Distributive Equations of Fuzzy Implications Based on Continuous Triangular Conorms Given as Ordinal...

IEEE TRANSACTIONS ON FUZZY SYSTEMS, VOL. 21, NO. 3, JUNE 2013 541

Distributive Equations of Fuzzy ImplicationsBased on Continuous Triangular Conorms

Given as Ordinal SumsAifang Xie, Cheng Li, and Huawen Liu

Abstract—Recently, the distributive equations of fuzzy impli-cations based on t-norms or t-conorms have become a focus ofresearch. The solutions to these equations can help people designthe structures of fuzzy systems in such a way that the number ofrules is largely reduced. This paper studies the distributive func-tional equation I(x, S1 (y, z)) = S2 (I(x, y), I(x, z)), whereS1 and S2 are two continuous t-conorms given as ordinal sums, andI : [0, 1]2 → [0, 1] is a binary function which is increasing withrespect to the second place. If there is no summand of S2 in theinterval [I(1, 0), I(1, 1)], we get its continuous solutions directly.If there are summands of S2 in the interval [I(1, 0), I(1, 1)], bydefining a new concept called feasible correspondence and usingthis concept, we describe the solvability of the distributive equationabove and characterize its general continuous solutions. When I isrestricted to fuzzy implications, it is showed that there is no contin-uous solution to this equation. We characterize its fuzzy implicationsolutions, which are continuous on (0, 1] × [0, 1].

Index Terms—Continuous t-conorms, distributive functionalequations, fuzzy implications, ordinal sums.

I. INTRODUCTION

DURING the past 30 years, fuzzy systems have been suc-cessfully used in many fields, such as in control sys-

tems [1]–[4], decision making, and signal processing. It is wellknown that fuzzy systems can approximate any continuous func-tion to any desired accuracy (see [5]–[7]). However, a large rulebase is required usually to achieve high accuracy. In many situ-ations, the problem of rule explosion has become a major draw-back which hinders the successful application of fuzzy systems.

To construct fuzzy systems using as less as possible fuzzyrules with guaranteed performance, some researchers have donea lot of work and have obtained encouraging results [8]–[11].For example, in order to avoid combinational rule explosion,Combs and Andrews [9] introduced the classical tautology(p ∧ q) → r ≡ (p → r) ∨ (q → r). They referred to the left-

Manuscript received September 20, 2011; revised May 4, 2012 and Septem-ber 3, 2012; accepted September 11, 2012. Date of publication October 2, 2012;date of current version May 29, 2013. This work was supported by the NationalNatural Science Foundation of China under Grant 61174099.

A. Xie and H. Liu are with the School of Mathematics, Shandong University,Jinan 250100, China (e-mail: [email protected]; [email protected]).

C. Li was with the School of Mathematics, Shandong University, Jinan250100, China. He is now with Northeastern University, Boston, MA 02115USA (e-mail: [email protected]).

Color versions of one or more of the figures in this paper are available onlineat http://ieeexplore.ieee.org.

Digital Object Identifier 10.1109/TFUZZ.2012.2221719

hand side of this equivalence as an intersection rule configura-tion (IRC) and to its right-hand side as a union rule configuration(URC). Later, some discussions appeared [12]–[15], and mostof them believed that it is necessary to theoretically investigatethis tautology before employing it. Dick and Kandel [14] wrote,“Future work on this issue will require an examination of theproperties of various combinations of fuzzy unions, intersec-tions and implications.” Mendel and Liang [15] also said, “Wethink that what this all means is that we have to look past themathematics of IRC⇔ URC and inquire whether what we aredoing when we replace IRC by URC makes sense.” Trillas andAlsina [16], in the standard fuzzy sets theory, first transformedthe above tautology into the following functional equation

I(T (x, y), z) = S(I(x, z), I(y, z)), x, y, z ∈ [0, 1] (1)

where I is a fuzzy implication, T is a t-norm, and S is a t-conorm.They also found out all the solutions T and S to (1) whenI is an R-implication, an S-implication, or a QL-implication,respectively. In [17]–[19], T, S, and I in (1) are generalizedinto uninorms and the implications derived from uninorms,respectively.

Balasubramaniam and Rao [20] proposed three dual equa-tions of (1). One of those dual equations is

I(x, S1(y, z)) = S2(I(x, y), I(x, z)), x, y, z ∈ [0, 1](2)

where I denotes a fuzzy implication, and S1 and S2 are t-conorms. Obviously, the above equation is a generalization ofthe classical tautology p → (q ∨ r) ≡ (p → q) ∨ (p → r). In[20], the authors also proved that if I is an R-implication derivedfrom a nilpotent t-norm or is an S-implication, then S1 = S2 =max.

Baczynski [21], [22] investigated (2) in the case that S1and S2 are continuous Archimedean t-conorms and obtainedits general fuzzy implication solutions. Following his work, Xieet al. [23] studied (2) under the condition that S2 is a continuousArchimedean t-conorm and S1 is a continuous t-conorm givenas an ordinal sum. It is well known that if S is a continuous t-conorm, then either S = SM , or S is a continuous Archimedeant-conorm, or S is an ordinal sum of continuous Archimedeant-conorms [24]. By far, (2) has not been considered when bothS1 and S2 are continuous t-conorms given as ordinal sums.

In this paper, we will study (2) with two continuous t-conorms given as ordinal sums. This paper generalizes the re-sults in [21]–[23] since a continuous Archimedean t-conormis an ordinal sum with one summand. In addition, it shouldbe pointed out that the method in this paper is different from

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542 IEEE TRANSACTIONS ON FUZZY SYSTEMS, VOL. 21, NO. 3, JUNE 2013

the previous ones in [21]–[23]. For example, in [21]–[23], theform of the vertical section I(x, ·) was given for every fixedx ∈ [0, 1], and then, these forms of vertical sections are used toderive the general continuous solutions to (2). However, in thispaper, there is no need to get the form of the vertical sectionI(x, ·) for every fixed x ∈ [0, 1]. In fact, if there is no summandof S2 in the interval [I(1, 0), I(1, 1)], we get directly its gen-eral continuous solutions; if there are summands of S2 in theinterval [I(1, 0), I(1, 1)], by introducing a new concept calledfeasible correspondence, we describe the solvability of (2) andcharacterize its general continuous solutions.

The rest of this paper is organized as follows. Some basic def-initions and theorems are reviewed in Section II. In Section III,we discuss (2) and obtain its general continuous solutionsand fuzzy implication solutions. We end this paper with theconclusions.

II. PRELIMINARIES

In this section, we briefly recall some of the concepts andresults used in the sequel.

Definition 1 [24], [25]: An associative, commutative, andincreasing operation S : [0, 1]2 → [0, 1] is called a triangularconorm (t-conorm for short) if it has the neutral element 0, i.e.,S(x, 0) = x for any x ∈ [0, 1].

Example 1 [24]: The following are the four basic t-conormsSM , SP , SL , and SD given by, respectively

SM (x, y) = max(x, y), SP (x, y) = x + y − xy

SL (x, y) = min(x + y, 1), and

SD (x, y) ={

max(x, y), x = 0 or y = 01, otherwise.

Definition 2 [24]: A t-conorm S is called a continuousArchimedean t-conorm if it is continuous and satisfies S(x, x) >x for all x ∈ (0, 1).

A continuous t-conorm S is said to be strict if it holds thatS(x, y) < S(x, z) for all x < 1 and y < z. A t-conorm S is saidto be nilpotent if it is continuous, and for any x ∈ (0, 1), there ex-ists some n ∈ N such that xn

S = 1, where xnS = S(x, x, . . . , x)︸︷︷︸

n times

.

For example, SP is strict and SL is nilpotent. We also knowthat any continuous Archimedean t-conorm is either strict ornilpotent.

Theorem 1 [24], [26]: A function S : [0, 1]2 → [0, 1] is a con-tinuous Archimedean t-conorm if and only if S has a continuousadditive generator, i.e., there exists a continuous and strictly in-creasing function s : [0, 1] → [0,∞] with s(0) = 0, which isuniquely determined up to a positive multiplicative constant,such that S(x, y) = s(−1)(s(x) + s(y)) for all x, y ∈ [0, 1],where s(−1) is the pseudoinverse of s, given by

s(−1)(x) ={

s−1(x), x ∈ [0, s(1)]1, x ∈ [s(1),∞].

Remark 1: 1) Without the pseudoinverse, the representationof a t-conorm S in Theorem 1 can be rewritten as S(x, y) =s−1(min(s(x) + s(y), s(1))) for all x, y ∈ [0, 1].

2) A t-conorm S is strict if and only if each continuous addi-tive generator s of S satisfies s(1) = ∞.

3) A t-conorm S is nilpotent if and only if each continuousadditive generator s of S satisfies s(1) < ∞.

Proposition 1 [24]: Let (Sm )m∈J be a family of t-conormsand ((am , bm ))m∈J be a family of nonempty, nonoverlapping,and open subintervals of [0, 1], where J is a finite or countableinfinite index set. Then the function S : [0, 1]2 → [0, 1] that isdefined by

S(x, y) =

⎧⎪⎪⎨⎪⎪⎩

am + (bm − am )Sm

(x − am

bm − am,

y − am

bm − am

)

(x, y) ∈ [am , bm ]2

max(x, y), otherwise

is a t-conorm which is called the ordinal sum of summands〈am , bm , Sm 〉m∈J , and we write S = (〈am , bm , Sm 〉)m∈J .

Remark 2 [24]: A t-conorm S is continuous if and only if Sis either SM , a continuous Archimedean t-conorm, or an ordi-nal sum (〈am , bm , Sm 〉)m∈J with each Sm being a continuousArchimedean t-conorm.

Definition 3 [27]: A function I : [0, 1]2 → [0, 1] is said to bea fuzzy implication if it satisfies the following:

I1) the first place antitonicity;I2) the second place isotonicity;I3) I(0, 0) = I(1, 1) = I(0, 1) = 1 and I(1, 0) = 0.From the definition above, it is clear that for any fuzzy impli-

cation I , it holds that I(0, x) = I(x, 1) = 1 for all x ∈ [0, 1].Now, let us recall some facts about the additive Cauchy func-

tional equation (see [28]) and some other similar equations.Theorem 2 [22]: For a function f : [0,∞] → [0,∞], the fol-

lowing statements are equivalent:1) f satisfies the additive Cauchy functional equation

f(x + y) = f(x) + f(y) for all x, y ∈ [0,∞].

2) Either

f = ∞, or f = 0, or f(x) ={

0, x = 0∞, x ∈ (0,∞]

or

f(x) ={

0, x ∈ [0,∞)∞, x = ∞

or there exists a unique constant p ∈ (0,∞) such that f(x) = pxfor all x ∈ [0,∞].

Clearly, if f is continuous, then f = ∞, or f = 0, or f(x) =px with a unique constant p ∈ (0,∞).

Theorem 3 [21]: Fix real b > 0. For a function f : [0,∞] →[0, b], the following are equivalent:

1) f satisfies the functional equation

f(x + y) = min(f(x) + f(y), b) for all x, y ∈ [0,∞].

XIE et al.: DISTRIBUTIVE EQUATIONS OF FUZZY IMPLICATIONS 543

2) Either

f = b, or f = 0, or f(x) ={

0, x = 0b, x ∈ (0,∞]

or

f(x) ={

0 x ∈ [0,∞)b x = ∞

or there exists a unique constant p ∈ (0,∞) such that f(x) =min(px, b) for all x ∈ [0,∞].

Theorem 4 [22]: Fix real a, b > 0. For a function f : [0, a] →[0, b], the following statements are equivalent:

1) f satisfies the functional equation

f(min(x + y, a)) = min(f(x) + f(y), b) for all x, y ∈ [0, a].

2) Either

f = b, or f = 0, or f(x) ={

0, x = 0b, x ∈ (0, a]

or there exists a unique constant p ∈ [ ba ,∞) such that f(x) =

min(px, b) for all x ∈ [0, a].

III. SOLUTIONS TO (2)

In this section, we discuss (2) with two continuous t-conormsgiven as ordinal sums. This section is divided into two sections.In Section III-A, we discuss the general continuous solutions to(2). In Section III-B, we discuss its fuzzy implication solutionswhich are continuous on (0, 1] × [0, 1].

A. Continuous Solutions to (2)

Lemma 1: Let S1 and S2 be two t-conorms and I : [0, 1]2 →[0, 1] be a binary function. If the triple (S1 , S2 , I) satisfies(2) and y is an idempotent element of S1 , then for everyx ∈ [0, 1], I(x, y) is an idempotent element of S2 .

Proof: Suppose y is an idempotent element of S1 . For ev-ery x ∈ [0, 1], we get from (2) that I(x, y) = I(x, S1(y, y)) =S2(I(x, y), I(x, y)). Therefore, I(x, y) is an idempotent ele-ment of S2 . �

Lemma 2: Let S1 and S2 be two continuous t-conorms givenby (〈am , bm , S ′

m 〉)m∈J1 and (〈cn , dn , S ′′n 〉)n∈J2 , respectively,

and I : [0, 1]2 → [0, 1] be a binary function which is increasingin its second place. Arbitrarily fix x ∈ [0, 1]. Suppose the triple(S1 , S2 , I) satisfies (2). If there exist an m0 ∈ J1 and a y0 ∈(am 0 , bm 0 ) such that I(x, y0) is an idempotent element of S2 ,then I(x, y) = I(x, y0) for all y ∈ [y0 , bm 0 ].

Proof: Case 1. Suppose the corresponding t-conorm S ′m 0

is anilpotent t-conorm with an additive generator s′m 0

.Because I(x, y0) is an idempotent element of S2 , then for

any y ∈ [y0 , bm 0 ], I(x, S1(y, y0)) = S2(I(x, y), I(x, y0)) =max(I(x, y), I(x, y0)) = I(x, y), namely,

I(x, S1(y, y0)) = I(x, y)

for any y ∈ [y0 , bm 0 ].Let us define function ϕ : [am 0 , bm 0 ] → [0, 1] by ϕ(x) =

x−am 0bm 0 −am 0

and function Ix : [am 0 , bm 0 ] → [0, 1] by Ix(y) =I(x, y).

Then the above equation can be rewritten as Ix ◦ ϕ−1 ◦s′−1

m 0◦ min(s′m 0

◦ ϕ(y) + s′m 0◦ ϕ(y0), s′m 0

(1)) = Ix(y) forany y ∈ [y0 , bm 0 ]. By routine substitutions, hx = Ix ◦ϕ−1 ◦ s′−1

m 0, u = s′m 0

◦ ϕ(y), and v0 = s′m 0◦ ϕ(y0), we get the

following equation:

hx(min(u + v0 , s′m 0

(1)) = hx(u) (3)

for any u ∈ [v0 , s′m 0

(1)], where hx : [0, s′m 0(1)] → [0, 1] is an

increasing function.Let a = ϕ−1 ◦ s′−1

m 0(s′m 0

(1) − s′m 0(ϕ(y0))). Clearly, it holds

that a ∈ (am 0 , bm 0 ).Case 1.1. Assume that y0 ≤ a, i.e., 2v0 ≤ s′m 0

(1).For any y ∈ [y0 , a], we have u = s′m 0

◦ ϕ(y) ∈ [v0 ,s′m 0

(1) − v0 ] and u + v0 ∈ [2v0 , s′m 0

(1)]. Thus, we get from(3) that

hx(u + v0) = hx(u) (4)

for any u ∈ [v0 , s′m 0

(1) − v0 ].Let P = {n|n is an integer and nv0 ∈ [v0 , s

′m 0

(1) − v0 ]}.Since 1 · v0 ∈ [v0 , s

′m 0

(1) − v0 ], it holds that P �= ∅. Becauses′m 0

(1) − v0 is a real number, there exists an n0 = max{n|nis an integer and nv0 ∈ [v0 , s

′m 0

(1) − v0 ]}. Therefore, n0v0 ∈[v0 , s

′m 0

(1) − v0 ] and (n0 + 1)v0 ∈ (s′m 0(1) − v0 , s

′m 0

(1)].From (4), we get that hx(v0) = hx(2v0) = · · · =

hx(n0v0) = hx((n0 + 1)v0). Observing the monotonicityof function hx , we have hx(t) = hx(v0) for any t ∈ [v0 ,(n0 + 1)v0 ] and then hx((n0 + 1)v0) = hx(v0).

On the other hand, we can obtain from (4) thathx(s′m 0

(1) − v0) = hx(s′m 0(1)). Hence, hx(t) = hx(s′m 0

(1)),t ∈ [s′m 0

(1) − v0 , s′m 0

(1)]. Since (n0 + 1)v0 ∈ (s′m 0(1) −

v0 , s′m 0

(1)], we have hx((n0 + 1)v0) = hx(s′m 0(1)).

From the above steps, we obtain that hx(v0) = hx(s′m 0(1)).

Consequently, hx(t) = hx(v0), t ∈ [v0 , s′m 0

(1)]. Therefore,I(x, y) = I(x, y0), y ∈ [y0 , bm 0 ].

Case 1.2. Assume that y0 > a, i.e., 2v0 > s′m 0(1).

For any y ∈ [y0 , bm 0 ], we get that u = s′m 0◦ ϕ(y) ∈

[v0 , s′m 0

(1)]. Therefore, u + v0 ∈ [2v0 , s′m 0

(1) + v0 ]. Because2v0 > s′m 0

(1), we obtain from (3) that hx(s′m 0(1)) = hx(u),

namely, I(x, y) = I(x, bm 0 ) for any y ∈ [y0 , bm 0 ]. Especially,I(x, y0) = I(x, bm 0 ). Hence, I(x, y) = I(x, y0) for all y ∈[y0 , bm 0 ].

Case 2. Suppose the corresponding t-conorm S ′m 0

isa strict t-conorm with an additive generator s′m 0

. It fol-lows from (2) that for any y ∈ [y0 , bm 0 ], I(x, S1(y, y0)) =S2(I(x, y), I(x, y0)) = max(I(x, y), I(x, y0)), i.e.,

I(x, S1(y, y0)) = I(x, y)

for any y ∈ [y0 , bm 0 ].Just as the proof in Case 1, we get the following equation:

hx(u + v0) = hx(u), for any u ∈ [v0 ,∞]

where hx : [0,∞] → [0, 1]. This equation shows that hx is aperiodic function on [v0 ,∞]. The increasing property of func-tion hx implies that hx must be a constant on [v0 ,∞]. Hence,hx(u) = hx(v0) for all u ∈ [v0 ,∞]. Therefore, I(x, y) =I(x, y0), y ∈ [y0 , bm 0 ]. �


Corollary 1: Let S1 and S2 be two continuous t-conormsgiven by (〈am , bm , S ′

m 〉)m∈J1 and (〈cn , dn , S ′′n 〉)n∈J2 , respec-

tively, and I : [0, 1]2 → [0, 1] be a binary function which isincreasing with respect to the second place. Arbitrarily fixx ∈ [0, 1]. Suppose the triple (S1 , S2 , I) satisfies (2). If I(x, ·)is continuous on [am 0 , bm 0 ] (some m0 ∈ J1) and I(x, y) isan idempotent element of S2 for any y ∈ (am 0 , bm 0 ), thenthere exists an idempotent element ex,m 0 of S2 such thatI(x, y) = ex,m 0 for all y ∈ [am 0 , bm 0 ].

Proof: It can be proved that I(x, y) equals each other forany y ∈ (am 0 , bm 0 ). Otherwise, suppose that there exist y1 andy2 (y1 < y2) in the interval (am 0 , bm 0 ) such that I(x, y1) �=I(x, y2). Since y1 < y2 and I(x, y1) is an idempotent elementof S2 , we obtain from Lemma 2 that I(x, y2) = I(x, y1). Thisproduces a contradiction. Consequently, there exists an idem-potent element ex,m 0 of S2 such that I(x, y) = ex,m 0 for ally ∈ (am 0 , bm 0 ). By the continuity of I(x, ·) on [am 0 , bm 0 ], weget the result. �

In this paper, for any binary function f and any set G, wedenote {f(x, y)|y ∈ G} as f(x,G).

Lemma 3: Let S1 and S2 be two continuous t-conorms givenby (〈am , bm , S ′


and I : [0, 1]2 → [0, 1] be a binary function which is increasingin its second place. Arbitrarily fix x ∈ [0, 1]. Suppose the triple(S1 , S2 , I) satisfies (2) and I(x, ·) is continuous on [am 0 , bm 0 ](some m0 ∈ J1). If there exists a y0 ∈ (am 0 , bm 0 ) such thatI(x, y0) is a non-idempotent element of S2 , then there existsan nx,m 0 ∈ J2 such that I(x, [am 0 , bm 0 ]) = [cnx , m 0

, dnx , m 0].

Especially,1) if there exists a y1 ∈ (am 0 , bm 0 ) such that I(x, y1) is an

idempotent element of S2 , then the corresponding t-conormS ′′

nx , m 0must be nilpotent;

2) if for any y ∈ (am 0 , bm 0 ), I(x, y) is a non-idempotent el-ement of S2 , then the corresponding t-conorm S ′′

nx , m 0is strict

if S ′m 0

is strict, and S ′′nx , m 0

is nilpotent if S ′m 0

is nilpotent.Proof: We get from Lemma 2 that I(x, y) is not an idempo-

tent element of S2 for any y ∈ (am 0 , y0). Let us define kx,m 0 =inf{y ∈ (am 0 , bm 0 )|I(x, y) is an idempotent element of S2}(with the convention that inf ∅ = bm 0 ). Clearly, am 0 < y0 ≤kx,m 0 ≤ bm 0 .

Denote T = {y ∈ (am 0 , bm 0 )|I(x, y) is an idempotent ele-ment of S2}. In the following, we will prove that I(x, kx,m 0 ) isan idempotent element of S2 .

In fact, according to the continuity of S2 and the continuityof I(x, ·) on [am 0 , bm 0 ], we have

S2(I(x, kx,m 0 ), I(x, kx,m 0 ))

= S2(I(x, inf T ), I(x, inf T ))

= S2(inf{I(x, y)|y ∈ T}, inf{I(x, y)|y ∈ T})= inf{S2(I(x, y), I(x, y))|y ∈ T}= inf{I(x, y)|y ∈ T}= I(x, inf{y|y ∈ T}) = I(x, kx,m 0 ).

Therefore, I(x, kx,m 0 ) is an idempotent element of S2 . Ac-cording to Lemma 2, we get that I(x, y) = I(x, kx,m 0 ) for any

y ∈ [kx,m 0 , bm 0 ]. In addition, it is easy to see that I(x, y) ∈[I(x, am 0 ), I(x, kx,m 0 )] for any y ∈ (am 0 , kx,m 0 ).

Considering that I(x, y) is not an idempotent element forany y ∈ (am 0 , kx,m 0 ), I(x, am 0 ) and I(x, kx,m 0 ) are idempo-tent elements of S2 , and I(x, ·) is continuous on [am 0 , bm 0 ],we obtain that there exists an nx,m 0 ∈ J2 such that I(x, y) ∈(cnx , m 0

, dnx , m 0) for all y ∈ (am 0 , kx,m 0 ), I(x, am 0 ) = cnx , m 0

,and I(x, y) = dnx , m 0

for all y ∈ [kx,m 0 , bm 0 ]. Therefore,I(x, [am 0 , bm 0 ]) = [cnx , m 0

, dnx , m 0].

Next we will prove 1). Clearly, it holds that kx,m 0 ≤ y1 <bm 0 .

Suppose S ′′nx , m 0

is strict. Because kx,m 0 is a non-idempotentelement of S1 , it holds that S1(kx,m 0 , kx,m 0 ) > kx,m 0 . SinceS1 is continuous, there must exist a y′ ∈ (am 0 , kx,m 0 ) suchthat S1(y′, y′) ∈ (kx,m 0 , S1(kx,m 0 , kx,m 0 )), which implies thatI(x, S1(y′, y′)) = dnx , m 0

. However, S2(I(x, y′), I(x, y′)) <dnx , m 0

because I(x, y′) ∈ (cnx , m 0, dnx , m 0

) and S ′′nx , m 0

is strict.This contradicts (2). Consequently, S ′′

nx , m 0must be nilpotent.

Finally, we will prove 2). Suppose the corresponding t-conorm S ′

m 0is strict. In the following, we will show that

S ′′nx , m 0

is strict. Otherwise, suppose S ′′nx , m 0

is nilpotent. Thenthere exists a z∗ ∈ (cnx , m 0

, dnx , m 0) such that S2(z∗, z∗) =

dnx , m 0. Because of the continuity of I(x, ·) on [am 0 , bm 0 ],

there exists a y′′ ∈ (am 0 , bm 0 ) such that I(x, y′′) = z∗. SinceS ′

m 0is strict, we have S1(y′′, y′′) ∈ (am 0 , bm 0 ), and then

I(x, S1(y′′, y′′)) ∈ (cnx , m 0, dnx , m 0

). This contradicts the factthat S2(I(x, y′′), I(x, y′′)) = S2(z∗, z∗) = dnx , m 0

.In a similar way, we can prove that if S ′

m 0is nilpotent, then

S ′′nx , m 0

is nilpotent. �Lemma 4: Let S1 and S2 be two continuous t-conorms

given by (〈am , bm , S ′m 〉)m∈J1 and (〈cn , dn , S ′′

n 〉)n∈J2 , respec-tively, and I : [0, 1]2 → [0, 1] be a binary function which isincreasing in its second place. Arbitrarily fix x ∈ [0, 1]. Sup-pose the triple (S1 , S2 , I) satisfies (2) and I(x, ·) is con-tinuous on [am , bm ] (any m ∈ J1). If there exist m1 ,m2 ∈J1 such that am 1 < am 2 , I(x, [am 1 , bm 1 ]) = [cnx , m 1

, dnx , m 1]

and I(x, [am 2 , bm 2 ]) = [cnx , m 2, dnx , m 2

] (some nx,m 1 , nx,m 2 ∈J2), then cnx , m 1

< cnx , m 2.

Proof: First, we prove that cnx , m 1�= cnx , m 2

. Other-wise, suppose cnx , m 1

= cnx , m 2. Then I(x, am 2 ) = cnx , m 1

andI(x, bm 1 ) = dnx , m 1

. Hence, I(x, am 2 ) < I(x, bm 1 ) becausecnx , m 1

< dnx , m 1. Obviously, this contradicts the increasing

property of I(x, ·). Therefore, cnx , m 1�= cnx , m 2

. The increasingproperty of I(x, ·) immediately implies that cnx , m 1

< cnx , m 2. �

Lemma 5: Let S1 and S2 be two continuous t-conormsgiven by (〈am , bm , S ′

m 〉)m∈J1 and (〈cn , dn , S ′′n 〉)n∈J2 , respec-

tively, and I : [0, 1]2 → [0, 1] be a continuous binary func-tion which is increasing in its second place. Suppose thetriple (S1 , S2 , I) satisfies (2). If there exist an x1 ∈ [0, 1]and some m0 ∈ J1 , n0 ∈ J2 such that I(x1 , [am 0 , bm 0 ]) =[cn0 , dn0 ], then I(x, [am 0 , bm 0 ]) = [cn0 , dn0 ] for any x ∈ [0, 1].

Proof: Suppose there exists an x2 ∈ [0, 1] (x1 �= x2) such thatI(x2 , [am 0 , bm 0 ]) �= [cn0 , dn0 ]. According to Corollary 1 andLemma 3, we have I(x2 , [am 0 , bm 0 ]) = ex2 ,m 0 (ex2 ,m 0 is anidempotent element of S2), or I(x2 , [am 0 , bm 0 ]) = [cn ′

0, dn ′

0].

Let us suppose x1 < x2 and cn ′0

< cn0 .


Case 1. Suppose I(x2 , [am 0 , bm 0 ]) = ex2 ,m 0 . ThenI(x1 , am 0 ) = cn0 and I(x2 , am 0 ) = ex2 ,m 0 . In this case,if ex2 ,m 0 ≥ dn0 , we have from the continuity of I(·, am 0 )that for any y0 ∈ (cn0 , dn0 ) ⊂ [cn0 , ex2 ,m 0 ], there existssome x0 ∈ (x1 , x2) such that I(x0 , am 0 ) = y0 . This showsthat I(x0 , am 0 ) is not an idempotent element of S2 , whichcontradicts Lemma 1.

If ex2 ,m 0 ≤ cn0 , then we have I(x1 , bm 0 ) = dn0 andI(x2 , bm 0 ) = ex2 ,m 0 . Similarly to the above, we can get acontradiction.

Case 2. Suppose I(x2 , [am 0 , bm 0 ]) = [cn ′0, dn ′

0]. Then we

obtain that I(x1 , am 0 ) = cn0 and I(x2 , am 0 ) = cn ′0. For any

y′ ∈ (cn ′0, dn ′

0) ⊂ (cn ′

0, cn0 ], there exists some x′ ∈ (x1 , x2)

satisfying that I(x′, am 0 ) = y′. This produces a contradictionwith Lemma 1. �

Theorem 5: Let S1 and S2 be two continuous t-conormsgiven by ordinal sums and I : [0, 1]2 → [0, 1] be a binaryfunction. Fix arbitrarily x ∈ [0, 1]. Suppose 〈am 0 , bm 0 , S

′m 0

〉and 〈cn0 , dn0 , S

′′n0〉 are two summands of S1 and S2 , re-

spectively. Then I(x, ·) is increasing and continuous on[am 0 , bm 0 ], I(x, [am 0 , bm 0 ]) = [cn0 , dn0 ], and I(x, S1(y, z)) =S2(I(x, y), I(x, z)) for all y, z ∈ [am 0 , bm 0 ] if and only if oneof the following statements holds.1) If S ′

m 0is a strict t-conorm with an additive generator s′m 0

andS ′′

n0is a nilpotent t-conorm with an additive generator s′′n0

, then

I(x, y) ={

Ex(y), y ∈ [am 0 , kx,m 0 )dn0 , y ∈ [kx,m 0 , bm 0 ]

for some kx,m 0 ∈ (am 0 , bm 0 ), where Ex(y) = cn0 + (dn0 −cn0 )s

′′−1n0

(s ′′

n 0(1)

s ′m 0

(k x , m 0 −a m 0

b m 0 −a m 0)s′m 0

( y−am 0bm 0 −am 0

)).

2) If S ′m 0

and S ′′n0

are strict t-conorms with s′m 0and s′′n0

beingtheir respective additive generators, then for any y ∈ [am 0 , bm 0 ],

I(x, y) = cn0 + (dn0 − cn0 )s′′−1n0

(px,m 0 s

′m 0

(y − am 0

bm 0 − am 0

))

where px,m 0 ∈ (0,∞) is a certain constant related to x and m0 ,uniquely determined up to a positive multiplicative constantdepending on constants for s′m 0

and s′′n0.

3) If S ′m 0

and S ′′n0

are nilpotent t-conorms with s′m 0and s′′n0

being their respective additive generators, then

I(x, y) ={


for some kx,m 0 ∈ (am 0 , bm 0 ], where Ex(y) is the same as instatement 1).

Proof: (⇒) Since I(x, [am 0 , bm 0 ]) = [cn0 , dn0 ], then we getfrom Lemma 3 that there are only three possible cases.

N1) S ′m 0

is a strict t-conorm, and S ′′n0

is a nilpotent t-conorm.N2) S ′

m 0and S ′′

n0are strict t-conorms.

N3) S ′m 0

and S ′′n0

are nilpotent t-conorms.The proof of 1). Suppose that S ′

m 0is strict and S ′′

n0is

nilpotent, i.e., N1 is satisfied. In line with Lemma 3, thereexist y1 , y2 ∈ (am 0 , bm 0 ) such that I(x, y1) is an idempo-tent element of S2 and I(x, y2) is not an idempotent el-ement of S2 . Let kx,m 0 = inf{y ∈ (am 0 , bm 0 )|I(x, y) is anidempotent element of S2}. Clearly, am 0 < y2 < kx,m 0 ≤

y1 < bm 0 . By the proof of Lemma 3, we get the factthat I(x, [kx,m 0 , bm 0 ]) = dn0 , I(x, am 0 ) = cn0 , and I(x, y) ∈[cn0 , dn0 ] if y ∈ (am 0 , kx,m 0 ).

Define functions ϕ : [am 0 , bm 0 ] → [0, 1] by ϕ(x) =x−am 0

bm 0 −am 0and ψ : [cn0 , dn0 ] → [0, 1] by ψ(x) = x−cn 0

dn 0 −cn 0. It fol-

lows from (2) that for any y, z ∈ (am 0 , kx,m 0 ),

I(x, ϕ−1(s′−1m 0

(s′m 0(ϕ(y)) + s′m 0

(ϕ(z)))))

=ψ−1(s′′−1n0

(min(s′′n0(ψ(I(x, y)))+s′′n0

(ψ(I(x, z))), s′′n0(1)))).

For the arbitrarily fixed x ∈ [0, 1], define a function Ix :[am 0 , bm 0 ] → [cn0 , dn0 ] given by Ix(y) = I(x, y). By routinesubstitutions, hx = s′′n0

◦ ψ ◦ Ix ◦ ϕ−1 ◦ s′−1m 0

, u = s′m 0◦ ϕ(y),

and v = s′m 0◦ ϕ(z), from the above equation, we obtain that

hx(u + v) = min(hx(u) + hx(v), s′′n0(1)) (5)

for any u, v ∈ (0, s′m 0(ϕ(kx,m 0 ))), where hx : [0,∞] →

[0, s′′n0(1)] is a continuous and increasing function.

By simple computation, we have that hx(0) = 0 andhx(s′m 0

(ϕ(kx,m 0 ))) = s′′n0(1). Then hx(t) = s′′n0

(1) for anyt ≥ s′m 0

(ϕ(kx,m 0 )) since hx is increasing. It is easy to checkthat hx satisfies the following functional equation:

hx(t1 + t2) = min(hx(t1) + hx(t2), s′′n0(1))

for all t1 , t2 ∈ [0,∞].From Theorem 3, we get that for any t ∈ [0,∞], hx(t) =

s′′n0(1); or hx(t) = 0; or hx(t) =

{0, t = 0s′′n0

(1), t ∈ (0,∞]; or

hx(t)={0, t ∈ (0,∞]s′′n0

(1), t = ∞; or hx(t)=min(px,m 0 t, s′′n0

(1))

with some constant px,m 0 ∈ (0,∞). Because of thecontinuity of hx in (0, s′m 0

(ϕ(kx,m 0 ))), then for anyt ∈ (0, s′m 0

(ϕ(kx,m 0 ))), hx(t) = s′′n0(1), hx(t) = 0, or hx(t)

= min(px,m 0 t, s′′n0

(1)). This indicates that for anyy ∈ (am 0 , kx,m 0 ), I(x, y) = dn0 , I(x, y) = cn0 , or I(x, y) =ψ−1 ◦ s′′−1

n0(min(px,m 0 s

′m 0

◦ ϕ(y), s′′n0(1))).

Again observing that I(x, y) is not an idempotent ele-ment of S2 for any y ∈ (am 0 , kx,m 0 ), we have I(x, y) =ψ−1 ◦ s′′−1

n0(px,m 0 s

′m 0

◦ ϕ(y)) and px,m 0 s′m 0

◦ ϕ(y) < s′′n0(1)

for any y ∈ (am 0 , kx,m 0 ). Therefore, px,m 0 ≤ s ′′n 0

(1)s ′

m 0◦ϕ(kx , m 0 ) . In

the following, we will prove that px,m 0 =s ′′

n 0(1)

s ′m 0

◦ϕ(kx , m 0 ) .

In fact, according to the continuity of I(x, ·) on [am 0 , bm 0 ],we get that limy→kx , m 0

− I(x, y) = I(x, kx,m 0 ), which shows

that ψ−1 ◦ s′′−1n0

(px,m 0 s′m 0

◦ ϕ(kx,m 0 )) = dn0 . Thus, px,m 0 =s ′′

n 0(1)

s ′m 0

◦ϕ(kx , m 0 ) . Therefore, 1) is derived.

The proof of 2). Suppose both S ′m 0

and S ′′n0

are strict, i.e.,N2 is satisfied. We get from Lemma 3 that I(x, y) is a non-idempotent element of S2 for any y ∈ (am 0 , bm 0 ). Obviously,I(x, am 0 ) = cn0 and I(x, bm 0 ) = dn0 .

For any y, z ∈ (am 0 , bm 0 ), similarly to the proof of 1), weget the equation

hx(u + v) = hx(u) + hx(v), u, v ∈ (0,∞)

where hx : [0,∞] → [0,∞] is given by hx = s′′n0◦ ψ ◦ Ix ◦

ϕ−1 ◦ s′−1m 0

, u = s′m 0◦ ϕ(y), and v = s′m 0

◦ ϕ(z). Similarly


to the proof of 1), we can use Theorem 2 to solvethe above equation and get that I(x, y) = cn0 + (dn0 −cn0 )s

′′−1n0

(px,m 0 s′m 0

( y−am 0bm 0 −am 0

)) with some constant px,m 0 ∈(0,∞).

In addition, similar to the proof of Theorem 10 in [22], it canbe proved that the constant px,m 0 is uniquely determined up to apositive multiplicative constant depending on constants for s′m 0

and s′′n0.

The proof of 3). Suppose S ′m 0

and S ′′n0

are nilpotent, i.e., N3is satisfied. According to Lemma 3, there exist two possiblecases: One case is that I(x, y) is not an idempotent element ofS2 for any y ∈ (am 0 , bm 0 ), and the other case is that there existsome y, z ∈ (am 0 , bm 0 ) such that I(x, y) is a non-idempotentelement of S2 and I(x, z) is an idempotent element of S2 .

If the case is the former, then just like the proofof 2), we can use Theorem 4 to solve the corre-sponding equation and then get I(x, y) = cn0 + (dn0 −cn0 )s

′′−1n0

(px,m 0 s′m 0

( y−am 0bm 0 −am 0

)), where px,m 0 =s ′′

n 0(1)

s ′m 0

(1) .

If the case is the latter, we also can obtain, similarly to theproof of 1), that

I(x, y) ={


for some kx,m 0 ∈ (am 0 , bm 0 ).By combining the two cases, we get 3).(⇐) Suppose the conditions in 1) are satisfied. It is clear

to get that I(x, ·) is increasing and continuous on [am 0 , bm 0 ]and I(x, [am 0 , bm 0 ]) = [cn0 , dn0 ]. In the following, we willshow that I(x, S1(y, z)) = S2(I(x, y), I(x, z)) for all y, z ∈[am 0 , bm 0 ].

Case 1. If at least one of y, z is an idempotent ele-ment of S1 , without loss of generality, we suppose y =am 0 . Therefore, I(x, y) = cn0 is an idempotent element ofS2 . Thus, we obtain that I(x, S1(y, z)) = I(x,max(y, z)) =max(I(x, y), I(x, z)) = S2(I(x, y), I(x, z)).

Case 2. If y, z ∈ (am 0 , bm 0 ), then it holds that S1(y, z) ∈(am 0 , bm 0 ) since S ′

m 0is strict.

Case 2.1. If S1(y, z) ∈ (am 0 , kx,m 0 ), then we obtainthat kx,m 0 > S1(y, z) ≥ max(y, z), and consequently, itholds that y, z < kx,m 0 . Therefore, I(x, y)=cn0 +(dn0 − cn0 )

s′′−1n0

(s ′′

n 0(1)

s ′m 0

(k x , m 0 −a m 0

b m 0 −a m 0)s′m 0

( y−am 0bm 0−am 0

))∈ [cn0 , dn0 ] and I(x, z)=

cn0 + (dn0 − cn0 )s′′−1n0

(s ′′

n 0(1)

s ′m 0

(k x , m 0 −a m 0

b m 0 −a m 0)s′m 0

( z−am 0bm 0 −am 0

)) ∈ [cn0 ,

dn0 ]. Then we can have forms (6) and (7), shown at the bottomof the page.

Moreover, we get from S1(y, z) < kx,m 0 that am 0 +(bm 0 −am 0 )(s

′−1m 0

(s′m 0( y−am 0

bm 0 −am 0)+s′m 0

( z−am 0bm 0 −am 0

)) < kx,m 0 , whichshows that

s′m 0

(y − am 0

bm 0 − am 0

)+ s′m 0

(z − am 0

bm 0 − am 0

)< s′m 0

(kx,m 0 − am 0

bm 0 − am 0

)

and then that

s′′n0(1)

s′m 0

(kx , m 0 −am 0bm 0 −am 0

)(

s′m 0

(y − am 0

bm 0 − am 0

)+s′m 0

(z − am 0

bm 0 − am 0

))

<s′′n0

(1)

s′m 0

(kx , m 0 −am 0bm 0 −am 0

)s′m 0

(kx,m 0 − am 0

bm 0 − am 0

)= s′′n0

(1).

Therefore, S2(I(x, y), I(x, z)) = cn0 + (dn0 − cn0 )(s′′−1n0

(s ′′

n 0(1)

s ′m 0

(k x , m 0 −a m 0

b m 0 −a m 0)(s′m 0

( y−am 0bm 0 −am 0

) + s′m 0( z−am 0

bm 0 −am 0))). Thus,

S2(I(x, y), I(x, z)) = I(x, S1(y, z)) holds.Case 2.2. If S1(y, z) ∈ [kx,m 0 , bm 0 ), it follows that I(x, S1

(y, z)) = dn0 . In this case, if y, z ∈ (am 0 , kx,m 0 ), thenS2(I(x, y), I(x, z)) has form (7). Again taking into account

I(x, S1(y, z)) = I

(x, am 0 + (bm 0 − am 0 )

(s′−1

m 0

(s′m 0

(y − am 0

bm 0 − am 0

)+ s′m 0

(z − am 0

bm 0 − am 0

))))

= cn0 + (dn0 − cn0 )(s′′−1n0

⎛⎝ s′′n0

(1)

s′m 0

(kx , m 0 −am 0bm 0 −am 0

)(

s′m 0

(y − am 0

bm 0 − am 0

)+ s′m 0

(z − am 0

bm 0 − am 0

))⎞⎠ (6)

S2(I(x, y), I(x, z)) = cn0 + (dn0 − cn0 )S′′n0

(I(x, y) − cn0

dn0 − cn0

,I(x, z) − cn0

dn0 − cn0

)

= cn0 + (dn0 − cn0 )(

s′′−1n0

(min

(s′′n0

(I(x, y) − cn0

dn0 − cn0

)+ s′′n0

(I(x, z) − cn0

dn0 − cn0

), s′′n0

(1))))

= cn0 + (dn0 − cn0 )

×

⎛⎝s′′−1

n0

⎛⎝min

⎛⎝ s′′n0

(1)

s′m 0

(kx , m 0 −am 0bm 0 −am 0

)(

s′m 0

(y − am 0

bm 0 − am 0

)+ s′m 0

(z − am 0

bm 0 − am 0

)), s′′n0

(1)

⎞⎠

⎞⎠

⎞⎠ . (7)


that S1(y, z) ≥ kx,m 0 , we can easily obtain

s′′n0(1)

s′m 0

(kx , m 0 −am 0bm 0 −am 0

)(s′m 0

(y − am 0

bm 0 − am 0

)+ s′m 0

(z − am 0

bm 0 − am 0

))

≥s′′n0

(1)

s′m 0

(kx , m 0 −am 0bm 0 −am 0

)s′m 0

(kx,m 0 − am 0

bm 0 − am 0

)= s′′n0

(1).

Therefore, it holds that S2(I(x, y), I(x, z)) = dn0 = I(x,S1(y, z)). In this case, if at least one of y and z is greaterthan kx,m 0 , without loss of generality, we suppose thaty ≥ kx,m 0 and then I(x, y) = dn0 . Hence, S2(I(x, y), I(x, z))= S2(dn0 , I(x, z)) = dn0 . As a consequence, S2(I(x, y),I(x, z)) = dn0 = I(x, S1(y, z)).

If the conditions in 2) or 3) are satisfied, we can similarly getthe result. �

Definition 4: Let I : [0, 1]2 → [0, 1] be a binary func-tion, and S1 and S2 be two continuous t-conorms given by(〈am , bm , S ′

m 〉)m∈J1 and (〈cn , dn , S ′′n 〉)n∈J2 , respectively. c, d

(c ≤ d) are two idempotent elements of S2 . Define ConditionM as follows.

M1) I is continuous on [0, 1]2 .M2) I is increasing with respect to the second place.M3) I(x, S1(y, z)) = S2(I(x, y), I(x, z)) for any x, y, z ∈

[0, 1].M4) I(1, 0) = c, I(1, 1) = d.Theorem 6: Let I, S1 , S2 , and M be the same as

above. Suppose that there is not any summand of S2in [c, d]. Then I satisfies Condition M if and only iffor any x, y ∈ [0, 1], I(x, y) = f(x, y), where f : [0, 1]2 →[sup{dn |dn ≤ c, n ∈ J2}, inf{cn |cn ≥ d, n ∈ J2}] is a con-tinuous function, which is increasing in its second placeand satisfies f(1, 0) = c and f(1, 1) = d. In addition, forany y ∈ [am , bm ] (m ∈ J1), f(x, y) = em (x), em : [0, 1] →[sup{dn |dn ≤ c, n ∈ J2}, inf{cn |cn ≥ d, n ∈ J2}] is a contin-uous function.

Proof: (⇒) We will show that Range(I)⊆ [sup{dn |dn ≤c, n ∈ J2}, inf{cn |cn ≥ d, n ∈ J2}]. In fact, it can be provedthat for any x ∈ [0, 1] and y ∈ [am , bm ] (any m ∈ J1), I(x, y)is an idempotent element of S2 . Otherwise, suppose thatthere exist some x0 ∈ [0, 1], y0 ∈ [am 0 , bm 0 ] (some m0 ∈ J1)such that I(x0 , y0) is a non-idempotent element of S2 .From Lemma 3, we can get that there exists some n0 ∈ J2such that I(x0 , [am 0 , bm 0 ]) = [cn0 , dn0 ]. Considering the con-tinuity of I and using the result of Lemma 5, we havethat I(x, [am 0 , bm 0 ]) = [cn0 , dn0 ] for any x ∈ [0, 1]. Espe-cially, I(1, [am 0 , bm 0 ]) = [cn0 , dn0 ]. Therefore, [cn0 , dn0 ] ⊆[c, d], which is a contradiction with the premise in the theo-rem. Because function I is continuous, we get that Range(I)⊆[sup{dn |dn ≤ c, n ∈ J2}, inf{cn |cn ≥ d, n ∈ J2}].

Denote f = I . f is continuous because I is continuous.f(1, 0) = c and f(1, 1) = d since I(1, 0) = c and I(1, 1) = d.In line with Corollary 1, we can get that f(x, y) = em (x) forany y ∈ [am , bm ] (m ∈ J1), where em : [0, 1] → [sup{dn |dn ≤c, n ∈ J2}, inf{cn |cn ≥ d, n ∈ J2}].

(⇐) It is obvious that I is continuous and is increasing in itssecond place and satisfies I(1, 0) = c, I(1, 1) = d. In the fol-

lowing, we will check (2). Clearly, the range of I only containsidempotent elements of S2 . Therefore, S2(I(x, y), I(x, z)) =max(I(x, y), I(x, z)).

If at least one of y and z is an idempotent elementof S1 , then it holds that I(x, S1(y, z)) = I(x,max(y, z)) =max(I(x, y), I(x, z)) = S2(I(x, y), I(x, z)).

If y, z are non-idempotent elements of S1 such thaty ∈ [am 1 , bm 1 ], z ∈ [am 2 , bm 2 ] (m1 �= m2 and m1 ,m2 ∈J1), then we have I(x, S1(y, z)) = I(x,max(y, z)) =max(I(x, y), I(x, z)) = S2(I(x, y), I(x, z)).

If y, z are non-idempotent elements of S1 such that y, z ∈[am 3 , bm 3 ] (some m3 ∈ J1), then S1(y, z) ∈ [am 3 , bm 3 ]. Con-sequently, I(x, S1(y, z)) = em 3 (x) and S2(I(x, y), I(x, z)) =S2(em 3 (x), em 3 (x)) = em 3 (x).

Thus, I(x, S1(y, z))= S2(I(x, y), I(x, z)) for any x, y, z ∈[0, 1]. �

If c = d in the above theorem, then I = c is a particularsolution to (2). The following corollary is immediately obtained.

Corollary 2: If c = d, then I(x, y) = c (x, y ∈ [0, 1]) is asolution to (2).

Theorem 6 only considers the case in which there is no sum-mand of S2 in the interval [I(1, 0), I(1, 1)]. In the following,we will discuss the case in which there exist summands of S2in the interval [I(1, 0), I(1, 1)]. First, we introduce a concept offeasible correspondence.

Definition 5: Let S1 and S2 be two continuous t-conormsgiven by (〈am , bm , S ′

m 〉)m∈J1 and (〈cn , dn , S ′′n 〉)n∈J2 , re-

spectively. Denote A = {〈am , bm , S ′m 〉|m ∈ J1} and B =

{〈cn , dn , S ′′n 〉|n ∈ J2}.

Define a partial order “ ≤1 ” on A as: 〈am 1 , bm 1 , S′m 1

〉 ≤1〈am 2 , bm 2 , S

′m 2

〉 if and only if am 1 ≤ am 2 (m1 ,m2 ∈ J1).Define a partial order “ ≤2 ” on B as: 〈cn1 , dn1 , S

′′n1〉 ≤2

〈cn2 , dn2 , S′′n2〉 if and only if cn1 ≤ cn2 (n1 , n2 ∈ J2).

Denote 〈am 1 , bm 1 , S′m 1

〉 <1 〈am 2 , bm 2 , S′m 2

〉 if 〈am 1 , bm 1 ,S ′

m 1〉 ≤1 〈am 2 , bm 2 , S

′m 2

〉 and 〈am 1 , bm 1 , S′m 1

〉 �= 〈am 2 , bm 2 ,S ′

m 2〉. Denote 〈cn1 , dn1 , S

′′n1〉 <2 〈cn2 , dn2 , S

′′n2〉 if 〈cn1 ,

dn1 , S′′n1〉 ≤2 〈cn2 , dn2 , S

′′n2〉 and 〈cn1 , dn1 , S

′′n1〉 �= 〈cn2 ,

dn2 , S′′n2〉.

Remark 3: Because (am , bm ) (m ∈ J1) are nonoverlappingopen intervals of [0, 1], (A,≤1) is a linearly ordered set. Simi-larly, (B,≤2) is a linearly ordered set.

Definition 6: Suppose that A′ is a subset of A and B′ is a sub-set of B. Two summands 〈am 1 , bm 1 , S

′m 1

〉, 〈am 2 , bm 2 , S′m 2

〉 ∈A satisfying 〈am 1 , bm 1 , S

′m 1

〉 <1 〈am 2 , bm 2 , S′m 2

〉 are calledconsecutive summands in A if there is no sum-mand 〈am 3 , bm 3 , S

′m 3

〉 ∈ A such that 〈am 1 , bm 1 , S′m 1

〉 <1〈am 3 , bm 3 , S

′m 3

〉 <1 〈am 2 , bm 2 , S′m 2

〉.Two summands 〈am 1 , bm 1 , S

′m 1

〉, 〈am 2 , bm 2 , S′m 2

〉 ∈ A′

satisfying 〈am 1 , bm 1 , S′m 1

〉 <1 〈am 2 , bm 2 , S′m 2

〉 are calledconsecutive summands in A′ if there is no sum-mand 〈am 3 , bm 3 , S

′m 3

〉 ∈ A′ such that 〈am 1 , bm 1 , S′m 1

〉 <1〈am 3 , bm 3 , S

′m 3

〉 <1 〈am 2 , bm 2 , S′m 2

〉.The consecutive summands in B and B′ are defined similarly.Remark 4: Two consecutive summands in A′ may be not

consecutive in A.Now, we introduce the concept of feasible correspondence

which plays a central role in describing the solvability of (2).


Definition 7: Suppose that c and d (c < d) are two idempotentelements of S2 . Let B′ = {〈cn , dn , S ′′

n 〉|[cn , dn ] ⊆ [c, d]} �= ∅be a subset of B and A′ = {〈am , bm , S ′

m 〉|m ∈ J ′1 ⊆ J1} be a

subset of A satisfying card(A′)=card(B′). Then a mapping F :A′ → B′ is called a feasible correspondence from A′ to B′ ifthe following conditions are satisfied.

C1) F is a one to one correspondence from A′ to B′.C2) 〈am 1 , bm 1 , S

′m 1

〉 <1 〈am 2 , bm 2 , S′m 2

〉 if and only ifF (〈am 1 , bm 1 , S

′m 1

〉) <2 F (〈am 2 , bm 2 , S′m 2

〉) (m1 ,m2 ∈ J ′1).

C3) Suppose 〈cn1 , dn1 , S′′n1〉 and 〈cn2 , dn2 , S

′′n2〉 are two

consecutive summands in B′ satisfying 〈cn1 , dn1 , S′′n1〉 <2

〈cn2 , dn2 , S′′n2〉, F (〈am 1 , bm 1 , S

′m 1

〉) = 〈cn1 , dn1 , S′′n1〉 and

F (〈am 2 , bm 2 , S′m 2

〉) = 〈cn2 , dn2 , S′′n2〉 (m1 ,m2 ∈ J ′

1). Ifdn1 < cn2 , then bm 1 < am 2 and ∪

〈am ,bm ,S ′m 〉∈M

[am , bm ] �=

[bm 1 , am 2 ], where M = {〈am , bm , S ′m 〉 ∈ A|〈am 1 , bm 1 , S

′m 1

〉<1 〈am , bm , S ′

m 〉 <1 〈am 2 , bm 2 , S′m 2

〉} and ∪〈am ,bm ,S ′

m 〉∈M

[am , bm ] is the union of the set [am , bm ]m∈J1 satisfying〈am , bm , S ′

m 〉 ∈ M .C4) If F (〈am , bm , S ′

m 〉) = 〈cn , dn , S ′′n 〉 and S ′′

n is a strict t-conorm, then S ′

m is a strict t-conorm (m ∈ J ′1 , 〈cn , dn , S ′′

n 〉 ∈B′).

Example 2:1) Suppose that S1 = (〈 1

5 , 25 , SP 〉, 〈 2

5 , 35 , SP 〉, 〈 2

3 , 34 ,

SP 〉, 〈 34 , 5

6 , SL 〉) and S2 = (〈 18 , 1

4 , SP 〉, 〈 13 , 1

2 , SP 〉, 〈 12 , 2

3 ,SL 〉). Choose c = 1

16 and d = 56 . Then there are only two fea-

sible correspondences F1 and F2 .1.1)

A′ ={⟨

15,25, SP

⟩,

⟨23,34, SP

⟩,

⟨34,56, SL

⟩}

F1

(⟨15,25, SP

⟩)=

⟨18,14, SP

⟩

F1

(⟨23,34, SP

⟩)=

⟨13,12, SP

⟩

and F1

(⟨34,56, SL

⟩)=

⟨12,23, SL

⟩.

1.2)

A′ ={⟨

25,35, SP

⟩,

⟨23,34, SP

⟩,

⟨34,56, SL

⟩}

F2

(⟨25,35, SP

⟩)=

⟨18,14, SP

⟩

F2

(⟨23,34, SP

⟩)=

⟨13,12, SP

⟩

and F2

(⟨34,56, SL

⟩)=

⟨12,23, SL

⟩.

The plots of F1 and F2 are presented in Fig. 1.2) If S1 = (〈 1

6 , 13 , SL 〉) and S2 = (〈 1

3 , 12 , SP 〉). Choose c = 0

and d = 1. Obviously, B′ = B = {〈 13 , 1

2 , SP 〉} and A′ = A ={〈 1

6 , 13 , SL 〉}. There does not exist any feasible correspondence

F from A to B since C4 cannot be satisfied for F .

SL

SP

SP

SP

SP

SP

SL

1/4

1/8

1/16

1/3

1/2

2/3

5/6

3/4

5/6

2/3

3/5

2/5

1/5

F1

SL

SP

SP

SP

SP

SP

SL

1/4

1/8

1/16

1/3

1/2

2/3

5/6

3/4

5/6

2/3

3/5

2/5

1/5

F2

Fig. 1. Plots of F1 and F2 .

The following theorem shows that the existence of feasiblecorrespondences is necessary for the solvability of (2).

Theorem 7: Let S1 and S2 be two continuous t-conorms givenby (〈am , bm , S ′


and I : [0, 1]2 → [0, 1] be a binary function with I(x, ·) be-ing increasing and continuous for each x ∈ [0, 1]. Suppose cand d (c < d) are two idempotent elements of S2 and B′ ={〈cn , dn , S ′′

n 〉|[cn , dn ] ⊆ [c, d]} �= ∅. If I satisfies I(1, 0) = cand I(1, 1) = d and (2) holds, then there exist a subset A′ of Asatisfying card(A′) = card(B′) and a feasible correspondence Ffrom A′ to B′.

Proof: Because I(1, 0) = c, I(1, 1) = d and I(1, ·) is contin-uous, then for any 〈cn , dn , S ′′

n 〉 ∈ B′, there exists a y ∈ [0, 1]such that I(1, y) ∈ (cn , dn ). It follows from Lemma 1 that yis not an idempotent element of S2 . Therefore, there existsan (am , bm ) (some m ∈ J1) such that y ∈ (am , bm ). FromLemma 3, we obtain that I(1, [am , bm ]) = [cn , dn ]. Just like


the proof of Lemma 4, it is easy to prove that [am ′ , bm ′ ] =[am , bm ] if I(1, [am ′ , bm ′ ]) = [cn , dn ] (m′ ∈ J1). Therefore, forany 〈cn , dn , S ′′

n 〉 ∈ B′, there is a unique [am , bm ](m ∈ J1) suchthat I(1, [am , bm ]) = [cn , dn ].

Let us define A′ = {〈am , bm , S ′m 〉 ∈ A| there exists a

〈cn , dn , S ′′n 〉 ∈ B′ such that I(1, [am , bm ]) = [cn , dn ]}.

Clearly, A′ ⊆ A and card(A′)=card(B′).Define a mapping F : A′ → B′ by

F (〈am , bm , S ′m 〉) = 〈cn , dn , S ′′

n 〉 if I(1, [am , bm ]) = [cn , dn ].

It is obvious that F is a one-to-one correspondence from A′

to B′. Therefore, C1 is satisfied.If 〈am 1 , bm 1 , S

′m 1

〉 and 〈am 2 , bm 2 , S′m 2

〉 are two sum-mands in A′ satisfying 〈am 1 , bm 1 , S

′m 1

〉 <1 〈am 2 , bm 2 ,S ′

m 2〉, F (〈am 1 , bm 1 , S

′m 1

〉) = 〈cn1 , dn1 , S′′n1〉 and F (〈am 2 ,

bm 2 , S′m 2

〉) = 〈cn2 , dn2 , S′′n2〉, then am 1 < am 2 , I(1, [am 1 ,

bm 1 ]) = [cn1 , dn1 ] and I(1, [am 2 , bm 2 ]) = [cn2 , dn2 ]. UsingLemma 4, we have cn1 < cn2 . Hence, F (〈am 1 , bm 1 , S

′m 1

〉) <2F (〈am 2 , bm 2 , S

′m 2

〉). Because F is a bijection, we haveF (〈am 1 , bm 1 , S

′m 1

〉) <2 F (〈am 2 , bm 2 , S′m 2

〉) if and only if〈am 1 , bm 1 , S

′m 1

〉 <1 〈am 2 , bm 2 , S′m 2

〉. Thus, C2 is satisfied.Now, we will prove C3. Suppose that 〈cn1 , dn1 , S

′′n1〉

and 〈cn2 , dn2 , S′′n2〉 are two consecutive summands in B′

with dn1 < cn2 , F (〈am 1 , bm 1 , S′m 1

〉) = 〈cn1 , dn1 , S′′n1〉, and

F (〈am 2 , bm 2 , S′m 2

〉) = 〈cn2 , dn2 , S′′n2〉. It holds that bm 1 ≤

am 2 because I(1, bm 1 ) = dn1 < cn2 = I(1, am 2 ) and I(1, ·)is increasing. Suppose bm 1 = am 2 . Then dn1 = I(1, bm 1 ) =I(1, am 2 ) = cn2 . This contradicts the assumption that dn1 <cn2 . Therefore, bm 1 < am 2 .

Moreover, if ∪〈am ,bm ,S ′

m 〉∈M[am , bm ] = [bm 1 , am 2 ], then for

each 〈am , bm , S ′m 〉 ∈ M , we have from Lemma 1

and Corollary 1 that there exists a unique e1,m ∈[dn1 , cn2 ] such that I(1, [am , bm ]) = e1,m . Therefore,I(1, ∪


[am , bm ]) = ∪〈am ,bm ,S ′

m 〉∈Me1,m is a finite

or countable infinite set. Then we have I(1, [bm 1 , am 2 ]) �=[dn1 , cn2 ]. This contradicts the continuity of I(1, ·). Hence, weobtain that ∪


[am , bm ] �= [bm 1 , am 2 ].

Lemma 3 implies that C4 is satisfied. �Definition 8: Let I : [0, 1]2 → [0, 1] be a binary func-

tion, and S1 and S2 be two continuous t-conorms given by(〈am , bm , S ′

m 〉)m∈J1 and (〈cn , dn , S ′′n 〉)n∈J2 , respectively. Sup-

pose A′ = {〈am , bm , S ′m 〉|m ∈ J ′

1 ⊆ J1} is a subset of A,B′ ={〈cn , dn , S ′′

n 〉|[cn , dn ] ⊆ [c, d]} �= ∅ (c and d are two idempotentelements of S2 with c < d) and F is a feasible correspondencefrom A′ to B′. Define Condition L as follows:

L1) I is continuous on [0, 1]2 .L2) I is increasing with respect to the second place.L3) I(x, S1(y, z)) = S2(I(x, y), I(x, z)) for any x, y, z ∈

[0, 1].L4) I(1, [am , bm ]) = [cn , dn ] if and only if

F (〈am , bm , S ′m 〉) = 〈cn , dn , S ′′

n 〉 (m ∈ J ′1 , 〈cn , dn , S ′′

n 〉 ∈ B′).L5) I(1, 0) = c, I(1, 1) = d.The following proposition shows that the existence of feasible

correspondences is also sufficient for the solvability of (2).

Proposition 2: Let I, S1 , S2 , and L be the same as in Defi-nition 8. If I satisfies Condition L, then we have the followingstatements.

1) For each 〈am , bm , S ′m 〉 ∈ A′ with F (〈am , bm , S ′

m 〉) =〈cn , dn , S ′′

n 〉, I has one of the following forms on [0, 1] ×(am , bm ):

1.1) If S ′m is a strict t-conorm and S ′′

n is a nilpotent t-conormwith s′m and s′′n being their respective additive generators, then

I(x, y) ={

E(x, y), y ∈ (am , km (x))dn , y ∈ [km (x), bm )

(8)

where km : [0, 1] → (am , bm ) is a continuous function, andE(x, y) = cn + (dn − cn )s′′n

−1( s ′′n (1)

s ′m ( k m (x )−a m

b m −a m)s′m ( y−am

bm −am)).

1.2) If S ′m and S ′′

n are nilpotent t-conorms with s′m and s′′nbeing their respective additive generators, then

I(x, y) ={


(9)

where km : [0, 1] → (am , bm ] is a continuous function, andE(x, y) is the same as in 1.1).

1.3) If S ′m and S ′′

n are strict t-conorms with s′m and s′′n beingtheir respective additive generators, then

I(x, y) = cn + (dn − cn )s′′n−1

(pm (x)s′m

(y − am

bm − am

))

(10)where pm : [0, 1] → (0,∞) is a continuous function.

2) Suppose that 〈am 1 , bm 1 , S′m 1

〉 and 〈am 2 , bm 2 , S′m 2

〉 aretwo consecutive summands in A′ satisfying 〈am 1 , bm 1 , S

′m 1

〉 <1〈am 2 , bm 2 , S

′m 2

〉, F (〈am 1 , bm 1 , S′m 1

〉) = 〈cn1 , dn1 , S′′n1〉 and

F (〈am 2 , bm 2 , S′m 2

〉) = 〈cn2 , dn2 , S′′n2〉. Then I has one of the

following forms on [0, 1] × [bm 1 , am 2 ].2.1) If dn1 = cn2 , then I(x, y) = dn1 .2.2) If dn1 < cn2 , then I(x, y) = fm 1 ,m 2 (x, y), where

fm 1 ,m 2 : [0, 1] × [bm 1 , am 2 ] → [dn1 , cn2 ] is a continuous func-tion. It is increasing in its second place and satisfiesfm 1 ,m 2 (x, bm 1 ) = dn1 and fm 1 ,m 2 (x, am 2 ) = cn2 for anyx ∈ [0, 1]. In addition, for any [am , bm ] (m ∈ J1) satisfy-ing 〈am 1 , bm 1 , S

′m 1

〉 <1 〈am , bm , S ′m 〉 <1 〈am 2 , bm 2 , S

′m 2

〉, itholds that fm 1 ,m 2 (x, y) = em (x) on [0, 1] × [am , bm ], whereem : [0, 1] → [dn1 , cn2 ] is a continuous function.

3) Define ia = inf{am |〈am , bm , S ′m 〉 ∈ A′} and ic =

inf{cn |〈cn , dn , S ′′n 〉 ∈ B′}. Then for any (x, y) ∈

[0, 1] × [0, ia ], I(x, y) = g(x, y), where g : [0, 1] × [0, ia ] →[sup{dn |[cn , dn ] ⊆ [0, c], 〈cn , dn , S ′′

n 〉 ∈ B}, ic ] is a con-tinuous function. It is increasing in its second place andsatisfies g(1, 0) = c and g(x, ia) = ic for any x ∈ [0, 1].In addition, for any [am , bm ] ⊆ [0, ia ] (〈am , bm , S ′

m 〉 ∈ A),it holds that g(x, y) = e′m (x) on [0, 1] × [am , bm ], wheree′m : [0, 1] → [sup{dn |[cn , dn ] ⊆ [0, c], 〈cn , dn , S ′′

n 〉 ∈ B}, ic ]is a continuous function.

4) Define sb = sup{bm |〈am , bm , S ′m 〉 ∈ A′} and sd =

sup{dn |〈cn , dn , S ′′n 〉 ∈ B′}. Then for any (x, y) ∈

[0, 1] × [sb , 1], I(x, y) = h(x, y), where h : [0, 1] × [sb , 1] →[sd, inf{cn |[cn , dn ] ⊆ [d, 1], 〈cn , dn , S ′′

n 〉 ∈ B}] is a con-tinuous function. It is increasing in its second place and


satisfies h(1, 1) = d and h(x, sb) = sd for any x ∈ [0, 1].In addition, for any [am , bm ] ⊆ [sb , 1] (〈am , bm , S ′

m 〉 ∈ A),it holds that h(x, y) = e′′m (x) on [0, 1] × [am , bm ], wheree′′m : [0, 1] → [sd, inf{cn |[cn , dn ] ⊆ [d, 1], 〈cn , dn , S ′′

n 〉 ∈ B}]is a continuous function.

Proof: The proof of 1). I(1, [am , bm ]) = [cn , dn ] sinceF (〈am , bm , S ′

m 〉) = 〈cn , dn , S ′′n 〉. By Lemma 5, we obtain that

I(x, [am , bm ]) = [cn , dn ] for any x ∈ [0, 1]. Theorem 5 and thecontinuity of function I immediately imply 1).

The proof of 2). Suppose that 〈am 1 , bm 1 , S′m 1

〉 and〈am 2 , bm 2 , S

′m 2

〉 ∈ A′ are two consecutive summandssatisfying 〈am 1 , bm 1 , S

′m 1

〉 <1 〈am 2 , bm 2 , S′m 2

〉, F (〈am 1 , bm 1 ,S ′

m 1〉) = 〈cn1 , dn1 , S

′′n1〉, and F (〈am 2 , bm 2 , S

′m 2

〉) =〈cn2 , dn2 , S

′′n2〉. Then we get that I(1, [am 1 , bm 1 ]) = [cn1 , dn1 ]

and I(1, [am 2 , bm 2 ]) = [cn2 , dn2 ], which together withLemma 5 indicates that I(x, [am 1 , bm 1 ]) = [cn1 , dn1 ] andI(x, [am 2 , bm 2 ]) = [cn2 , dn2 ] for any x ∈ [0, 1]. As a con-sequence, I(x, am 2 ) = cn2 and I(x, bm 1 ) = dn1 for anyx ∈ [0, 1]. By the continuity of I , we have I(x, y) ∈ [dn1 , cn2 ]for any x ∈ [0, 1] and y ∈ [bm 1 , am 2 ].

In this case, if dn1 = cn2 , then clearly it holds that I(x, y) =dn1 , x ∈ [0, 1], y ∈ [bm 1 , am 2 ].

If dn1 < cn2 , then denote fm 1 ,m 2 = I |[0,1]×[bm 1 ,am 2 ] . Theproperties of I imply that fm 1 ,m 2 : [0, 1] × [bm 1 , am 2 ] →[dn1 , cn2 ] is a continuous function which is increasing inits second place and satisfies fm 1 ,m 2 (x, bm 1 ) = dn1 andfm 1 ,m 2 (x, am 2 ) = cn2 for any x ∈ [0, 1].

We can prove that [dn1 , cn2 ] only contains idempotent el-ements of S2 . In fact, according to C1 and C2, we knowthat 〈cn1 , dn1 , S

′′n1〉 and 〈cn2 , dn2 , S

′′n2〉 are two consecu-

tive summands in B′. By the definition of B′, we have〈cn1 , dn1 , S

′′n1〉 and 〈cn2 , dn2 , S

′′n2〉 are also two consecutive

summands in B. Therefore, [dn1 , cn2 ] contains only idempo-tent elements of S2 . Then for any [am , bm ] (m ∈ J1) satis-fying 〈am 1 , bm 1 , S

′m 1

〉 <1 〈am , bm , S ′m 〉 <1 〈am 2 , bm 2 , S

′m 2

〉,we have from Corollary 1 that I(x, [am , bm ]) = em (x) ∈[dn1 , cn2 ]. Therefore, fm 1 ,m 2 (x, y) = em (x) on [0, 1] ×[am , bm ], where em : [0, 1] → [dn1 , cn2 ] is a continuous func-tion. Thus, 2) is derived.

The proof of 3). It is clear that ic ≥ c ≥ sup{dn |[cn , dn ] ⊆[0, c], 〈cn , dn , S ′′

n 〉 ∈ B}.For any 〈am , bm , S ′

m 〉 ∈ A′, there exists some〈cn , dn , S ′′

n 〉 ∈ B′ such that F (〈am , bm , S ′m 〉) = 〈cn , dn , S ′′

n 〉.Thus, I(1, [am , bm ]) = [cn , dn ]. Lemma 5 indicates thatI(x, [am , bm ]) = [cn , dn ] and I(x, am ) = cn for any x ∈ [0, 1].Because I is continuous and F is a one to one mapping, wehave that

I(x, ia) = I(x, inf{am |〈am , bm , S ′m 〉 ∈ A′})

= inf{I(x, am )|〈am , bm , S ′m 〉 ∈ A′}

= inf{cn |〈cn , dn , S ′′n 〉 ∈ B′} = ic

which indicates that I(x, ia) = ic and I(x, y) ≤ I(x, ia) = icfor any x ∈ [0, 1] and y ∈ [0, ia ].

We will show that if x ∈ [0, 1] and y ∈ [0, ia ], then I(x, y) ≥sup{dn |[cn , dn ] ⊆ [0, c], 〈cn , dn , S ′′

n 〉 ∈ B}. First, let us provethat for any x ∈ [0, 1] and y ∈ [0, ia ], I(x, y) is an idempotentelement of S2 .

Otherwise, suppose that there exist some x0 ∈ [0, 1] andy0 ∈ [0, ia ] such that I(x0 , y0) is not an idempotent element ofS2 . Obviously, there exists some [am ′ , bm ′ ] (m′ ∈ J1) such thaty0 ∈ (am ′ , bm ′). According to Lemma 3, there exists a unique[cn ′ , dn ′ ] such that I(x0 , [am ′ , bm ′ ]) = [cn ′ , dn ′ ]. It follows fromLemma 5 and the continuity of I that I(x, [am ′ , bm ′ ]) =[cn ′ , dn ′ ] for any x ∈ [0, 1]. Especially, I(1, [am ′ , bm ′ ]) =[cn ′ , dn ′ ]. Considering that I(1, 0) = c and I(1, 1) = d,we get that [cn ′ , dn ′ ] ⊆ [c, d] and then 〈cn ′ , dn ′ , S ′′

n ′ 〉 ∈ B′,which means that cn ′ ≥ ic . Then I(x0 , y0) > cn ′ ≥ ic sinceI(x0 , y0) ∈ [cn ′ , dn ′ ] is not an idempotent element of S2 .However, we have proved that I(x0 , y) ≤ I(x0 , ia) = ic forany y ∈ [0, ia ]. This produces a contradiction. Therefore,I(x, y) is an idempotent element of S2 . Therefore, I(x, y) ≥sup{dn |[cn , dn ] ⊆ [0, c], 〈cn , dn , S ′′

n 〉 ∈ B} if x ∈ [0, 1] andy ∈ [0, ia ].

From the above steps, one can see that for anyx ∈ [0, 1] and any y ∈ [0, ia ], I(x, y) ∈ [sup{dn |[cn , dn ] ⊆[0, c], 〈cn , dn , S ′′

n 〉 ∈ B}, ic ]. Denote g = I |[0,1]×[0,ia ] . Theng : [0, 1] × [0, ia ] → [sup{dn |[cn , dn ] ⊆ [0, c], 〈cn , dn , S ′′

n 〉 ∈B}, ic ] is a continuous function which is increas-ing in its second place. g(1, 0) = c and g(x, ia) = ic(x ∈ [0, 1]) since I(1, 0) = c and I(x, ia) = ic . Accord-ing to Corollary 1, it holds that for any [am , bm ] ⊆ [0, ia ](〈am , bm , S ′

m 〉 ∈ A), g(x, y) = e′m (x) on [0, 1] × [am , bm ],where e′m : [0, 1] → [sup{dn |[cn , dn ] ⊆ [0, c], 〈cn , dn , S ′′

n 〉 ∈B}, ic ] is a continuous function. Thus, we have proved 3).

The proof of 4) is similar to that of 3). �Theorem 8: Let I, S1 , S2 , and L be the same as in Definition

8. Then I satisfies Condition L if and only if I has the forms asin 1)–4) of Proposition 2.

Proof: The necessary proof is the one of Proposition 2. Onlythe sufficient proof remains to be considered.

First, we can prove that I is well defined on [0, 1]2 . In fact,if y ≤ ia or y ≥ sb , then I(x, y) can be calculated by 3) or 4)in Proposition 2. If ia < y < sb , by the definition of ia and sb ,there exist am 1 , bm 2 (m1 ,m2 ∈ J ′

1), such that am 1 < y < bm 2 ,then I(x, y) can be calculated by 1) or 2) in Proposition 2.

It is easy to check L1, L2, and L5. In order to prove L3 andL4, we first prove that the ranges of fm 1 ,m 2 , g and h in 2)–4) ofProposition 2 only contain idempotent elements of S2 .

Suppose 〈am 1 , bm 1 , S′m 1

〉 and 〈am 2 , bm 2 , S′m 2

〉 are twoconsecutive summands in A′ satisfying 〈am 1 , bm 1 ,S ′

m 1〉 <1 〈am 2 , bm 2 , S

′m 2

〉, F (〈am 1 , bm 1 , S′m 1

〉) = 〈cn1 , dn1 ,S ′′

n1〉, F (〈am 2 , bm 2 , S

′m 2

〉) = 〈cn2 , dn2 , S′′n2〉, and dn1 < cn2 .

Just as the proof of 2) in Proposition 2, we can get that[dn1 , cn2 ] contains only idempotent elements of S2 . Therefore,the range of fm 1 ,m 2 contains only idempotent elements of S2 .In addition, it is easy to check that the ranges of g and h containonly idempotent elements of S2 . Consequently, if I(x, y)has one of the forms in 2)–4), then I(x, y) is an idempotentelement of S2 .


In the following, we will verify L4. SupposeF (〈am , bm , S ′

m 〉) = 〈cn , dn , S ′′n 〉 (m ∈ J1). Then I has

form (8), form (9), or form (10) on [0, 1] × (am , bm ). Whilefrom forms (8)–(10), we get that I(1, (am , bm )) = (cn , dn ] (or(cn , dn )). Considering the continuity of I , we can have thatI(1, [am , bm ]) = [cn , dn ].

Conversely, if I(1, [am , bm ]) = [cn , dn ], then it can provedthat there exists some n′ ∈ J2 such that F (〈am , bm , S ′

m 〉) =〈cn ′ , dn ′ , S ′′

n ′ 〉. Otherwise, I(1, y) (y ∈ (am , bm )) has one of theforms in 2)–4). Therefore, I(1, y) is an idempotent element ofS2 for any y ∈ (am , bm ), which contradicts I(1, [am , bm ]) =[cn , dn ]. Similarly to the previous proof of F (〈am , bm , S ′

m 〉) =〈cn , dn , S ′′

n 〉 ⇒ I(1, [am , bm ]) = [cn , dn ], we can obtain thatI(1, [am , bm ]) = [cn ′ , dn ′ ]. Therefore, cn = cn ′ and dn = dn ′ .Thus, L4 has been proved.

Now, we will focus on the proof of L3, i.e., the proof of (2).Case 1. Suppose at least one of y and z is an idempo-

tent element of S1 . Without loss of generality, assume that yis an idempotent element of S1 . Thus, y �∈ (am , bm ) for any〈am , bm , S ′

m 〉 ∈ A′. Therefore, for any x ∈ [0, 1], I(x, y) hasone of the forms in 2)–4), and then I(x, y) is an idempotentelement of S2 . Consequently, I(x, S1(y, z)) = I(x,max(y, z))= max(I(x, y), I(x, z)) = S2(I(x, y), I(x, z)).

Case 2. Suppose both y and z are non-idempotent elementsof S1 with y ∈ (am ′ , bm ′) and z ∈ (am ′′ , bm ′′) (m′ �= m′′). ThenS1(y, z) = max(y, z) and I(x, S1(y, z)) = I(x,max(y, z)) =max(I(x, y), I(x, z)).

If 〈am ′ , bm ′ , S ′m ′ 〉, 〈am ′′ , bm ′′ , S ′

m ′′ 〉 ∈ A′, then F (〈am ′ , bm ′ ,S ′

m ′ 〉)=〈cn ′ , dn ′ , S ′′n ′ 〉 �=F (〈am ′′ , bm ′′ , S ′

m ′′ 〉)=〈cn ′′ , dn ′′ , S ′′n ′′ 〉.

Hence, I(x, y) ∈ [cn ′ , dn ′ ] and I(x, z) ∈ [cn ′′ , dn ′′ ]. As aconsequence, S2(I(x, y), I(x, z)) = max(I(x, y), I(x, z)).

If 〈am ′ , bm ′ , S ′m ′ 〉 �∈ A′ or 〈am ′′ , bm ′′ , S ′

m ′′ 〉 �∈ A′, withoutloss of generality, then we suppose that 〈am ′ , bm ′ , S ′

m ′ 〉 �∈ A′.This shows that I(x, y) = em ′(x) (or e′m ′(x), or e′′m ′(x)).Then I(x, y) is an idempotent element of S2 , whichimplies that S2(I(x, y), I(x, z)) = max(I(x, y), I(x, z)) =I(x, S1(y, z)).

Case 3. Suppose there exists some m0 ∈ J1 such that y, z ∈(am 0 , bm 0 ). It is easy to get that S1(y, z) ∈ [am 0 , bm 0 ]. If〈am 0 , bm 0 , S

′m 0

〉 ∈ A′, then just as the proof of Theorem 5,we can get the result. If 〈am 0 , bm 0 , S

′m 0

〉 �∈ A′, then I(x, y) =I(x, z) = I(x, S1(y, z)) = em 0 (x) (or e′m 0

(x), or e′′m 0(x)).

Therefore, we have S2(I(x, y), I(x, z)) = I(x, S1(y, z)) sinceem 0 (x), e′m 0

(x) and e′′m 0(x) are idempotent elements of

S2 . �Remark 5: In the theorem above, by fixing I(1, 0) = c

and I(1, 1) = d, we get the continuous solutions to (2). Thisapproach is different from the previous ones in [21]–[23],where the authors first derived the form of the vertical sec-tion I(x, ·) for each fixed x, and then used the forms ofthe vertical sections to get its continuous solutions. How-ever, the theorem above shows that we can get the continu-ous solutions to (2) by just fixing I(1, 0) and I(1, 1). There-fore, the method of finding the form of the vertical sectionI(x, ·) for each fixed x in [21]–[23] is unnecessary in thisstudy.

B. Fuzzy Implication Solutions to (2)

The general continuous solutions to (2) have been fully char-acterized in Theorems 6 and 8. Now, we turn to the fuzzy im-plication solutions.

Proposition 3: Let S1 and S2 be two t-conorms and I be acontinuous fuzzy implication. If (2) holds, then S2 = max.

Proof: According to Lemma 1 and the fact that 0 is an idempo-tent element of S1 , we have that I(x, 0) is an idempotent elementof S2 for any x ∈ [0, 1]. Considering I(0, 0) = 1, I(1, 0) = 0and the continuity of I , we obtain that for any z ∈ [0, 1],there exists some x ∈ [0, 1] such that I(x, 0) = z. There-fore, it holds that S2(z, z) = S2(I(x, 0), I(x, 0)) = I(x, 0) =z, namely, S2(z, z) = z for any z ∈ [0, 1]. Then we obtain thatS2 = max. �

From Proposition 3, we see that there is no continuous fuzzyimplication solution to (2) when S2 is an ordinal sum. There-fore, if one wants to maintain the boundary conditions I1–I3, thecontinuity of implications on [0, 1] × [0, 1] should be dropped.Usually, we wish the continuous interval of implications to be aslarge as possible since the continuity can provide us with someconvenience and robustness in practical applications. Therefore,in the sequel, we will focus on finding fuzzy implication solu-tions which are continuous on (0, 1] × [0, 1].

Letting c = 0, d = 1 (and thus B′ = B) in Theorem 8, usingthe following definition of Condition R and considering theantitonicity of I with respect to the first place, we will obtainthe fuzzy implication solutions to (2) (its fuzzy implicationsolutions can not be obtained from Theorem 6 because S2 is anordinal sum and there exist summands of S2 in [0, 1]).

Definition 9: Let I be a fuzzy implication, and S1and S2 be two continuous t-conorms given by (〈am , bm ,S ′

m 〉)m∈J1 and (〈cn , dn , S ′′n 〉)n∈J2 , respectively. Suppose A′ =

{〈am , bm , S ′m 〉|m ∈ J ′

1 ⊆ J1} is a subset of A, and F is a feasi-ble correspondence from A′ to B. Define Condition R as follows.

R1) I is continuous on (0, 1] × [0, 1].R2) I(x, S1(y, z)) = S2(I(x, y), I(x, z)) for any x, y, z ∈

[0, 1].R3) I(1, [am , bm ]) = [cn , dn ] if and only if F (〈am ,

bm , S ′m 〉) = 〈cn , dn , S ′′

n 〉 (m ∈ J ′1 , 〈cn , dn , S ′′

n 〉 ∈ B).Theorem 9: Let I, S1 , S2 , and R be the same as in Definition

9. Then fuzzy implication I satisfies Condition R if and only ifI has forms as follows.

1) For any 〈am , bm , S ′m 〉 ∈ A′ with F (〈am , bm , S ′

m 〉) =〈cn , dn , S ′′

n 〉, I has one of the following forms on (0, 1] ×(am , bm ):

1.1) If S ′m is a strict t-conorm and S ′′

n is a nilpotent t-conormwith s′m and s′′n being their respective additive generators, then

I(x, y) ={


(11)

where km : (0, 1] → (am , bm ) is a continuous andincreasing function, and E(x, y) = cn + (dn −cn )s′′n

−1( s ′′n (1)

s ′m ( k m (x )−a m

b m −a m)s′m ( y−am

bm −am)).


1.2) If S ′m and S ′′

n are nilpotent t-conorms with s′m and s′′nbeing their respective additive generators, then

I(x, y) =

⎧⎨⎩

E(x, y), y ∈ (am , km (x))

dn , y ∈ [km (x), bm )(12)

where km : (0, 1] → (am , bm ] is a continuous and increasingfunction, and E(x, y) is the same as in 1.1).

1.3) If S ′m and S ′′

n are strict t-conorms with s′m and s′′n beingtheir respective additive generators, then

I(x, y) = cn + (dn − cn )s′′n−1

(pm (x)s′m

(y − am

bm − am

))

(13)where pm : (0, 1] → (0,∞) is a continuous and decreasingfunction.

2) Suppose that 〈am 1 , bm 1 , S′m 1

〉 and 〈am 2 , bm 2 , S′m 2

〉 aretwo consecutive summands in A′ satisfying 〈am 1 , bm 1 , S

′m 1

〉 <1〈am 2 , bm 2 , S

′m 2

〉, F (〈am 1 , bm 1 , S′m 1

〉) = 〈cn1 , dn1 , S′′n1〉 and

F (〈am 2 , bm 2 , S′m 2

〉) = 〈cn2 , dn2 , S′′n2〉. Then I has one of the

following forms on (0, 1] × [bm 1 , am 2 ].2.1) If dn1 = cn2 , then I(x, y) = dn1 .2.2) If dn1 < cn2 , then I(x, y) = fm 1 ,m 2 (x, y),

where fm 1 ,m 2 : (0, 1] × [bm 1 , am 2 ] → [dn1 , cn2 ] is a continu-ous function. It is decreasing in its first place and increasingin its second place and satisfies fm 1 ,m 2 (x, bm 1 ) = dn1 andfm 1 ,m 2 (x, am 2 ) = cn2 for any x ∈ (0, 1]. For any [am , bm ](m ∈ J1) satisfying 〈am 1 , bm 1 , S

′m 1

〉 <1 〈am , bm , S ′m 〉 <1

〈am 2 , bm 2 , S′m 2

〉, it holds that fm 1 ,m 2 (x, y) = em (x) on(0, 1] × [am , bm ], where em : (0, 1] → [dn1 , cn2 ] is a contin-uous and deceasing function.

3) Define ia = inf{am |〈am , bm , S ′m 〉 ∈ A′} and ic =

inf{cn |〈cn , dn , S ′′n 〉 ∈ B}. Then for any (x, y) ∈ (0, 1] ×

[0, ia ], I(x, y) = g(x, y), where g : (0, 1] × [0, ia ] → [0, ic ] isa continuous function. It is deceasing in its first place andincreasing in its second place and satisfies g(1, 0) = 0 andg(x, ia) = ic for any x ∈ (0, 1]. In addition, for any [am , bm ] ⊆[0, ia ] (〈am , bm , S ′

m 〉 ∈ A), it holds that g(x, y) = e′m (x) on(0, 1] × [am , bm ], where e′m : (0, 1] → [0, ic ] is a continuousand deceasing function.

4) Define sb = sup{bm |〈am , bm , S ′m 〉 ∈ A′} and sd =

sup{dn |〈cn , dn , S ′′n 〉 ∈ B}. Then for any (x, y) ∈ (0, 1] ×

[sb , 1], I(x, y) = h(x, y), where h : (0, 1] × [sb , 1] → [sd, 1] isa continuous function. It is deceasing in its first place andincreasing in its second place and satisfies h(1, 1) = 1 andh(x, sb) = sd for any x ∈ (0, 1]. In addition, for any [am , bm ] ⊆[sb , 1] (〈am , bm , S ′

m 〉 ∈ A), it holds that h(x, y) = e′′m (x) on(0, 1] × [am , bm ], where e′′m : (0, 1] → [sd, 1] is a continuousand deceasing function.

5) I(0, y) = 1 for any y ∈ [0, 1].Proof: (⇒) The proofs of 1)–4) are similar to the ones of

Proposition 2. Property 5) can be got directly from Definition 3.(⇐) If (x, y) ∈ (0, 1] × [0, 1], then the corresponding proofs

are similar to the ones of Theorem 8. If x = 0 and y ∈ [0, 1],then (2) clearly holds since I(0, y) = 1. �

Algorithm of finding fuzzy implication solutions to (2): Ac-cording to Theorems 7 and 9, we can summarize the method offinding fuzzy implication solutions to (2).

1) Find A′ ⊆ A and the corresponding F : A′ → B such thatF is a feasible correspondence from A′ to B. If there is no suchfeasible correspondence, then, according to Theorem 7, there isno fuzzy implication solution to (2).

2) For each F , according to Theorem 9, find the fuzzy impli-cation solutions with parameters to (2).

3) (optional) Choose proper parameters to construct the de-sired particular fuzzy implication solutions.

Example 3: Suppose that S1 = (〈 16 , 1

3 , SL 〉) and S2 =(〈 1

3 , 12 , SP 〉). According to 2) of Example 2, there is no feasible

correspondence. Hence, (2) has no fuzzy implication solutions.Example 4: Suppose that S1 = (〈 1

8 , 14 , SL 〉, 〈 1

4 , 12 , SP 〉,

〈 12 , 2

3 , SL 〉, 〈 56 , 11

12 , SP 〉) and S2 = (〈0, 13 , SP 〉, 〈 1

2 , 23 , SL 〉) with

s(x) = − ln(1 − x) and s(x) = x being two additive generatorsof SP and SL , respectively.

1) We can find A′ and the feasible correspondence F3 fromA′ to B.

A′={〈 14 , 1

2 , SP 〉, 〈 56 , 11

12 , SP 〉}, F3(〈 14 , 1

2 , SP 〉)=〈0, 13 , SP 〉

andF3(〈 56 , 11

12 , SP 〉) = 〈 12 , 2

3 , SL 〉.The plot of F3 can be seen in Fig. 2.2) Now, we find fuzzy implication solutions with parameters.For feasible correspondence F3 , we can get the representation

of fuzzy implication I

I(x, y) =⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

1, x = 0, y ∈ [0, 1]

0, x ∈ (0, 1], y ∈[0,

14

]

13− 1

3(2 − 4y)p2 (x) , x ∈ (0, 1], y ∈

(14,12

)

f2,4(x, y), x ∈ (0, 1], y ∈[12,56

]

12

+16

ln(11 − 12y)ln(11 − 12k4(x))

, x ∈ (0, 1], y ∈(

56, k4(x)

)

23, x ∈ (0, 1], y ∈

[k4(x),

1112

)

h(x, y), x ∈ (0, 1], y ∈[1112

, 1]

(14)

where h : (0, 1] × [ 1112 , 1] → [ 2

3 , 1] and f2,4 : (0, 1] × [ 12 , 5

6 ] →[ 13 , 1

2 ] are continuous functions which are increasing in theirsecond places and decreasing in their first places and satisfyh(x, 11

12 ) = 23 , h(1, 1) = 1, f2,4(x, 1

2 ) = 13 , f2,4(x, 5

6 ) = 12 , and

f2,4(x, y) = e3(x) on (0, 1] × [ 12 , 2

3 ] with e3 : (0, 1] → [ 13 , 1

2 ]being a continuous and decreasing function; p2 : (0, 1] →(0,∞) is a decreasing and continuous functions; k4 : (0, 1] →( 5

6 , 1112 ) is an increasing and continuous function.

3) We can choose some desired parameters to get some par-ticular solutions.


SP

SP

1/2

2/3

1/3

0

SL

SL

SP

SL

11/12

2/3

5/6

F3

1/2

1/4

1/8

00.2

0.40.6

0.81

00.2

0.40.6

0.810

0.5

1

XY

Z

Fig. 2. Plots of F3 and the particular fuzzy implication solution I in Example4. The bold line is the line I(0, y) = 1.

For feasible correspondence F3 , take

f2,4(x, y) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

13, y ∈

[12,23

]

y − 13, y ∈

(23,56

] (x, y) ∈ (0, 1] ×[12,56

]

h(x, y) =13(12y − 11)x +

23, (x, y) ∈ (0, 1] ×

[1112

, 1]

p2(x) =1x

, x ∈ (0, 1]

k4(x) =124

x +56, x ∈ (0, 1].

Then we get a particular solution which is presented in form(15)

I(x, y) =⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

1, x = 0, y ∈ [0, 1]

0, x ∈ (0, 1], y ∈[0,

14

]

13− 1

3(2 − 4y)

1x , x ∈ (0, 1], y ∈

(14,12

)

13, x ∈ (0, 1], y ∈

[12,23

]

y − 13, x ∈ (0, 1], y ∈

[23,56

]

12

+16

ln(11 − 12y)ln

(1 − 1

2 x) , x ∈ (0, 1], y ∈

(56,

124

x +56

)

23, x ∈ (0, 1], y ∈

[124

x +56,1112

)

13(12y − 11)x +

23, x ∈ (0, 1], y ∈

[1112

, 1]

.

(15)

Its plot is presented in Fig. 2.

IV. CONCLUSION

Baczynski and Jayaram [22] investigated (2) in the case whereS1 and S2 are continuous Archimedean t-conorms and pointedout that, “In our future works, we will try to concentrate onsome cases that are not considered in this paper, for example,when S1 is a strict t-conorm and S2 is a nilpotent t-conorm, andvice versa. Also, the situation when S1 and S2 are continuoust-conorms is still unsolved.” In [21], Baczynski studied (2) in thecase that S1 is a strict t-conorm and S2 is a nilpotent t-conorm,and vice versa. Xie et al. [23] solved (2) under the conditionthat S2 is a continuous Archimedean t-conorm and S1 is acontinuous t-conorm given as an ordinal sum. In this paper, wediscussed (2) with two continuous t-conorms given as ordinalsums. Clearly, this paper together with [21]–[23] completelyresolved the problems in [22].

If there is no summand of S2 in the interval [I(1, 0), I(1, 1)],we get directly its general continuous solutions. In order tocharacterize the solvability of (2) if there are summands of S2in the interval [I(1, 0), I(1, 1)], we introduced a new conceptcalled feasible correspondence. Using feasible correspondences,we obtained the general continuous solutions to (2) and thengot its fuzzy implication solutions which are continuous on(0, 1] × [0, 1]. It should be pointed out that the method usedin this paper can also be applied to the other three distributiveequations which are closely related to (2).

The approach introduced in the paper can help reduce thenumber of rules in fuzzy systems. However, the approach seemsto be a bit complicated, because it only gives the algorithm offinding fuzzy implication solutions to (2) but not its concretefuzzy implication solutions. Our future work will focus on find-ing simpler methods to reduce the number of rules in fuzzysystems.


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Aifang Xie received the M.S. degree from JiangxiNormal University, Nanchang, China, in 2010. Sheis currently working toward the Ph.D. degree withthe Mathematics Department, Shandong University,Jinan, China.

Her research interests include fuzzy logic, fuzzyfunctional equations, and fuzzy systems.

Cheng Li received the M.S. degree from ShandongUniversity, Jinan, China in 2012. He is currentlyworking toward the Ph.D. degree with NortheasternUniversity, Boston, MA.

Huawen Liu received the Ph.D. degree in mathemat-ics from Shandong University, Jinan, China, in 2005.

She is currently a Professor of mathematics andsystem science with Shandong University. Her re-search interests include fuzzy systems, fuzzy logicconnectives, and approximate reasoning.

Distributive Equations of Fuzzy Implications Based on Continuous Triangular Conorms Given as Ordinal...

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