Constructive methods for intuitionistic fuzzy implication operators

Abstract

In this paper, two novel constructive methods for fuzzy implication operators are proposed. And they can be used to resolve the questions about how to construct intuitionistic fuzzy implication operators to inference sentence “If x is A, then y is B,” where A and B are intuitionistic fuzzy sets over X and Y, respectively. In the first method, cut sets and representation theorem of intuitionistic fuzzy sets are utilized through triple valued fuzzy implications as well as fuzzy relations to obtain seven intuitionistic fuzzy implication operators. In the second method, the Lebesgue measure of the fuzzy implications with value one and zero is used, and one intuitionistic fuzzy implication operator can be obtained. Finally, properties of those operators are discussed.

Appendix

Property 1

$I_{Li}((a, b),(c, d)) (i=1,...,4)$ satisfy the following:

• (1) $I_{Li}((a, b),(c, d))$ satisfy (A1)–(A6), (P1)–(P4), (P12), (P14), $(i=1,,4)$;

• (2) If $a + d>1$, then $I_{Li}((a, b),(c, d))$ satisfy (P6), $(i=1,3)$;

• (3) $I_{L2}((a, b),(c, d))$ satisfy (P7);

• (4)

  • (a) If $a=c$, then $I_{L1}((a, b),(c, d))$ satisfies (P8);

  • (b) $I_{L2}((a, b),(c, d))$ satisfies (P8);

• (5) $I_{Li}((a, b),(c, d))$ satisfy (P9), $(i=1,3)$;

• (6) $I_{Li}((a, b),(c, d))$ satisfy (P10), $(i=3,4)$;

• (7) $I_{Li}((a, b),(c, d))$ satisfy (P11), $(i=1,2,4)$;

• (8) $I_{Li}((a, b),(c, d))$ satisfy (P15), $(i=2,4)$;

• (9) If $a> (b \wedge 1/2)$, then $I_{Li}((a, b),(c, d))$ satisfy (P16), $(i=1,3)$;

• (10)

  • (a) $I_{Li}((a, b),(c$, d)) satisfy (P17), ($i=1, 2, 4$);

  • (b) If $a \le d, c\le b$, then $I_{L3}((a, b),(c, d))$ satisfies (P17).

Proof

To (A1)–(A4): When $i=1$, and by Eq. (1), it can be easily concluded that (A1)–(A4) are true. The proofs for other cases are similar.

To (A5): If $(a, b) \le (a', b')$, then $a\le a'$, $b \ge b'$. When $i=1$, by Eq. (1), we have the following:

• (i) When $a\le c, b<d$, we have $I_{L1}((a, b)$, $(c, d))=(1-d, 0)$ and $b'\le b<d$, we have

  $$\begin{aligned}&I_{L1} (({a}',{b}'),(c,d))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1-d,0)}&{} {{a}'\le c\;\;\hbox {and}\;{b}'<d} \\ {(c,0)}&{} {{a}'>c\;\;\hbox {and}\;{a}'+d\le 1} \\ {(c,d)}&{} {{a}'>c\;\;\hbox {and}\;{a}'+d>1} \\ \end{array} }} \right. \quad . \end{aligned}$$

  Therefore, $I_{L1}((a', b'), (c, d))\le I_{L1}((a, b), (c, d))$;

• (ii) When $a \le c$, $b \ge d$, we have $I_{L1}((a, b), (c, d))=(1,0)$, then $\forall (a, b) \le (a',b')$, we have $I_{L1}((a', b'), (c, d))\le I_{L1}((a, b)$, (c, d));

• (iii) When $a >c$ and $a\le 1-d$, we can draw that $I_{L1}((a, b), (c, d))=(c,0)$ and

  $$\begin{aligned}&I_{L1} (({a}',{b}'),(c,d))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(c,0)}&{} {{a}'>c\;\;\hbox {and}\;{a}'+d\le 1} \\ {(c,d)}&{} {{a}'>c\;\;\hbox {and}\;{a}'+d>1} \\ \end{array} }} \right. , \end{aligned}$$

  so $I_{L1}((a', b')$, ($c, d))\le I_{L1}((a, b)$, (c, d)).

• (iv) When $a > c$ and $a> 1-d$, we have $I _{L1}((a,b)$, $(c,d))=(c,d)$. If ($a,b) \le (a', b')$, then $a'> c$ and $a'+d >1$, therefore we have

  $$\begin{aligned} I _{L1}((a',b'), (c,d))=I _{L1}((a,b), (c,d)). \end{aligned}$$

In summary, we have if $(a,b) \le (a', b')$, then $I _{L1}((a',b'), (c,d))\le I _{L1}((a,b)$, (c, d)).

The proofs for other cases are similar. $\square$

To (A6): If $(c,d) \le (c', d')$, then $c\le c'$, $d \ge d'$. When $i=1$, by Eq. (1), we can draw that

(i) When $a \le c$ and $b < d$, we have $I _{L1}((a,b), (c,d))=(1-d, 0)$ and $a \le c'$, so

$$\begin{aligned} I_{L1} ((a,b),({c}',{d}'))=\left\{ {{\begin{array}{l@{\quad }l} {(1-{d}',0)}&{} {a\le {c}'\;\hbox {and}\;b<{d}'} \\ {(1,0)}&{} {a\le {c}'\;\hbox {and}\;b\ge {d}'}, \\ \end{array} }} \right. \end{aligned}$$

Therefore, $I_{L1}((a,b)$, $(c', d'))\ge I_{L1}((a, b), (c, d))$;

(ii) When $a \le c$ and $b\ge d$, we have $I_{L1}((a,b)$, $(c,d))=(1, 0)$.

$\forall (c,d) \le (c', d')$, we have $I_{L1}((a,b), (c',d')) = I_{L1}((a,b)$, $(c,d))=(1, 0)$;.

(iii) When $a \le c$ and $a+ d\le 1$, we have $I_{L1}((a,b)$, $(c,d))=(c, 0)$, that is to say

$$\begin{aligned}&I_{L1} ((a,b),({c}',{d}'))=\left\{ {{\begin{array}{l@{\quad }l} {(1-{d}',0)}&{} {a\le {c}'\;\hbox {and}\;b<{d}';} \\ {(1,0)}&{} {a\le {c}'\;\;\hbox {and}\;b\ge {d}';} \\ {({c}',0)}&{} {a>{c}'\;\;\hbox {and}\;a+{d}'\le 1.} \\ \end{array} }} \right. \end{aligned}$$

Therefore, $I_{L1}((a, b)$, $(c', d'))\ge I_{L1}((a, b)$, (c, d));

(iv) When $a > c$ and $a +d >1$, we have $I_{L1}((a, b)$, $(c, d))=(c, d)$ and

$$\begin{aligned} I_{L1} ((a,b),({c}',{d}'))=\left\{ {{\begin{array}{l@{\quad }l} {(1-{d}',0)}&{} {a\le {c}'\;\hbox {and}\;b<{d}';} \\ {(1,0)}&{} {a\le {c}'\;\hbox {and}\;b\ge {d}';} \\ {({c}',0)}&{} {a>{c}'\;\hbox {and}\;a+{d}'\le 1;} \\ {({c}',{d}')}&{} {a>{c}'\;\hbox {and}\;a+{d}'>1.} \\ \end{array} }} \right. \end{aligned}$$

Therefore, $I_{L1}((a, b)$, $(c', d'))\ge (c, d)=I_{L1}((a, b), (c, d))$.

In summary, we have if $(c, d) \le (c', d')$, then $I_{L1}((a, b), (c', d')) \ge I_{L1}((a, b), (c, d))$.

The proofs for other cases are similar.

To (P1): When $i=1$, If $a +b=1$, $c +d=1$, we have

$$\begin{aligned} I_{L1} ((a,b),(c,d))=\left\{ {{\begin{array}{l@{\quad }l} {(1,0)}&{} {a\le c\;\;\hbox {and}\;b\ge d} \\ {(c,d)}&{} {a>c\;\;} \\ \end{array} }\;} \right. , \end{aligned}$$

so

$$\begin{aligned} \pi _{I_{L1} ((a,b),(c,d))} =0. \end{aligned}$$

The proofs for other cases are similar. $\square$

To (P2)–(P3): By Eqs. (1)–(4), it can be easily concluded that (P2)–(P3) are true.

To (P4): When i=1, by Eq. (1), we have

$$\begin{aligned} \pi _{I_{L1} ((a,b),(c,d))} =\left\{ {{\begin{array}{l@{\quad }l} d&{} {a\le c\;\hbox {and}\;b<d;} \\ 0&{} {a\le c\;\hbox {and}\;b\ge d;} \\ {1-c}&{} {a>c\;\hbox {and}\;a+d\le 1;} \\ {1-c-d}&{} {a>c\;\hbox {and}\;a+d>1,} \\ \end{array} }} \right. \end{aligned}$$

Therefore,

$$\begin{aligned} \pi _{I_{L1} ((a,b),(c,d))}\le & {} 1-c\le (1-a)\vee (1-b)\\&\vee \, (1-c)\vee (1-d) \end{aligned}$$

The proofs for other cases are similar.

To (P6): When $i=1$, by Eq. (1), we have

$$\begin{aligned} \pi _{I_{L1} ((a,b),(c,d))} =\left\{ {{\begin{array}{l@{\quad }l} d&{} {a\le c\;\hbox {and}\;b<d} \\ 0&{} {a\le c\;\hbox {and}\;b\ge d} \\ {1-c}&{} {a>c\;\hbox {and}\;a+d\le 1} \\ {1-c-d}&{} {a>c\;\hbox {and}\;a+d>1} \\ \end{array} }} \right. \end{aligned}$$

If $\pi _{(a,b)} =\pi _{(c,d)}$, that is to say $a +b=c +d$. When $a +d> 1$, we have

$$\begin{aligned} \pi _{I_{L1} ((a,b),(c,d))} =1-c-d=\pi _{(c,d)} . \end{aligned}$$

The proofs for other cases are similar.

To (P7): By Eq. (2), we have $I_{L2}((1, 0), (c, d)) =(c, d)$.

To (P8): When $i=2$, by Eq. (2), we have

• (i) If $d>0$, then $I_{L2}((c, d), (e, f)) =(1, 0)$ and we have

  $$\begin{aligned} I_{L2} ((a,b),I_{L2} ((c,d),(e,f)))=(1,0) \end{aligned}$$

• (ii) If $d=0$ and $c +f<1$, then $I_{L2}((c, d)$, $(e, f)) =(1-f, 0)$, so we have

  $$\begin{aligned}&I_{L2} ((a,b),I_{L2} ((c,d),(e,f)))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1,0)}&{} {(b>0)\;\hbox {or}\;(b=0\;\hbox {and}\;a<1)} \\ {(1-f,0)}&{} {b=0\;\;\hbox {and}\;a=1} \\ \end{array} }} \right. \end{aligned}$$

• (iii) If $d=0$ and $c +f \ge 1$ and $c<1$, then $I_{L2}((c, d)$, $(e, f)) =(e, 0)$, so we have

  $$\begin{aligned}&I_{L2} ((a,b),I_{L2} ((c,d),(e,f)))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1,0)}&{} {(b>0)\;\hbox {or}\;(b=0\;\hbox {and}\;a<1)} \\ {(e,0)}&{} {b=0\;\;\hbox {and}\;a=1} \\ \end{array} }} \right. \end{aligned}$$

• (iv) If $d=0$ and $c=1$, then $I_{L2}((c, d)$, $(e, f)) = (e, f)$, so we have

  $$\begin{aligned}&I_{L2} ((a,b),I_{L2} ((c,d),(e,f)))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1,0)}&{} {b>0;}\\ {(1-f,0)}&{} {b=0\;\;\hbox {and}\;a+f<1;}\\ {(e,0)}&{} {b=0\;\;\hbox {and}\;a+f\ge 1\;\hbox {and}\;a<1;} \\ {(e,f)}&{} {b=0\;\;\hbox {and}\;a=1.}\\ \end{array} }} \right. \end{aligned}$$

In summary, we can draw that

$$\begin{aligned}&I_{L2} ((a,b),I_{L2} ((c,d),(e,f)))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1,0)}&{} {\hbox {else};} \\ {(e,0)}&{} {\begin{array}{l} (d=0\;\hbox {and}\;c=1\;\hbox {and}\;b=0\;\hbox {and}\;1>a\ge 1-f) \\ \hbox {or}\;(b=0\;\hbox {and}\;a=1\;\hbox {and}\;d=0\;\hbox {and}\;1-f\le c<1); \\ \end{array}}\\ {(1-f,0)}&{} {\begin{array}{l} (d=0\;\hbox {and}\;c=1\;\hbox {and}\;b=0\;\hbox {and}\;a+f<1) \\ \hbox {or}\;(b=0\;\hbox {and}\;a=1\;\hbox {and}\;d=0\;\;\hbox {and}\;f+c<1); \\ \end{array}} \\ {(e,f)}&{} {b=0\;\hbox {and}\;a=1\;\hbox {and}\;d=0\;\hbox {and}\;c=1.}\\ \end{array} }} \right. \end{aligned}$$

Similarly,

$$\begin{aligned}&I_{L2} ((a,b),I_{L2} ((c,d),(e,f)))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1,0)}&{} {\hbox {else};} \\ {(e,0)}&{} {\begin{array}{l} (d=0\;\hbox {and}\;c=1\;\hbox {and}\;b=0\;\hbox {and}\;1>a\ge 1-f) \\ \hbox {or}\;(b=0\;\hbox {and}\;a=1\;\hbox {and}\;d=0\;\hbox {and}\;1-f\le c<1); \\ \end{array}} \\ {(1-f,0)}&{} {\begin{array}{l} (d=0\;\hbox {and}\;c=1\;\hbox {and}\;b=0\;\hbox {and}\;a+f<1) \\ \hbox {or}\;(b=0\;\hbox {and}\;a=1\;\hbox {and}\;d=0\;\;\hbox {and}\;f+c<1); \\ \end{array}} \\ {(e,f)}&{} {b=0\;\hbox {and}\;a=1\;\hbox {and}\;d=0\;\hbox {and}\;c=1.} \\ \end{array} }} \right. \end{aligned}$$

That is to say

$$\begin{aligned}&I_{L2} ((a,b),I_{L2} ((c,d),(e,f)))\\&\quad =I_{L2} ((c,d),I_{L2} ((a,b),\\&\qquad (e,f)))\,\forall (a,b),(c,d),(e,f)\in L. \end{aligned}$$

The proofs for other cases are similar.

To (P9): When $i=1$, by Eq. (1), we have

$$\begin{aligned}&I_{L1} ((a,b),(c,d))\\&\quad =(1,0)\Leftrightarrow a\le c,b\ge d\Leftrightarrow (a,b)\le (c,d). \end{aligned}$$

The proofs for other cases are similar.

To (P10): When $i=3$, by Eq. (3), we have

$$\begin{aligned} I_{L3} ((a,b),(0,1))=\left\{ {{\begin{array}{l@{\quad }l} {(1,0)}&{} {b=1,a=0} \\ {(b,a)}&{} {b<1,a>0} \\ \end{array} }} \right. , \end{aligned}$$

so $I_{L3}((a, b)$, $(0, 1)) =(b, a)$.

The proofs for other cases are similar.

To (P11): When $i=1$, by Eq. (1), we have $(c, d) \le (c, 0) \le (1-d, 0)\le (1, 0)$. So

$$\begin{aligned} I_{L1}((a, b), (c, d))\ge (c, d). \end{aligned}$$

The proofs for other cases are similar.

To (P12): When $i=1$, by Eq. (1), we have $I_{L1}((a, b)$, $(a, b))=(1, 0)$.

The proofs for other cases are similar.

To (P14): It can be easily concluded.

To (P15): When $i=2$, by Eq. (2), we have $I_{L2}((1, 0), (1, 0))=(0, 1)$.

The proofs for other cases are similar.

To (P16): When $i=1$, by Eq. (1), we have

$$\begin{aligned} I_{L1} ((a,b),(b,a))=\left\{ {{\begin{array}{l@{\quad }l} {(1,0)}&{} {b\ge a;} \\ {(b,0)}&{} {b<a,a\le 0.5;} \\ {(b,a)}&{} {b<a,a>0.5,} \\ \end{array} }} \right. \end{aligned}$$

So if $a> (b\wedge 1/2)$, then $I_{L1}((a, b)$, $(b, a))=(b, a)$.

The proofs for other cases are similar.

To (P17): (a) When $i=1$, by (P13), (P14), (P5), we have $I_{L1}((c, d)$, $(a, b))\ge (a, b)$, and

$$\begin{aligned}&I_{L1} ((a,b),I_{L1} ((c,d),(a,b)))\\&\quad \ge I_{L1} ((a,b),(a,b))=(1,0) \end{aligned}$$

The proofs for other cases are similar.

(b) By Eq. (3), we have

$$\begin{aligned} I_{L3} ((c,d),(a,b))=\left\{ {{\begin{array}{l@{\quad }l} {(d,0)}&{} {d<b\;\hbox {and}\;c+b\le 1}\\ {(1,0)}&{} {d\ge b\;\hbox {and}\;c\le a} \\ {(d,c)}&{} {d<b\;\hbox {and}\;c+b>1}\\ {(1-c,0)}&{} {d\ge b\;\hbox {and}\;c>a}\\ \end{array} }} \right. \end{aligned}$$

• (i) If $d< b$ and $c +b\le 1$ and $a \le d$, then $I_{L3}((a, b)$, $I_{L3}((c, d)$, $(a, b)))= (1, 0)$;

• (ii) If $b \le d$ and $c \le a$, then $I_{L3}((a, b)$, $I_{L3}((c, d), (a, b)))= (1, 0)$;

• (iii) If $d< b$ and $c +b>1$ and $a \le d$, $c \le b$, then $I_{L3}((a, b)$, $I_{L3}((c, d)$, $(a, b)))= (1, 0)$;

• (iv) If $b \le d$ and $c > a$, then $I_{L3}((a, b)$, $I_{L3}((c, d)$, $(a, b)))= (1, 0)$;.

In summary, we can draw that if $a\le d$ and $c\le b$, then

$$\begin{aligned} I_{L3}((a, b), I_{L3}((c, d), (a, b)))= (1, 0). \end{aligned}$$

Property 2

$I_{Li}((a,b)$,$(c, d)) (i=5, 6, 7)$ satisfy the following:

• (1) $I_{Li}((a, b)$,(c, d)) satisfy (A1)–(A4), (P1), (P4)–(P7), (P10), (P15), (P16), ($i=5,6,7)$;

• (2) $I_{Li}((a, b),(c, d))$ satisfy (A6), (P2), $(i=5,7$);

• (3) $I_{Li}((a, b)$, (c, d)) satisfy (A5), (P3), (P11), $(i=6,7)$;

• (4)

  • (a) If $a \vee b=c \vee d$ and $a \wedge b=c \wedge d$, then $I_{Li}((a, b)$, (c, d)) satisfy (P8), (P13), $(i=5,6)$;

  • (b) $I_{L7}((a,b)$, (c, d)) satisfies (P8), (P13);

• (5) $I_{L7}((a, b)$, (c, d)) satisfies (P14).

Proof

To (A1)–(A4): When $i=5$, by Eq. (5), it can be easily concluded that (A1)–(A4) are true. The proofs for other cases are similar.

To (A5): If $(a,b) \le (a', b')$, then $a\le a'$, $b \ge b'$, so $b\vee c \ge b \vee c'$, $a\wedge d \le a' \wedge d$ .

When $i=7$ and by Eq. (7), we have

$$\begin{aligned} I_{L7}((a', b'), (c, d))\le I_{L7}((a, b), (c, d)). \end{aligned}$$

The proof for $i=6$ is similar. $\square$

To (A6): When $i=7$ and by Eq. (7), we have if $(c, d)\le (c', d')$, then $c\le c'$, $d \ge d'$, so $b \vee c \le b \vee c'$, $a\wedge d \ge a \wedge d'$, that is to say

$$\begin{aligned} I_{L7}((a, b), (c, d))\le I_{L7}((a, b), (c', d')). \end{aligned}$$

The proof for $i=5$ is similar.

To (P1): When $i=5$ and if $a +b=1$, $c +d=1$, we can draw that

$$\begin{aligned} I_{L5}((a, b),(c, d))= (b\vee (a \wedge c), a\vee (b \wedge d)). \end{aligned}$$

So $1-[b\vee (a \wedge c)] = a\wedge (b \vee d)$, that is to say

$$\begin{aligned} \pi _{I_{L5} ((a,b),(c,d))} =0. \end{aligned}$$

The proofs for other cases are similar.

To (P2)–(P3): By Eqs. (5)–(7), it can be easily concluded that (P2)–(P3) are true.

To (P4): It can be concluded by Eqs. (5)–(7).

To (P5): When $i=5$, by Eq. (5), we have $I_{L5}((a, b)$, $(a, b)) = (b \wedge a, a \wedge b)$, therefore

$$\begin{aligned} \pi _{I_{L5} ((a,b),(c,d))} =\pi _{(a,b)} \end{aligned}$$

The proofs for other cases are similar.

To (P6): If $\pi _{(a, b)} = \pi _{(c, d)}$, then $a +b=c +d$, when $i=7$ and by Eq. (7), we have

(i) When $a \ge d$, we have $b \le c$, then $I_{L7}((a, b), (c, d)) = (b \vee c, a \wedge d)= (c, d)$, therefore

$$\begin{aligned} \pi _{I_{L7} ((a,b),(c,d))} =\pi _{(a,b)} . \end{aligned}$$

(ii) When $a \le d$, we have $b\ge c$, then $I_{L7}((a, b)$, $(c, d)) = (b \vee c, a \wedge d)= (b, a)$ and

$$\begin{aligned} \pi _{I_{L7} ((a,b),(c,d))} =\pi _{(a,b)} . \end{aligned}$$

In summary, we have if $\pi _{(a, b)} = \pi _{(c, d)}$, then $\pi _{I_{L7} ((a,b),(c,d))}=\pi _{(a,b)}$.

The proofs for other cases are similar.

To (P7): When $i=5$ and by Eq. (5), we have $I_{L5}((1, 0), (c, d)) = (c, d)$

The proofs for other cases are similar.

To (P8): When $i=5$ and by Eq. (5), we have

$$\begin{aligned}&I_{L5} ((c,d),(e,f))=(d\vee (c\wedge e),c\wedge (d\vee f))\\&I_{L5} ((a,b),I_{L5} ((c,d),(e,f)))\\&\quad =(b\vee (a\wedge [d\vee (c\wedge e)]),a\wedge (b\vee [c\wedge (d\vee f)])) \\&\quad =(b\vee (a\wedge d)\vee (a\wedge c\wedge e),a\wedge (b\vee c)\wedge (b\vee d\vee f)) \\&\quad =([b\vee (a\wedge d)]\vee (a\wedge c\wedge e),[a\wedge (b\vee c)]\wedge (b\vee d\vee f)) \\&\quad =([(b\vee a)\wedge (b\vee d)]\vee (a\wedge c\wedge e),\\&\qquad [(a\wedge b)\vee (a\wedge c)]\wedge (b\vee d\vee f)). \end{aligned}$$

On the other hand, we have

$$\begin{aligned}&I_{L5} ((a,b),(e,f))=(b\vee (a\wedge e),a\wedge (b\vee f))\\&I_{L5} ((c,d),I_{L5} ((a,b),(e,f)))\nonumber \\&\quad =(d\vee (c\wedge [b\vee (a\wedge e)]), c\wedge (d\vee [a\wedge (b\vee f)])) \\&\quad =(d\vee (c\wedge b)\vee (c\wedge a\wedge e),c\wedge (d\vee a)\wedge (d\vee b\vee f)) \\&\quad =([d\vee (c\wedge b)]\vee (c\wedge a\wedge e),[c\wedge (d\vee a)]\wedge (d\vee b\vee f)) \\&\quad =([(d\vee c)\wedge (d\vee b)]\vee (c\wedge a\wedge e),[(c\wedge d)\vee (c\wedge a)]\\&\qquad \wedge (d\vee b\vee f)) \end{aligned}$$

In summary, we have if $a \wedge b= c \wedge d$ and $a \wedge b= c \wedge d$, then

$$\begin{aligned}&I_{L5} ((a,b),I_{L5} ((c,d),(e,f)))\\&\quad =I_{L5} ((c,d),I_{L5} ((a,b),(e,f))) \end{aligned}$$

The proofs for other cases are similar.

To (P10): When $i=5$ and by Eq. (5), we have $I_{L5}((a, b), (0, 1)) = (b, a)$.

The proofs for other cases are similar.

To (P11): When $i=6$ and by Eq. (6), we have $I_{L6}((a, b),(c, d))= (c\vee (b \wedge d)$, $d\wedge (a \vee c))$.

For the reason that $c\vee (b \wedge d) \ge c$, $d \wedge (a \vee c) \le d$, we can draw that

$$\begin{aligned} I_{L6}((a, b),(c, d))\ge (c, d). \end{aligned}$$

The proofs for other cases are similar.

To (P13): When i=5 and by Eq. (5), we have if $a \vee b= c \vee d$ and $a \wedge b= c \wedge d$, then

$$\begin{aligned}&I_{L5} ((d,c),(b,a))\\&\quad =(c\vee (d\wedge b),d\wedge (c\vee a))\\&\quad =((c\vee d)\wedge (c\vee b)),(d\wedge c)\vee (d\wedge a))) \\&\quad =((a\vee b)\wedge (c\vee b)),(b\wedge a)\vee (d\wedge a))) \\&\quad =(b\vee (a\wedge c),a\wedge (b\vee d)) \\&\quad =I_{L5} ((a,b),(c,d)). \end{aligned}$$

The proofs for other cases are similar.

To (P15): When $i=5$ and by Eq. (5), we have

$$\begin{aligned} I_{L5} ((a,b),(c,d))&=(b\vee (a\wedge c),a\wedge (b\vee d))=(0,1) \\&\Leftrightarrow b\vee (a\wedge c)=0,a\wedge (b\vee d)=1 \\&\Leftrightarrow b=\;a\wedge c=0\;and\;a=\;b\vee d=1 \\&\Leftrightarrow (a,b)=(1,0),(c,d)=(0,1) \end{aligned}$$

The proofs for other cases are similar.

To (P16): When $i=5$ and by Eq. (5), we have

$$\begin{aligned} I_{L5}((a, b), (b, a)) = (b \vee (a \wedge b), a \wedge (b \vee a))= (b, a). \end{aligned}$$

The proofs for other cases are similar.

Property 3

$I_{L8}((a, b), (c, d))$ satisfies the following:

• (1) (A1)–(A6), (P1)–(P4), P(7), (P9)–(P15) and (P17) hold;

• (2) If $a \vee c \le e$, then (P8) holds.

Proof

By Eq. (8), it can be easily concluded that (A1)–(A4) are true.

To (A5): If $(a, b)\le (a', b')$, then $a \le a'$, $b \ge b'$. By Eq. (8), we can draw that

$$\begin{aligned}&I_{L8} ((a,b),(c,d))\nonumber \\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(c+b,d+a-1)}&{}\; {a+d>1} \\ {(1-\max (a-c,0)-\max (d-b,0),0)}&{}\; {a+d\le 1} \\ \end{array} }} \right. . \end{aligned}$$

• (i) When $a +d > 1$, we have $I_{L8}((a, b)$, $(c, d))= (b +c, d +a -1)$ and

  $$\begin{aligned} I_{L8} (({a}',{b}'),(c,d))=(c+{b}',d+{a}'-1). \end{aligned}$$

  For the reason that $c + b \ge c + b', d+a -1\le d+ a'-1$, therefore

  $$\begin{aligned} I_{L8} ((a,b),(c,d))\ge I_{L8} (({a}',{b}'),(c,d)). \end{aligned}$$

• (ii) When $a +d\le 1$, we have $I_{L8}((a, b)$, $(c, d))= (1- \max (a - c, 0) - \max (d - b, 0), 0)$.

  • (a) If $a \le c$, $b \ge d$, then $I_{L8}((a, b)$, $(c, d))= (1, 0)$. Therefore,

    $$\begin{aligned}&\forall (a,b)\le ({a}',{b}'),I_{L8} ((a,b),(c,d))\\&\quad \ge I_{L8} (({a}',{b}'),(c,d)). \end{aligned}$$

  • (b) If $a \le c$, $b < d$, then $I_{L8}((a, b)$, $(c, d))= (1- d+b, 0)$, because $b'<b$, we have

    $$\begin{aligned}&I_{L8} (({a}',{b}'),(c,d))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1-d+{b}'-\max ({a}' -c,0),0)}&{}\; {{a}'+d\le 1;} \\ {(c+{b}',d+{a}'-1)}&{}\; {{a}'+d>1,} \\ \end{array} }} \right. \end{aligned}$$

    and $(1- \max (a - c, 0) -d +b', 0)\le (1 - d +b, 0)$ as well as

    $$\begin{aligned} (c +b', d +a'-1)\le (1- d +b, 0). \end{aligned}$$

    That is to say

    $$\begin{aligned} I_{L8} ((a,b),(c,d))\ge I_{L8} (({a}',{b}'),(c,d)); \end{aligned}$$

  • (c) If $a > c$, $b \ge d$, then $I_{L8}((a, b)$, $(c, d))= (1- a+c, 0)$, and

    $$\begin{aligned}&I_{L8} (({a}',{b}'),(c,d))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1-{a}'+c-\max (d-{b}',0),0)}&{}\; {{a}'+d\le 1;} \\ {(c+{b}',d+{a}'-1)}&{}\; {{a}'+d>1.} \\ \end{array} }} \right. \end{aligned}$$

    For $a \le a$’, we have

    $$\begin{aligned}&(1- \max (d - b', 0) -a' +c, 0) \le (1 - a +c, 0)\hbox { and }\\&(c +b', d +a'-1)\le (1 - a +c, 0). \end{aligned}$$

    That is to say

    $$\begin{aligned} I_{L8} ((a,b),(c,d))\ge I_{L8} (({a}',{b}'),(c,d)). \end{aligned}$$

  • (d) If $a > c$, $b < d$, then $I_{L8}((a, b)$, $(c, d))= (1- a+c- d +b, 0)$. And

    $$\begin{aligned}&I_{L8} (({a}',{b}'),(c,d))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(1-{a}'+c-d+{b}',0)}&{} {{a}'+d\le 1;} \\ {(c+{b}',d+{a}'-1)}&{} {{a}'+d>1.} \\ \end{array} }} \right. \end{aligned}$$

For the reason that $a \le a'$, we have $(1-d +b'-a' +c, 0) \le (1- d +b-a +c, 0)$, and for $b' \le b$, we have $(c +b', d +a'-1)\le (1 - a + c -d + b, 0)$. That is to say

$$\begin{aligned} I_{L8} ((a,b),(c,d))\ge I_{L8} (({a}',{b}'),(c,d)). \end{aligned}$$

In summary, we have if $(a, b)\le (a', b')$, then $I_{L8}((a, b)$, $(c, d))\ge I_{L8}((a', b')$, (c, d)). $\square$

To (A6): If $(c, d)\le (c', d')$, then $c \le c'$, $d \ge d'$. By Eq. (8), we have

$$\begin{aligned}&I_{L8} ((a,b),(c,d))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(c+b,d+a-1)}&{}\; {a+d>1} \\ {(1-\max (a-c,0)-\max (d-b,0),0)}&{}\; {a+d\le 1} \\ \end{array} }} \right. \end{aligned}$$

• (i) When $a +d \le$ 1, we have

  $$\begin{aligned}&I_{L8} ((a,b),(c,d))\\&\quad =(1-\max (a-c,0)-\max (d-b,0),0). \end{aligned}$$

  If $d \ge \quad d'$, then $a+d' \le a+d \le 1$, therefore

  $$\begin{aligned}&(1-\max (a-{c}',0)-\max ({d}'-b,0),0)\\&\quad \ge (1-\max (a-c,0)-\max (d-b,0),0), \end{aligned}$$

  and we have

  $$\begin{aligned} I_{L8}((a, b), (c, d))\le I_{L8}((a, b), (c', d')). \end{aligned}$$

• (ii) When $a +d > 1$, we have $I_{L8}((a, b)$, $(c, d))= (b+c, a+d-1)$.

  • (a) If $a +d' > 1$, then $I_{L8}((a, b)$, $(c', d'))= (b+c', a+d'-1) \ge (b+c, a+d-1)$;

  • (b) If $a +d' \le$ 1, then $I_{L8}((a, b), (c', d'))= (1- \max (a -c', 0)- \max (d'- b, 0), 0)$;

When $a \le c', b \ge d'$, we have $I_{L8}((a, b)$, $(c', d'))= (1, 0)$. Therefore,

$$\begin{aligned} \forall (c,d)\le ({c}',{d}'),I_{L8} ((a,b),(c,d))\le I_{L8} ((a,b),({c}',{d}')); \end{aligned}$$

When $a \le c'$, $b < d'$, we have $I_{L8}((a, b), (c', d'))= ( - d'+b, 0)$ and

$$\begin{aligned} (1 - d'+b, 0)\ge & {} (b+c', 0)\ge (b+c, 0) \\\ge & {} (b+c, a+d -1). \end{aligned}$$

Therefore,

$$\begin{aligned} I_{L8} ((a,b),(c,d))\le I_{L8} ((a,b),({c}',{d}')). \end{aligned}$$

When $a > c'$, $b \ge d'$, we have $I_{L8}((a, b)$, $(c', d'))= (1 - a+c', 0)$ and

$$\begin{aligned} (1 - a+c', 0)\ge & {} (b+c', 0) \ge (b+c, 0) \\\ge & {} (b+c, a+d -1). \end{aligned}$$

Therefore,

$$\begin{aligned} I_{L8} ((a,b),(c,d))\le I_{L8} ((a,b),({c}',{d}')); \end{aligned}$$

When $a > c'$, $b < d'$, we have $I_{L8}((a, b), (c', d'))= (1 - a+c'+b- d' 0)$.

For $a + d' \le$ 1, we have

$$\begin{aligned} (1 - a+c'+b - d', 0)\ge & {} (b+c', 0) \ge (b+c, 0) \\\ge & {} (b+c, a+d -1); \end{aligned}$$

Therefore,

$$\begin{aligned} I_{L8} ((a,b),(c,d))\le I_{L8} ((a,b),({c}',{d}')). \end{aligned}$$

In summary, we have if ($c, d) \le (c', d')$, then $I_{L8}((a, b)$, $(c', d')) \ge I_{L8}((a, b)$, (c, d)).

To (P1): When $a +b$=1, $c +d$=1, we can draw that

• (i) If $a \le c$, then $a \le 1-d$ and $b \ge d$, therefore $I_{L8}((a, b)$, $(c, d))= (1, 0)$. That is to say

  $$\begin{aligned} \pi _{I_{L8} ((a,b),(c,d))} =0; \end{aligned}$$

• (ii) If $a > c$, then $I_{L8}((a, b)$, $(c, d))= (1-a+c, a-c)$. That is to say

  $$\begin{aligned} \pi _{I_{L8} ((a,b),(c,d))} =0; \end{aligned}$$

  In summary, we have (P1) holds.

To (P2)–(P3): By Eq. (8), it can be easily concluded that (P2)–(P3) are true.

To (P4): By Eq. (8), we have

$$\begin{aligned}&\pi _{I_{L8} ((a,b),(c,d))}\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {2-c-b-d-a}&{} {a+d\ge 1;} \\ {\max (a-c,0)+\max (d-b,0)}&{} {a+d<1.} \\ \end{array} }} \right. \end{aligned}$$

If $a + d \ge$ 1, then

$$\begin{aligned} \pi _{I_{L8} ((a,b),(c,d))} =2-a-b-c-d\le 1-c-b\le 1-c. \end{aligned}$$

If $a + d < 1$, then

$$\begin{aligned} \pi _{I_{L8} ((a,b),(c,d))}= & {} \max (a-c,0)+\max (d-b,0)\\\le & {} a-c+d-b\le 1-c-b\le 1-c. \end{aligned}$$

In summary, we have

$$\begin{aligned} \pi _{I_{L8} ((a,b),(c,d))} \le (1-a)\vee (1-b)\vee (1-c)\vee (1-d). \end{aligned}$$

To (P7): By Eq. (8), we have $I_{L8}((1, 0), (c, d))= (c, d)$.

To (P8): Suppose that $a \le e$, we have $a+f \le 1$ and

$$\begin{aligned} I_{L8} ((a,b),(e,f))=(1-\max (f-b,0),0). \end{aligned}$$

(i) When $b \ge f$, we have $I_{L8}((a, b), (e, f))= (1, 0)$, therefore

$$\begin{aligned} I_{L8} ((c,d)),I_{L8} ((a,b),(e,f)))= & {} I_{L8} ((c,d)),(1,0))\\= & {} (1,0) \end{aligned}$$

On the other hand, by (P10), we have $I_{L8}((c, d)$, (e, f)) $\ge (e, f)$, so

$$\begin{aligned} I_{L8} ((a,b),I_{L8} ((c,d),(e,f)))\ge & {} I_{L8} ((a,b),(e,f))\\= & {} (1,0). \end{aligned}$$

Therefore, we have $I_{L8}$((a, b),$I_{L8}$((c, d), (e, f)))= (1, 0).

That is to say

$$\begin{aligned}&I_{L8} ((a,b),I_{L8} ((c,d),(e,f)))\\&\quad =I_{L8} ((c,d)),I_{L8} ((a,b),(e,f))). \end{aligned}$$

(ii) When $b \le f$, we have $I_{L8}((a, b)$, $(e, f))= (1 - f+b, 0)$, therefore

$$\begin{aligned}&I_{L8} ((c,d),I_{L8} ((a,b),(e,f)))\\&\quad =(1-\max (c-1+f-b,0),0) \end{aligned}$$

(a) If $c > 1-f + b$, then $I_{L8}((c, d)$, $I_{L8}((a, b)$, $(e, f)))= (2- c -f + b, 0)$. On the other hand, for the reason that $c > 1-f + b$, we have

$$\begin{aligned} I_{L8}((c, d), (e, f))= (d+e, c+f - 1). \end{aligned}$$

And we can draw that $I_{L8}((a, b), I_{L8}((c, d), (e, f)))= (2 - c-f + b, 0)$.

In summary, we have

$$\begin{aligned}&I_{L8} ((a,b),I_{L8} ((c,d),(e,f)))\\&\quad =I_{L8} ((c,d),I_{L8} ((a,b),(e,f))). \end{aligned}$$

(b) If $c < 1-f + b$, then $I_{L8}((c, d)$, $I_{L8}((a, b)$, $(e, f)))= (1, 0)$.

On the other hand,

$$\begin{aligned}&I_{L8} ((c,d),(e,f))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(e+d,f+c-1)}&{} {f+c>1} \\ {(1-\max (c-e,0)-\max (f-d,0),0)}&{} {c+f\le 1} \\ \end{array} }} \right. . \end{aligned}$$

For the reason that $I_{L8}((c, d), (e, f)))\ge (e, f)$, we have

$$\begin{aligned} 1- \max (c-e, 0)-\max (f -d, 0)\ge e \ge a; \end{aligned}$$

so we can conclude that $I_{L8}((c, d), (e, f)))\ge (a, b)$. That is to say

$$\begin{aligned}&I_{L8} ((a,b),I_{L8} ((c,d),(e,f)))\\&\quad \ge I_{L8} ((a,b),(a,b))=(1,0). \end{aligned}$$

Therefore,

$$\begin{aligned}&I_{L8} ((a,b),I_{L8} ((c,d),(e,f)))\\&\quad =I_{L8} ((c,d),I_{L8}((a,b),(e,f)))=(1,0). \end{aligned}$$

In summary, we have if $a \le e$, then

$$\begin{aligned}&I_{L8} ((a,b),I_{L8} ((c,d),(e,f)))\\&\quad =I_{L8} ((c,d),I_{L8} ((a,b),(e,f))). \end{aligned}$$

To (P9):

$$\begin{aligned} (a,b)\le & {} (c,d)\Leftrightarrow a\le c,b\ge d\Leftrightarrow a\le c\le 1-d,b\\\ge & {} d\Leftrightarrow I_{L8} ((a,b),(c,d))=(1,0). \end{aligned}$$

To (P10): By Eq. (8), $I_{L8}((a, b), (0, 1))= (b, a)$.

To (P11): By Eq. (8), we can draw that

$$\begin{aligned}&I_{L8} ((a,b),(c,d))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(b+c,a+d-1)}&{} {d+a>1} \\ {(1-\max (d-b,0)-\max (a-c,0),0)}&{} {d+a\le 1} \\ \end{array} }} \right. \end{aligned}$$

• (i) When $a+d > 1$, we have $I_{L8}((a, b), (c, d)) = (c + b, d + a - 1) \ge (c, d)$;

• (ii) When $a+d \le 1$, we have $I_{L8}((a, b)$, $(c, d)) = (1-\max (a- c, 0) - \max (d- b, 0), 0)$;

  • (a) If $a \le c, b \ge d$, then $I_{L8}((a, b), (c, d)) = (1, 0)\ge (c, d)$;

  • (b) If $a \le c, b < d$, then $I_{L8}((a, b), (c, d))= (1-d + b, 0)\ge (c, d)$;

  • (c) If $a > c, b \ge d$, then $I_{L8}((a, b)$, $(c, d)) = (1-a + c, 0) \ge (c, d)$;

  • (d) If $a> c, b < d$, then $I_{L8}((a, b), (c, d)) = (1- a + c -d + b, 0)$;

For the reason that $a+d \le 1$, then we have $1- a + c -d + b \ge c + b \ge c$ and $d \ge 0$. So we can draw that

$$\begin{aligned} I_{L8} ((a,b),(c,d))\ge (c,d). \end{aligned}$$

In summary, we can draw that

$$\begin{aligned} I_{L8} ((a,b),(c,d))\ge (c,d). \end{aligned}$$

To (P12): By Eq. (8), we have $I_{L8}((a, b), (a, b)) = (1, 0)$

To (P13): By Eq. (8),

$$\begin{aligned}&I_{L8} ((d,c),(b,a))\\&\quad =\left\{ {{\begin{array}{l@{\quad }l} {(b+c,a+d-1)}&{} {d+a\ge 1} \\ {(1-\max (d-b,0)-\max (a-c,0),0)}&{} {d+a<1} \\ \end{array} }} \right. =I_{L8} ((a,b),(c,d)). \end{aligned}$$

To (P15): It can be drawn by Eq. (8).

To (P17): By (P13), we have $I_{L8}((c, d)$, $(a, b)) \ge (a, b)$. And by (P11), (P2), we have

$$\begin{aligned}&I_{L8} ((a,b),I_{L8} ((c,d),(a,b)))\\&\quad \ge I_{L8} ((a,b),(a,b))=(1,0). \end{aligned}$$

That is to say

$$\begin{aligned} I_{L8} ((a,b),I_{L8} ((c,d),(a,b)))=(1,0). \end{aligned}$$