Quintuple Implication Principle on interval-valued intuitionistic fuzzy sets

Abstract

This paper mainly aims to introduce Quintuple Implication Principle (QIP) on interval-valued intuitionistic fuzzy sets (IVIFSs). Firstly, some algebraic properties of a class of interval-valued intuitionistic triangular norms are discussed in detail. In particular, a unified expression of residual interval-valued intuitionistic fuzzy implications generated by left-continuous triangular norms is presented. Secondly, Triple Implication Principles (TIPs) of both interval-valued intuitionistic fuzzy modus ponens (IVIFMP) and fuzzy modus tollens (IVIFMT) based on residual interval-valued intuitionistic fuzzy implications are analyzed. It is shown that the TIP solution of IVIFMP is recoverable, and the TIP solution of IVIFMT is only weakly local recoverable. Moreover, it sees by an illustrated example that the TIP method sometimes makes the computed solutions for IVIFMP and IVIFMT meaningless or misleading. To avoid the above shortcoming and enhance the recovery property of TIP solution of IVIFMT, QIP and $\alpha$-QIP for IVIFMP and IVIFMT are investigated and the corresponding expressions of solutions of them are also given, respectively. In addition, the QIP methods for IVIFMP and IVIFMT are recoverable and sound. Finally, QIP solutions of IVIFMP for multiple fuzzy rules are provided. An application example for medical diagnosis is given to illustrate the feasibility and effectiveness of the QIP of IVIFMP.

Appendices

Appendix 1

The proof of Proposition 7 is shown as follows:

• (1) Since $(\otimes _{SL^*},\rightarrow _{SL^*})$ is an interval-valued intuitionistic adjoint pair, then $\gamma \rightarrow _{SL^*} \eta =\mathbf{1 }^*$ if and only if $\mathbf{1 }^*\otimes _{SL^*}\gamma \le \eta$ if and only if $\gamma \le \eta$.

• (2) $\eta \le \gamma \rightarrow _{SL^*}\beta$ if and only if $\eta \otimes _{SL^*}\gamma \le \beta$ if and only if $\gamma \otimes _{SL^*}\eta \le \beta$ if and only if $\gamma \le \eta \rightarrow _{SL^*} \beta$.

• (3) $\forall \alpha \in SL^*$, $\alpha \le \gamma \rightarrow _{SL^*}(\eta \rightarrow _{SL^*}\beta )$ if and only if $\alpha \otimes _{SL^*}\gamma \le \eta \rightarrow _{SL^*}\beta$ if and only if $\alpha \otimes _{SL^*}\gamma \otimes _{SL^*}\eta \le \beta$ if and only if $\alpha \otimes _{SL^*}\eta \le \gamma \rightarrow _{SL^*}\beta$ if and only if $\alpha \le \eta \rightarrow _{SL^*} (\gamma \rightarrow _{SL^*}\beta )$, that is, $\gamma \rightarrow _{SL^*}(\eta \rightarrow _{SL^*}\beta )=\eta \rightarrow _{SL^*}(\gamma \rightarrow _{SL^*}\beta )$.

• (4) $\mathbf{1 }^*\rightarrow _{SL^*}\eta =\vee \{\gamma \in SL^*|\gamma \otimes _{SL^*}\mathbf{1 }^*\le \eta \}=\eta$.

• (5) $\forall \alpha \in SL^*$, $\alpha \le \gamma \rightarrow _{SL^*}(\wedge _{i\in I}\beta _i)$ if and only if $\alpha \otimes _{SL^*}\gamma \le \wedge _{i\in I}\beta _i$ if and only if $\forall i\in I$, $\alpha \otimes _{SL^*}\gamma \le \beta _i$ if and only if $\forall i\in I$, $\alpha \le \gamma \rightarrow _{SL^*}\beta _i$ if and only if $\alpha \le \wedge _{i\in I}(\gamma \rightarrow _{SL^*}\beta _i)$, that is, $\gamma \rightarrow _{SL^*}(\wedge _{i\in I}\beta _i)=\wedge _{i\in I}(\gamma \rightarrow _{SL^*}\beta _i)$.

• (6) Since $\otimes _{SL^*}$ is left continuous, then $\forall \alpha \in SL^*$, $\alpha \le (\vee _{i\in I}\gamma _i)\rightarrow _{SL^*} \beta$ if and only if $\alpha \otimes _{SL^*}(\vee _{i\in I}\gamma _i)\le \beta$ if and only if $\vee _{i\in I}(\alpha \otimes _{SL^*}\gamma _i)\le \beta$ if and only if $\forall i\in I$,$\alpha \otimes _{SL^*}\gamma _i\le \beta$ if and only if $\forall i\in I$,$\alpha \le \gamma _i \rightarrow _{SL^*}\beta$ if and only if $\alpha \le \wedge _{i\in I}(\gamma _i\rightarrow _{SL^*}\beta )$, that is, $(\vee _{i\in I}\gamma _i)\rightarrow _{SL^*} \beta =\wedge _{i\in I}(\gamma _i\rightarrow _{SL^*}\beta )$.

• (7) Let $\forall \alpha _1, \alpha _2\in SL^*$ and $\alpha _1\le \alpha _2$, then $\alpha _1\wedge \alpha _2=\alpha _1$, it follows from Proposition 7 (5) that $\beta \rightarrow _{SL^*}\alpha _1=\beta \rightarrow _{SL^*}(\alpha _1\wedge \alpha _2)=(\beta \rightarrow _{SL^*}\alpha _1)\wedge (\beta \rightarrow _{SL^*}\alpha _2)\le \beta \rightarrow _{SL^*}\alpha _2$, that is, $\rightarrow _{SL^*}$ is isotone in the second variable. Similarly, it follows from Proposition 7 (6) that $\rightarrow _{SL^*}$ is antitone in the first variable.

Appendix 2

The proof of Theorem 2 is shown as follows:

Let $\gamma =([a,b],[c,d])=\alpha \rightarrow _{SL^*}\beta$,

$\eta _i=([e_i,f_i],[h_i,k_i])\in SL^*$. From Theorem 1, it follows that

$$\begin{aligned}&\eta =([a,b],[c,d])=\alpha \rightarrow _{SL^*}\beta \\&\quad =\bigvee \{\eta _i\in SL^*\mid \eta _i\otimes _{SL^*} \alpha \le \beta \}\\&\quad =\bigvee \{([e_i,f_i],[h_i,k_i])\mid e_i\otimes a_1\le a_2,\\&\qquad f_i\otimes b_1\le b_2, c_2\le h_i\oplus c_1, d_2\le k_i\oplus d_1, \\&\qquad f_i+h_i\le 1, f_i+k_i\le 1\}\\&\quad =(\vee [e_i,f_i],\wedge [h_i,k_i]). \end{aligned}$$

We firstly compute the second element [c, d] of $\gamma$, it could be given by the following form:

$[c,d]=\bigwedge \{[h_i,k_i]\mid c_2\le h_i\oplus c_1, d_2\le k_i\oplus d_1\}=\bigwedge \{[h_i,k_i]\mid [c_2,d_2]\le [h_i,k_i]\oplus _{SI}[c_1,d_1] \}=[c_2\ominus c_1,(c_2\ominus c_1)\vee (d_2\ominus d_1)]$.

In fact, suppose that $M=\{[h_i,k_i]\mid c_2\le h_i\oplus c_1, d_2\le k_i\oplus d_1\}$, if $\forall [h_i,k_i]\in M$, then $c_2\le h_i\oplus c_1$ and $d_2\le k_i\oplus d_1$, therefore $c_2\ominus c_1\le h_i\le k_i$ and $d_2\ominus d_1\le k_i$, hence $[c_2\ominus c_1,(c_2\ominus c_1)\vee (d_2\ominus d_1)]\le [h_i,k_i]$.

Therefore,

Secondly,

$[c_2\ominus c_1,(c_2\ominus c_1)\vee (d_2\ominus d_1)]\oplus _{SI} [c_1,d_1] =[(c_2\ominus c_1)\oplus c_1,((c_2\ominus c_1)\vee (d_2\ominus d_1))\oplus d_1] \ge [(c_2\ominus c_1)\oplus c_1,(d_2\ominus d_1)\oplus d_1]\ge [c_2,d_2]$, then

Finally, from (I) and (II),

For the first argument [a, b] of $\gamma$, $[a,b]=\bigvee \{[e_i,f_i]\mid e_i\otimes a_1\le a_2, f_i\otimes b_1\le b_2, c_2\le h_i\oplus c_1, d_2\le k_i\oplus d_1, f_i\le 1-h_i, f_i\le 1-k_i\}$, so we can obtain that

Similarly, we have

From (III), (IV) and (V), it follows that

Moreover,

$$\begin{aligned}&([(a_1\rightarrow a_2)\wedge (b_1\rightarrow b_2)\wedge (1-c_2\ominus c_1)\\&\qquad \wedge (1-d_2\ominus d_1),(b_1\rightarrow b_2)\wedge (1-c_2\ominus c_1)\\&\qquad \wedge (1-d_2\ominus d_1)],[c_2\ominus c_1,(c_2\ominus c_1)\\&\qquad \vee (d_2\ominus d_1)])\otimes _{SL^*}\alpha \\&\quad =([(a_1\rightarrow a_2)\wedge (b_1\rightarrow b_2)\wedge (1-c_2\ominus c_1)\\&\qquad \wedge (1-d_2\ominus d_1),(b_1\rightarrow b_2)\wedge (1-c_2\ominus c_1)\\&\qquad \wedge (1-d_2\ominus d_1)],[c_2\ominus c_1,(c_2\ominus c_1)\\&\qquad \vee (d_2\ominus d_1)])\otimes _{SL^*}([a_1,b_1],[c_1,d_1])\\&\quad =([((a_1\rightarrow a_2)\wedge (b_1\rightarrow b_2)\wedge (1-c_2\ominus c_1)\\&\qquad \wedge (1-d_2\ominus d_1))\otimes a_1,((b_1\rightarrow b_2)\wedge (1-c_2\ominus c_1)\\&\qquad \wedge (1-d_2\ominus d_1))\otimes b_1],[(c_2\ominus c_1)\oplus c_1,((c_2\ominus c_1)\\&\qquad \vee (d_2\ominus d_1))\oplus d_1])\\&\quad \le ([(a_1\rightarrow a_2)\otimes a_1,(b_1\rightarrow b_2)\otimes b_1], \\&\qquad [(c_2\ominus c_1)\oplus c_1,(d_2\ominus d_1)\oplus d_1])\\&\quad \le ([a_2,b_2],[c_2,d_2])\\&\quad =\beta . \end{aligned}$$

From Eq. (8),

Hence, this completes the proof from (VI) and (VII).

Appendix 3

Example 6

Let $\otimes =\otimes _G$. Then,

$$\begin{aligned} a_1\rightarrow _G a_2= & {} \left\{ \begin{array}{ll} 1, &{} a_1\le a_2;\\ a_2, &{} a_1>a_2.\\ \end{array} \right. \\ b_1\rightarrow _G b_2= & {} \left\{ \begin{array}{ll} 1, &{} b_1\le b_2;\\ b_2, &{} b_1>b_2.\\ \end{array} \right. \\ c_2\ominus _G c_1= & {} \left\{ \begin{array}{ll} 0, &{} c_2\le c_1;\\ c_2, &{} c_2>c_1.\\ \end{array} \right. \\ 1-c_2\ominus _G c_1= & {} \left\{ \begin{array}{ll} 1, &{} c_2\le c_1;\\ 1-c_2, &{} c_2>c_1.\\ \end{array} \right. \\ d_2\ominus _G d_1= & {} \left\{ \begin{array}{ll} 0, &{} d_2\le d_1;\\ d_2, &{} d_2>d_1.\\ \end{array} \right. \\ 1-d_2\ominus _G d_1= & {} \left\{ \begin{array}{ll} 1, &{} d_2\le d_1;\\ 1-d_2, &{} d_2>d_1.\\ \end{array} \right. \end{aligned}$$

Therefore, the result of implication $\alpha \rightarrow _{SL^*}\beta$ could be computed and obtained as follows:

If $a_1\le a_2, b_1\le b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([1,1],[0,0])$;

If $a_1\le a_2, b_1\le b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([1-d_2,1-d_2],[0,d_2])$;

If $a_1\le a_2, b_1\le b_2, c_2> c_1, d_2\le d_1$, $\alpha \rightarrow _{SL^*}\beta =([1-c_2,1-c_2],[c_2,c_2])$;

If $a_1\le a_2, b_1\le b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([1-d_2,1-d_2],[c_2,d_2])$;

If $a_1\le a_2, b_1> b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([b_2,b_2],[0,0])$;

If $a_1\le a_2, b_1> b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([b_2,b_2],[0,d_2])$;

If $a_1\le a_2, b_1> b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([b_2,b_2],[c_2,c_2])$;

If $a_1\le a_2, b_1> b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([b_2,b_2],[c_2,d_2])$ ;

If $a_1> a_2, b_1\le b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([a_2,1],[0,0])$;

If $a_1> a_2, b_1\le b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([a_2,1-d_2],[0,d_2])$;

If $a_1> a_2, b_1\le b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([a_2,1-c_2],[c_2,c_2])$;

If $a_1> a_2, b_1\le b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([a_2,1-d_2],[c_2,d_2])$;

If $a_1> a_2, b_1> b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([a_2,b_2],[0,0])$;

If $a_1> a_2, b_1> b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([a_2,b_2],[0,d_2])$;

If $a_1> a_2, b_1> b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([a_2,b_2],[c_2,c_2])$;

If $a_1> a_2, b_1> b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([a_2,b_2],[c_2,d_2])$.

Example 7

Let $\otimes =\otimes _{Lu}$. Then,

$$\begin{aligned}&a_1\rightarrow _{Lu} a_2=(1-a_1+a_2)\wedge 1, \\&b_1\rightarrow _{Lu} b_2=(1-b_1+b_2)\wedge 1;\\&c_2\ominus _{Lu} c_1=(c_2-c_1)\vee 0, \\&1-c_2\ominus _{Lu} c_1=(1-c_2+c_1)\wedge 1;\\&d_2\ominus _{Lu} d_1=(d_2-d_1)\vee 0, \\&1-d_2\ominus _{Lu} d_1=(1-d_2+d_1)\wedge 1; \end{aligned}$$

Thus, $a=(1-a_1+a_2)\wedge (1-b_1+b_2)\wedge (1-c_2+c_1)\wedge (1-d_2+d_1)\wedge 1$,

$$\begin{aligned}&b=(1-b_1+b_2)\wedge (1-c_2+c_1)\wedge (1-d_2+d_1)\wedge 1,\\&c=(c_2-c_1)\vee 0,\\&d=(c_2-c_1)\vee (d_2-d_1)\vee 0. \end{aligned}$$

Therefore,

$$\begin{aligned}&\alpha \rightarrow _{SL^*}\beta =([(1-a_1+a_2)\wedge (1-b_1+b_2)\\&\quad \wedge (1-c_2+c_1)\wedge (1-d_2+d_1)\wedge 1,(1-b_1+b_2)\\&\quad \wedge (1-c_2+c_1)\wedge (1-d_2+d_1)\wedge 1],[(c_2-c_1)\\&\quad \vee 0,(c_2-c_1)\vee (d_2-d_1)\vee 0]). \end{aligned}$$

Example 8

Let $\otimes =\otimes _\pi$. Then,

$$\begin{aligned} a_1\rightarrow _\pi a_2= & {} \left\{ \begin{array}{ll} 1, &{} a_1\le a_2;\\ \frac{a_2}{a_1}, &{} a_1>a_2.\\ \end{array} \right. \\ b_1\rightarrow _\pi b_2= & {} \left\{ \begin{array}{ll} 1, &{} b_1\le b_2;\\ \frac{b_2}{b_1}, &{} b_1>b_2.\\ \end{array} \right. \\ c_2\ominus _\pi c_1= & {} \left\{ \begin{array}{ll} 0, &{} c_2\le c_1;\\ \frac{c_2-c_1}{1-c_1}, &{} c_2>c_1.\\ \end{array} \right. \\ 1-c_2\ominus _\pi c_1= & {} \left\{ \begin{array}{ll} 1, &{} c_2\le c_1;\\ \frac{1-c_2}{1-c_1}, &{} c_2>c_1.\\ \end{array} \right. \\ d_2\ominus _\pi d_1= & {} \left\{ \begin{array}{ll} 0, &{} d_2\le d_1;\\ \frac{d_2-d_1}{1-d_1}, &{} d_2>d_1.\\ \end{array} \right. \\ 1-d_2\ominus _\pi d_1= & {} \left\{ \begin{array}{ll} 1, &{} d_2\le d_1;\\ \frac{1-d_2}{1-d_1}, &{} d_2>d_1.\\ \end{array} \right. \end{aligned}$$

Therefore, the result of implication $\alpha \rightarrow _{SL^*}\beta$ could be computed and obtained as follows:

If $a_1\le a_2, b_1\le b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([1,1],[0,0])$;

$If a_1\le a_2, b_1\le b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{1-d_2}{1-d_1}, \frac{1-d_2}{1-d_1}],[0,\frac{d_2-d_1}{1-d_1}])$;

If $a_1\le a_2, b_1\le b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{1-c_2}{1-c_1}, \frac{1-c_2}{1-c_1}],[\frac{c_2-c_1}{c_2-c_1},\frac{c_2-c_1}{c_2-c_1}])$;

If $a_1\le a_2, b_1\le b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{1-c_2}{1-c_1} \wedge \frac{1-d_2}{1-d_1},\frac{1-c_2}{1-c_1}\wedge \frac{1-d_2}{1-d_1}], [\frac{c_2-c_1}{c_2-c_1},\frac{c_2-c_1}{c_2-c_1} \vee \frac{d_2-d_1}{1-d_1}])$;

If $a_1\le a_2, b_1> b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{b_2}{b_1},\frac{b_2}{b_1}],[0,0])$;

If $a_1\le a_2, b_1> b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{b_2}{b_1}\wedge \frac{1-d_2}{1-d_1},\frac{b_2}{b_1}\wedge \frac{1-d_2}{1-d_1}],[0,\frac{d_2-d_1}{1-d_1}])$;

If $a_1\le a_2, b_1> b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{b_2}{b_1}\wedge \frac{1-c_2}{1-c_1},\frac{b_2}{b_1}\wedge \frac{1-c_2}{1-c_1}], {[\frac{c_2-c_1}{1-c_1},\frac{c_2-c_1}{1-c_1}]})$;

If $a_1\le a_2, b_1> b_2, c_2> c_1, d_2> d_1$, $\alpha \rightarrow _{SL^*}\beta =([\frac{b_2}{b_1}\wedge \frac{1-c_2}{1-c_1}\wedge \frac{1-d_2}{1-d_1},\frac{b_2}{b_1} \wedge \frac{1-c_2}{1-c_1} \wedge \frac{1-d_2}{1-d_1}],[\frac{c_2-c_1}{1-c_1}, \frac{c_2-c_1}{1-c_1}\vee \frac{d_2-d_1}{1-d_1}])$;

If $a_1> a_2, b_1\le b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{a_2}{a_1},1],[0,0])$;

If $a_1> a_2, b_1\le b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{a_2}{a_1}\wedge \frac{1-d_2}{1-d_1},\frac{1-d_2}{1-d_1}], {[0,\frac{d_2-d_1}{1-d_1}]})$;

If $a_1> a_2, b_1\le b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{a_2}{a_1}\wedge \frac{1-c_2}{1-c_1},\frac{1-c_2}{1-c_1}],[\frac{c_2-c_1}{1-c_1}, \frac{c_2-c_1}{1-c_1}])$;

If $a_1> a_2, b_1\le b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{a_2}{a_1}\wedge \frac{1-c_2}{1-c_1}\wedge \frac{1-d_2}{1-d_1},\frac{1-c_2}{1-c_1} \wedge \frac{1-d_2}{1-d_1}], [\frac{c_2-c_1}{1-c_1},\frac{c_2-c_1}{1-c_1}\vee \frac{d_2-d_1}{1-d_1}])$;

If $a_1> a_2, b_1> b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{a_2}{a_1}\wedge \frac{b_2}{b_1},\frac{b_2}{b_1}],[0,0])$;

If $a_1> a_2, b_1> b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{a_2}{a_1}\wedge \frac{b_2}{b_1}\wedge \frac{1-d_2}{1-d_1},\frac{b_2}{b_1} \wedge \frac{1-d_2}{1-d_1}], [0,\frac{d_2-d_1}{1-d_1}])$;

If $a_1> a_2, b_1> b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{a_2}{a_1}\wedge \frac{b_2}{b_1}\wedge \frac{1-c_2}{1-c_1},\frac{b_2}{b_1}\wedge \frac{1-c_2}{1-c_1}], [\frac{c_2-c_1}{1-c_1},\frac{c_2-c_1}{1-c_1}])$;

If $a_1> a_2, b_1> b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([\frac{a_2}{a_1}\wedge \frac{b_2}{b_1}\wedge \frac{1-c_2}{1-c_1}\wedge \frac{1-d_2}{1-d_1},\frac{b_2}{b_1}\wedge \frac{1-c_2}{1-c_1} \wedge \frac{1-d_2}{1-d_1}],[\frac{c_2-c_1}{1-c_1}, \frac{c_2-c_1}{1-c_1}\vee \frac{d_2-d_1}{1-d_1}])$.

Example 9

Let $\otimes =\otimes _0$. Then,

$$\begin{aligned} a_1\rightarrow _0 a_2= & {} \left\{ \begin{array}{ll} 1, &{} a_1\le a_2;\\ (1-a_1)\vee a_2, &{} a_1>a_2.\\ \end{array} \right. \\ b_1\rightarrow _0 b_2= & {} \left\{ \begin{array}{ll} 1, &{} b_1\le b_2;\\ (1-b_1)\vee b_2, &{} b_1>b_2.\\ \end{array} \right. \\ c_2\ominus _0 c_1= & {} \left\{ \begin{array}{ll} 0, &{} c_2\le c_1;\\ c_2\wedge (1-c_1), &{} c_2>c_1.\\ \end{array} \right. \\ 1-c_2\ominus _0 c_1= & {} \left\{ \begin{array}{ll} 1, &{} c_2\le c_1;\\ c_1\vee (1-c_2), &{} c_2>c_1.\\ \end{array} \right. \\ d_2\ominus _0 d_1= & {} \left\{ \begin{array}{ll} 0, &{} d_2\le d_1;\\ d_2\wedge (1-d_1), &{} d_2>d_1.\\ \end{array} \right. \\ 1-d_2\ominus _0 d_1= & {} \left\{ \begin{array}{ll} 1, &{} d_2\le d_1;\\ d_1\vee (1-d_2), &{} d_2>d_1.\\ \end{array} \right. \end{aligned}$$

Therefore, the result of implication $\alpha \rightarrow _{SL^*}\beta$ could be computed and obtained as follows:

If $a_1\le a_2, b_1\le b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([1,1],[0,0])$;

If $a_1\le a_2, b_1\le b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([d_1\vee (1-d_2),d_1 \vee (1-d_2)],[0,d_2\wedge (1-d_1)])$;

If $a_1\le a_2, b_1\le b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([(c_1\vee (1-c_2)),(c_1\vee (1-c_2))], [c_2\wedge (1-c_1),c_2\wedge (1-c_1)])$;

If $a_1\le a_2, b_1\le b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([(c_1\vee (1-c_2))\wedge (d_1 \vee (1-d_2)),(c_1\vee (1-c_2))\wedge (d_1\vee (1-d_2))], [c_2\wedge (1-c_1),(c_2\wedge (1-c_1))\vee (d_2\wedge (1-d_1))])$;

If $a_1\le a_2, b_1> b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([(1-b_1)\vee b_2,(1-b_1)\vee b_2],[0,0])$,

If $a_1\le a_2, b_1> b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([((1-b_1)\vee b_2)\wedge (d_1\vee (1-d_2)), ((1-b_1)\vee b_2)\wedge (d_1\vee (1-d_2))],[0,d_2\wedge (1-d_1)])$;

If $a_1\le a_2, b_1> b_2, c_2> c_1, d_2\le d_1$, $\alpha \rightarrow _{SL^*}\beta =([((1-b_1)\vee b_2)\wedge (c_1\vee (1-c_2)),((1-b_1)\vee b_2)\wedge (c_1\vee (1-c_2))],[c_2\wedge (1-c_1),c_2\wedge (1-c_1)])$;

If $a_1\le a_2, b_1> b_2, c_2> c_1, d_2> d_1$, $\alpha \rightarrow _{SL^*}\beta =([((1-b_1)\vee b_2)\wedge (c_1\vee (1-c_2)) \wedge (d_1\vee (1-d_2)),((1-b_1)\vee b_2)\wedge (c_1\vee (1-c_2))\wedge (d_1\vee (1-d_2))],[c_2 \wedge (1-c_1),(c_2\wedge (1-c_1)) \vee (d_2\wedge (1-d_1))])$;

If $a_1> a_2, b_1\le b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([(1-a_1)\vee a_2,1],[0,0])$;

If $a_1> a_2, b_1\le b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([((1-a_1)\vee a_2)\wedge (d_1\vee (1-d_2)), d_1\vee (1-d_2)],[0,d_2\wedge (1-d_1)])$;

If $a_1> a_2, b_1\le b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([((1-a_1)\vee a_2)\wedge (c_1\vee (1-c_2)), c_1\vee (1-c_2)],[c_2\wedge (1-c_1),c_2\wedge (1-c_1)])$;

If $a_1> a_2, b_1\le b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([((1-a_1)\vee a_2)\wedge (c_1\vee (1-c_2)) \wedge (d_1\vee (1-d_2)),(c_1\vee (1-c_2))\wedge (d_1 \vee (1-d_2))],[c_2\wedge (1-c_1), (c_2\wedge (1-c_1))\vee (d_2\wedge (1-d_1))])$;

If $a_1> a_2, b_1> b_2, c_2\le c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([((1-a_1)\vee a_2)\wedge ((1-b_1)\vee b_2), (1-b_1)\vee b_2],[0,0])$;

If $a_1> a_2, b_1> b_2, c_2\le c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([((1-a_1)\vee a_2)\wedge ((1-b_1)\vee b_2) \wedge (d_1\vee (1-d_2)),((1-b_1)\vee b_2)\wedge (d_1\vee (1-d_2))],[0,d_2\wedge (1-d_1)])$;

If $a_1> a_2, b_1> b_2, c_2> c_1, d_2\le d_1$, then $\alpha \rightarrow _{SL^*}\beta =([((1-a_1)\vee a_2)\wedge ((1-b_1)\vee b_2)\wedge (c_1\vee (1-c_2)),((1-b_1)\vee b_2)\wedge (c_1\vee (1-c_2))],[c_2\wedge (1-c_1),c_2\wedge (1-c_1)])$;

If $a_1> a_2, b_1> b_2, c_2> c_1, d_2> d_1$, then $\alpha \rightarrow _{SL^*}\beta =([((1-a_1)\vee a_2)\wedge ((1-b_1)\vee b_2) \wedge (c_1\vee (1-c_2))\wedge (d_1\vee (1-d_2)),((1-b_1)\vee b_2)\wedge (c_1\vee (1-c_2))\wedge (d_1\vee (1-d_2))], [c_2\wedge (1-c_1),(c_2\wedge (1-c_1))\vee (d_2\wedge (1-d_1))])$.