne \emptyset$ by Definition 2. Using Definition 7, we only need to check the conditions (T1), (T2) and (T3) to be satisfied by $\overline{\mathcal {T}_{(A,B)}}(R)$, respectively.

First to check (T1) for $\overline{\mathcal {T}_{(A,B)}}(R)$.

Using Lemma 2(1), we receive $\overline{R}(\emptyset , B)=(\emptyset , \emptyset ^\prime \cup B)=(\emptyset , P)$ since $\emptyset ^\prime =P$ for $\emptyset \subseteq O$. Hence, $(\emptyset , B)\in \overline{\mathcal {T}_{(A,B)}}(R)$ holds since $\emptyset \subseteq A$ and $P\supseteq B$ together follow $(\emptyset , P)\sqsubseteq (A,B)$. Thus, $\overline{\mathcal {T}_{(A,B)}}(R)\ne \emptyset$ holds.

Second to check (T2) for $\overline{\mathcal {T}_{(A,B)}}(R)$.

Let $(X_2,Y_2)\sqsubseteq (X_1,Y_1)\in \overline{\mathcal {T}_{(A,B)}}(R)$. Then, we may easily find $X_2\subseteq X_1$ and $Y_2\supseteq Y_1$ by the definition of $\sqsubseteq$. In addition, we also find the following two facts.

$$\begin{aligned}&\mathrm {Fact} \;1:\;(X_1,Y_1)\in \overline{\mathcal {T}_{(A,B)}}(R)\Rightarrow \overline{R}(X_1,Y_1)\sqsubseteq (A,B)\\&\quad ( \mathrm {definition}\, \mathrm {of}\, \overline{\mathcal {T}_{(A,B)}}(R))\\&\quad \Rightarrow (X_1,X_1^\prime \cup Y_1)\sqsubseteq (A,B)\quad ( \mathrm {Lemma} \,2(1))\\&\Rightarrow X_1\subseteq A \,\mathrm {and}\, X_1^\prime \cup Y_1\supseteq B. \quad \quad (\mathrm {definition}\,\mathrm {of} \,\sqsubseteq )\\&\mathrm {Fact}\; 2: \;(X_2,Y_2)\sqsubseteq (X_1,Y_1)\\&\quad \Rightarrow X_2\subseteq X_1\,\mathrm {and}\,Y_2\supseteq Y_1 \quad ( \mathrm {definition}\,\mathrm {of}\, \sqsubseteq )\\&\quad \Rightarrow X_2^\prime \supseteq X_1^\prime \,\\&\quad \mathrm {and}\, Y_2\supseteq Y_1 \quad ( \mathrm {Lemma}\,1(1))\\&\quad \Rightarrow X_2^\prime \cup \\&\quad Y_2\supseteq X_1^\prime \cup Y_1 \end{aligned}$$

Combining the above two facts and $X_2\subseteq X_1\subseteq A$, we obtain $X_2^\prime \cup Y_2\supseteq X_1^\prime \cup Y_1\supseteq B$. Thus, $\overline{R}(X_2,Y_2)=(X_2,X_2^\prime \cup Y_2)\sqsubseteq (A,B)$ holds. So, $(X_2,Y_2)\in \overline{\mathcal {T}_{(A,B)}}(R)$ is followed.

Third to check (T3) for $\overline{\mathcal {T}_{(A,B)}}(R)$.

Let $(X_1,Y_1), (X_2,Y_2)\in \overline{\mathcal {T}_{(A,B)}}(R)$ satisfy $|(X_2,Y_2)|<|(X_1,Y_1)|$ and $(X_1,Y_1)\ne (\emptyset ,\emptyset )$. Then, we get the following fact:

$$\begin{aligned}&(X_j,Y_j)\in \overline{\mathcal {T}_{(A,B)}}(R)\\&\quad \Rightarrow \overline{R}(X_j,Y_j)\sqsubseteq (A,B)\, (j=1,2) \,\,\,\\&\quad ( \mathrm {definition}\,\mathrm {of} \,\overline{\mathcal {T}_{(A,B)}}(R))\\&\quad \Rightarrow (X_j,X_j^\prime \cup Y_j)\sqsubseteq (A,B)\, (j=1,2)\,\,\, \\&\quad ( \mathrm {Lemma}\,2(1))\\&\quad \Rightarrow X_j\subseteq A\,\mathrm {and}\,X_j^\prime \cup Y_j\supseteq B\, (j=1,2)\,\,\, \\&\quad ( \mathrm {definition}\,\mathrm {of}\,\sqsubseteq ) \end{aligned}$$

Via Lemma 4, we confirm $(X_1,Y_1){\setminus }(X_2,Y_2)\ne (\emptyset ,\emptyset )$. Then, we select $(a,b)\in (X_1,Y_1)\setminus (X_2,Y_2)$. We divide two cases to finish the proof.

Case 1. $X_1\setminus X_2=\emptyset$.

It gets $a=\emptyset$ and $b\in Y_1{\setminus } Y_2\ne \emptyset$. Then, we obtain $\overline{R}((X_2,Y_2)\cup (a,b))=\overline{R}(X_2, Y_2\cup b )= (X_2,(X_2^\prime \cup Y_2)\cup b )$ holds according to Lemma 2(1). Using $X_2^\prime \cup Y_2\supseteq B$, we obtain $(X_2^\prime \cup Y_2)\cup b \supseteq B$. Thus, combining $X_2\subseteq A$ with the results mentioned above, we confirm $\overline{R}((X_2,Y_2)\cup (a,b))\sqsubseteq (A,B)$. Hence, we receive $(X_2,Y_2)\cup (a,b)\in \overline{\mathcal {T}_{(A,B)}}(R)$.

Case 2. $X_1\setminus X_2\ne \emptyset$.

Select $a\in X_1{\setminus } X_2$ and $b\in Y_1{\setminus } Y_2$. If $Y_1{\setminus } Y_2=\emptyset$ (or $Y_1{\setminus } Y_2\ne \emptyset$), then $b=\emptyset$ (or $b\ne \emptyset$). No matter which situation happens, it follows

$$\begin{aligned} \overline{R}((X_2,Y_2)\cup (a,b))&=\overline{R}(X_2\cup a ,Y_2\cup b )\\&\quad (\mathrm {definition}\,\mathrm {of}\,\cup )\\&\quad =(X_2\cup a ,(X_2\cup a )^\prime \cup (Y_2\cup b ))\\&\quad (\mathrm {Lemma } \,2(1))\\&\quad =(X_2\cup a ,(X_2^\prime \cap a^\prime )\cup (Y_2\cup b ))\\&\quad (\mathrm {Lemma}\, 1(2))\end{aligned}$$

Combining $a\in X_1\subseteq A$ and $X_2\subseteq A$, we obtain $X_2\cup a \subseteq X_2\cup X_1\subseteq A$. So, $(X_2\cup a )^\prime =X_2^\prime \cap a^\prime \supseteq A^\prime$ in light of Lemma 1. Thus, we confirm $(X_2^\prime \cap a^\prime )\cup (Y_2\cup b )\supseteq A^\prime \cup (Y_2\cup b)\supseteq A^\prime \supseteq B$ since $B\subseteq A^\prime$. Furthermore, we get $(X_2,Y_2)\cup (a,b)\in \overline{\mathcal {T}_{(A,B)}}(R)$.

Summing up the above two cases, (T3) is correct for $\overline{\mathcal {T}_{(A,B)}}(R)$.

4. Proofs of items (2), (3) and (4) in Theorem 4

Proof

The proof of item (2) will be finished by two parts.

$$\begin{aligned} \mathrm {Part }\,1.\,&(X_1,Y_1)\sqsubseteq (X_2,Y_2)\\&\quad \Rightarrow X_1\subseteq X_2\,\mathrm {and}\, Y_1\supseteq Y_2\quad (\mathrm {definition}\, \mathrm {of} \,\sqsubseteq )\\&\quad \Rightarrow X_1\subseteq X_2,X_1^\prime \supseteq X_2^\prime \,\mathrm {and}\, Y_1\supseteq Y_2\quad (\mathrm {Lemma}\,1(1))\\&\quad \Rightarrow X_1\subseteq X_2\,\mathrm {and}\, X_1^\prime \cap \\&\quad Y_1\supseteq X_2^\prime \cap Y_2\\&\quad \Rightarrow (X_1,X_1^\prime \cap Y_1)\sqsubseteq (X_2,X_2^\prime \cap Y_2)\\&\quad (\mathrm {defition}\,\mathrm {of}\,\sqsubseteq )\\&\Rightarrow \underline{R}(X_1,Y_1)\sqsubseteq \underline{R}(X_2,Y_2)\\&\quad ( \mathrm {Lemma}\,2(1)) \end{aligned}$$

$$\begin{aligned}\mathrm {Part}\,2.\,&(X_1,Y_1)\sqsubseteq (X_2,Y_2)&\\&\Rightarrow X_1\subseteq X_2\,\mathrm {and}\, Y_1\supseteq Y_2&(\mathrm {definition}\, \mathrm {of} \,\sqsubseteq )\\&\Rightarrow X_1\subseteq X_2,X_1^\prime \supseteq X_2^\prime \,\mathrm {and}\, Y_1\supseteq Y_2&(\mathrm {Lemma}\,1(1))\\&\Rightarrow X_1\subseteq X_2\,\mathrm {and}\,X_1^\prime \cup Y_1\supseteq X_2^\prime \cup Y_2\\&\Rightarrow (X_1,X_1^\prime \cup Y_1)\sqsubseteq (X_2,X_2^\prime \cup Y_2)&(\mathrm {definition}\,\mathrm {of}\,\sqsubseteq )\\&\Rightarrow \overline{R}(X_1,Y_1)\sqsubseteq \overline{R}(X_2,Y_2)&(\mathrm {Lemma} \,2(1) ) \end{aligned}$$

To prove item (3).

Combining Remark 2, we know $\emptyset ^\prime =P$ for $\emptyset \subseteq O$. This implies $(\emptyset ,P)\in \mathcal {B}(\mathbb {K})$ by Definition 2(2) and Remark 2. So, it infers $\underline{R}(\emptyset ,P)=\overline{R}(\emptyset ,P)=(\emptyset ,P)$ using Lemma 2(4).

It is clear $(\emptyset , P)\sqsubseteq (A,B)$. Moreover, we receive $(\emptyset , P)\in \underline{\mathcal {T}_{(A,B)}}(R)$ and $(\emptyset , P)\in \overline{\mathcal {T}_{(A,B)}}(R)$. Additionally, both $(\emptyset ,P)\sqsubseteq (X_j,Y_j)$ and $(\emptyset ,P)\sqsubseteq (X_p,Y_p)$ evidently hold $(j\in \mathcal {J};p\in \mathcal {P})$. Considered item (1), we can indicate $(\emptyset ,P)$ to be the minimum element in the poset $(\underline{\mathcal {T}_{A,B)}}(R),\sqsubseteq )$, and also the minimum element in the poset $(\overline{\mathcal {T}_{(A,B)}}(R),\sqsubseteq )$.

Additionally, we obtain the following expression Part I.

$$\begin{aligned} \mathrm {Part}\, \mathrm {I}.\,&(X_j,Y_j)\in \underline{\mathcal {T}_{(A,B)}}(R)\,(j\in \mathcal {J})\\&\quad \Rightarrow \underline{R}(X_j,Y_j)\sqsubseteq (A,B)\,(j\in \mathcal {J})\\&\quad \quad (\mathrm {definition}\,\mathrm {of}\,\underline{\mathcal {T}_{(A,B) }}(R))\\&\Rightarrow (X_j,X_j^\prime \cap Y_j)\sqsubseteq (A,B)\,(j\in \mathcal {J})\\&\quad (\mathrm {Lemma}\,2(1))\\&\Rightarrow X_j\subseteq A\,\mathrm {and}\,X_j^\prime \cap Y_j\supseteq B\,(j\in \mathcal {J})\\&\quad (\mathrm {definition}\, \mathrm {of}\, \sqsubseteq )\\&\Rightarrow \cup _{j\in \mathcal {J}}X_j\subseteq A\\&\mathrm {and} (\cup _{j\in \mathcal {J}}X_j)^\prime \cap (\cap _{j\in \mathcal {J}}Y_j) =\cap _{j\in \mathcal {J}}(X_j^\prime \cap Y_j)\supseteq B\\&\quad (\mathrm {Lemma}\,1(2))&\\&\Rightarrow (\cup _{j\in \mathcal {J}}X_j, (\cup _{j\in \mathcal {J}}X_j)^\prime \cap (\cap _{j\in \mathcal {J}}Y_j))\sqsubseteq (A,B)\\&\quad \quad (\mathrm {definition}\,\mathrm {of}\,\sqsubseteq )\\&\Rightarrow \underline{R}(\cup _{j\in \mathcal {J}}X_j, \cap _{j\in \mathcal {J}}Y_j)\sqsubseteq (A,B)\\&\quad \,\,\, (\mathrm {Lemma}\,2(1) )\\&\Rightarrow (\cup _{j\in \mathcal {J}}X_j,\cap _{j\in \mathcal {J}}Y_j)\in \underline{ \mathcal {T}_{(A,B)}}(R)\\&\quad (\mathrm {definition} \;\mathrm {of} \;\underline{\mathcal {T}_{(A,B)}}(R)) \end{aligned}$$

Using the definition of $\sqsubseteq$, we may easily obtain “$X_j\subseteq \cup _{j\in \mathcal {J}}X_j$ and $Y_j\supseteq \cap _{j\in \mathcal {J}}Y_j$”$\Rightarrow ``(X_j,Y_j)\sqsubseteq (\cup _{j\in \mathcal {J}}X_j,\cap _{j\in \mathcal {J}}Y_j)$” for any $j\in \mathcal {J}$. Considered this result with the result in Part I, we find $(\cup _{j\in \mathcal {J}}X_j,\cap _{j\in \mathcal {J}}Y_j)$ to be the maximum element in $(\underline{\mathcal {T}_{(A,B)}}(R),\sqsubseteq )$.

We can obtain the following expression Part II.

$$\begin{aligned} \mathrm {Part}\,\mathrm {II}.\,&(X_p,Y_p)\in \overline{\mathcal {T}_{(A,B)}}(R)\,(p\in \mathcal {P})&\\&\quad \Rightarrow \overline{R}(X_p,Y_p)\sqsubseteq (A,B)\,(p\in \mathcal {P})\\&\quad \,\,\, (\mathrm {definition}\,\mathrm {of}\,\overline{\mathcal {T}_{(A,B)}}(R))\\&\quad \Rightarrow (X_p,X_p^\prime \cup Y_p)\sqsubseteq (A,B)\,(p\in \mathcal {P})\\&\quad (\mathrm {Lemma}\,2(1)) \\&\Rightarrow X_p\subseteq A\,(p\in \mathcal {P})\quad \quad \quad \,\,\\&\quad \,\, (\mathrm {definition}\,\mathrm {of}\,\sqsubseteq )\\&\Rightarrow \cup _{p\in \mathcal {P}}X_p\subseteq A\\&\Rightarrow (\cup _{p\in \mathcal {P}}X_p)^\prime \cup (\cap _{p\in \mathcal {P}}Y_p)\\&\quad \supseteq (\cup _{p\in \mathcal {P}}X_p)^\prime \supseteq A^\prime \supseteq B \\&\quad ( \mathrm {Lemma}\,1(1), \mathrm {and} \; A^\prime \supseteq B )\\&\Rightarrow (\cup _{p\in \mathcal {P}}X_p,(\cup _{p\in \mathcal {P}}X_p)^\prime \cup (\cap _{p\in \mathcal {P}}\\&\quad Y_p))\sqsubseteq (A,B)\quad \quad (\mathrm {definition}\,\mathrm {of}\,\sqsubseteq )\\&\Rightarrow \overline{R}(\cup _{p\in \mathcal {P}} X_p,\cap _{p\in \mathcal {P}}Y_p)\sqsubseteq (A,B)\\&\quad \quad \,\, \,(\mathrm {Lemma}\,2(1))\\&\Rightarrow (\cup _{p\in \mathcal {P}}X_p,\cap _{p\in \mathcal {P}}Y_p)\in \overline{\mathcal {T}_{(A,B)}}(R)\\&\quad (\mathrm {definition}\; \mathrm {of} \;\overline{\mathcal {T}_{(A,B)}}(R)) \end{aligned}$$

Using the definition of $\sqsubseteq$, we may easily obtain “$X_p\subseteq \cup _{p\in \mathcal {P}}X_p$ and $Y_p\supseteq \cap _{p\in \mathcal {P}}Y_p$”$\Rightarrow ``(X_p,Y_p)\sqsubseteq (\cup _{p\in \mathcal {P}}X_p,\cap _{p\in \mathcal {P}}Y_p)$” for any $p\in \mathcal {P}$. Combining the result of Part II with the above result, we find $(\cup _{p\in \mathcal {P}}X_p,\cap _{p\in \mathcal {P}}Y_p)$ to be the maximum element in $(\overline{\mathcal {T}_{(A,B)}}(R),\sqsubseteq )$.

To prove item (4).

We will use two parts to finish the proof.

$$\begin{aligned} \mathrm {Part}\,1.\,&(X,Y)\in \underline{\mathcal {T}_{(A_1,B_1)}}(R) \cap \underline{\mathcal {T}_{(A_2,B_2)}}(R)\\&\quad \Leftrightarrow (X,Y)\in \underline{\mathcal {T}_{(A_i,B_i)}}(R)\, (i=1,2)\quad \quad \\&\quad \Leftrightarrow \underline{R}(X,Y)\sqsubseteq (A_i,B_i)\,(i=1,2)\\&\quad \quad (\mathrm {definition}\,\mathrm {of}\,\underline{\mathcal {T}_{(A_i,B_i)}}(R))\\&\quad \Leftrightarrow X\subseteq A_i\,\mathrm {and}\,X^\prime \cap Y\supseteq B_i\,(i=1,2)\,\\&\quad (\mathrm {Lemma}\,2(1),\mathrm {definition}\,\mathrm {of}\,\sqsubseteq )\\&\quad \Leftrightarrow X\subseteq A_1\cap A_2 \,\mathrm {and}\, X^\prime \cap Y\supseteq B_1\cup B_2\\&\quad \Leftrightarrow \underline{R}(X,Y)\sqsubseteq (A_1\cap A_2,B_1\cup B_2)\,\,\\&\quad (\mathrm {Lemma}\, 2(1))\\&\quad \Leftrightarrow (X,Y)\in \underline{\mathcal {T}_{(A_1\cap A_2,B_1\cup B_2)}}(R)\,\,\,\,\,\\&\quad (\mathrm {definition} \;\mathrm {of}\; \underline{\mathcal {T}_{(A_1\cap A_2,B_1\cup B_2)}}(R)) \\ \mathrm {Part}\,2.\,&(X,Y)\in \overline{\mathcal {T}_{(A_1,B_1)}}(R)\cap \overline{\mathcal {T}_{(A_2,B_2)}}(R)\\&\quad \Leftrightarrow \overline{R}(X,Y)\sqsubseteq (A_i,B_i)\,(i=1,2)\,\\&\quad (\mathrm {definition}\,\mathrm {of}\,\overline{\mathcal {T}_{(A_i,B_i)}}(R))\\&\quad \Leftrightarrow X\subseteq A_i\,\mathrm {and}\, X^\prime \cup Y\supseteq B_i\,(i=1,2)\, \\&\quad (\mathrm {Lemma}\,2(1) ,\mathrm {definition}\, \mathrm {of} \,\sqsubseteq )\\&\quad \Leftrightarrow X\subseteq A_1\cap A_2\,\mathrm {and}\,X^\prime \cup Y\supseteq B_1\cup B_2 \\&\quad \Leftrightarrow \overline{R}(X,Y)\sqsubseteq (A_1\cap A_2, B_1\cup B_2)\,\\&\quad (\mathrm {Lemma} \,2(1),\mathrm {definition}\, \mathrm {of} \,\sqsubseteq )\\&\quad \Leftrightarrow (X,Y)\in \overline{\mathcal {T}_{(A_1\cap A_2,B_1\cup B_2)}}(R)\\&\quad \,(\mathrm {definition}\,\mathrm {of}\; \overline{\mathcal {T}_{(A_1\cap A_2,B_1\cup B_2)}}(R)) \end{aligned}$$