Distributavity

590 IEEE TRANSACTIONS ON FUZZY SYSTEMS, VOL. 17, NO. 3, JUNE 2009
On the Distributivity of Fuzzy Implications Over
Nilpotent or Strict Triangular Conorms
Michał Baczyński and Balasubramaniam Jayaram, Member, IEEE
Abstract—Recently, many works have appeared in this very
journal dealing with the distributivity of fuzzy implications over t-
norms and t-conorms. These equations have a very important role
to play in efficient inferencing in approximate reasoning, especially
fuzzy control systems. Of all the four equations considered, the
equation I(x, S1 (y, z)) = S2 (I(x, y), I(x, z)), when S1 , S2
are both t-conorms and I is an R-implication obtained from a
strict t-norm, was not solved. In this paper, we characterize func-
tions I that satisfy the previous functional equation when S1 , S2
are either both strict or nilpotent t-conorms. Using the obtained
characterizations, we show that the previous equation does not hold
when S1 , S2 are either both strict or nilpotent t-conorms, and I
is a continuous fuzzy implication. Moreover, the previous equation
does not hold when I is an R-implication obtained from a strict
t-norm, and S1 , S2 are both strict t-conorms, while it holds for an
R-implication I obtained from a strict t-norm T if and only if the
t-conorms S1 = S2 are Φ-conjugate to the Łukasiewicz t-conorm
for some increasing bijection ϕ of the unit interval, which is also a
multiplicative generator of T .
Index Terms—Combs methods, functional equations, fuzzy im-
plication, R-implication, t-conorm, t-norm.
I. INTRODUCTION
DISTRIBUTIVITY of fuzzy implication operations over
different fuzzy logic connectives has been studied in the
recent past by many authors. This interest, perhaps, was kick
started by Combs and Andrews in [13], wherein they exploit the
following classical tautology:
(p ∧ q) → r ≡ (p → r) ∨ (q → r)
in their inference mechanism toward reduction in the complex-
ity of fuzzy “If–Then” rules. They refer to the left-hand side of
this equivalence as an intersection rule configuration (IRC) and
to its right-hand side as a union rule configuration (URC). Sub-
sequently, there were many discussions [14]–[16], [24], most
of them pointing out the need for a theoretical investigation re-
quired for employing such equations, as concluded by Dick and
Kandel [16], “Future work on this issue will require an examina-
tion of the properties of various combinations of fuzzy unions,
intersections and implications” or by Mendel and Liang [24],
“We think that what this all means is that we have to look past
the mathematics of IRC⇔URC and inquire whether what we
are doing when we replace IRC by URC makes sense.” It was
Manuscript received May 4, 2007; accepted July 17, 2007. First published
April 30, 2008; current version published June 11, 2009.
M. Baczyński is with the Institute of Mathematics, University of Silesia,
Katowice 40-007, Poland (e-mail: michal.baczynski@us.edu.pl).
B. Jayaram is with the Department of Mathematics and Computer Sciences,
Sri Sathya Sai University, Anantapur, Andhra Pradesh 515134, India (e-mail:
jbala@ieee.org).
Digital Object Identifier 10.1109/TFUZZ.2008.924201
Trillas and Alsina [32] who were the first to investigate the
generalized version of the previous law, viz.,
I(T(x, y), z) = S(I(x, z), I(y, z)), x, y, z ∈ [0, 1] (1)
where T, S are a t-norm and a t-conorm, respectively, general-
izing the ∧, ∨ operators, respectively, and I is a fuzzy implica-
tion. From their investigations of (1) for the three main families
of fuzzy implications, viz., S-implications, R-implications, and
QL-implications, it was shown that in the case of R-implications
obtained from left-continuous t-norms and S-implications, (1)
holds if and only if T = min and S = max. Also along the
previous lines, Balasubramaniam and Rao [10] considered the
following dual equations of (1):
I(S(x, y), z) = T(I(x, z), I(y, z)) (2)
I(x, T1(y, z)) = T2(I(x, y), I(x, z)) (3)
I(x, S1(y, z)) = S2(I(x, y), I(x, z)) (4)
where again, T, T1, T2 and S, S1, S2 are t-norms and t-conorms,
respectively, and I is a fuzzy implication. Similarly, it was
shown that when I is either an R-implication obtained from
a left-continuous t-norm or an S-implication, in almost all the
cases, the distributivity holds only when T = T1 = T2 = min
and S = S1 = S2 = max, while (4) for the case when I is an
R-implication obtained from a strict t-norm was left unsolved
(cf. [10, Th. 4]). This forms the main motivation of this paper.
Meanwhile, Baczyński in [2] and [3] considered the func-
tional equation (3), both independently and along with other
equations, and characterized fuzzy implications I in the case
when T1 = T2 is a strict t-norm. Some partial studies regarding
distributivity of fuzzy implications over maximum and mini-
mum were presented by Bustince et al. in [12]. It may be worth
recalling that (3) with T1 = T2 is one of the characterizing prop-
erties of A-implications proposed by Türksen et al. in [33].
As we mentioned earlier, the previous equations (1)–(4) have
an important role to play in inference invariant rule reduction in
fuzzy inference systems (see also [8], [9], and [30]). The very
fact that about half-a-dozen works have appeared in this very
journal dealing with these distributive equations is a pointer
to the importance of these equations. That more recent works
dealing with distributivity of fuzzy implications over uninorms
(see [27] and [28]) have appeared is an indication of the sus-
tained interest in the previous equations.
This paper differs from the previous works that have ap-
peared in this journal on these equations, in that, we attempt to
solve the problem in a more general setting, by characterizing
functions I that satisfy the functional equation (4) when S1, S2
are either both strict or nilpotent t-conorms. Then, using these
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BACZY ŃSKI AND JAYARAM: DISTRIBUTIVITY OF FUZZY IMPLICATIONS OVER NILPOTENT OR STRICT TRIANGULAR CONORMS 591
characterizations, we also investigate the conditions under
which (4) holds when I is an R-implication obtained from a
strict t-norm.
The paper is organized as follows. In Section II, we give some
results concerning basic fuzzy logic connectives and functional
equations that will be employed extensively in the sequel. In
Section III, we study (4), when I is a binary operation on [0, 1],
while S1, S2 are both strict t-conorms. Based on the obtained
characterization, we show that there exists no continuous so-
lution I for (4) that is a fuzzy implication. Subsequently, we
obtain the characterization of noncontinuous fuzzy implications
I that are solutions for (4). In Section IV, we mimic the ap-
proach taken in Section III, except that in this case, S1, S2 are
both nilpotent t-conorms. Again in this case, the results parallel
those of Section III. In Section V, we study (4) in the case when
I is an R-implication obtained from a strict t-norm T. We show
that (4) does not hold when S1, S2 are strict t-conorms, while
using one of the characterization obtained in the previous sec-
tion, we show that it holds if and only if the t-conorms S1 = S2
are Φ-conjugate to the Łukasiewicz t-conorm for some increas-
ing bijection ϕ, which is a multiplicative generator of the strict
t-norm T.
II. PRELIMINARIES
A. Basic Fuzzy Logic Connectives
First, we recall some basic notations and results that will
be useful in the sequel. We start with the notation of conju-
gacy (see [21, p. 156]). By Φ, we denote the family of all
increasing bijections ϕ: [0, 1] → [0, 1]. We say that functions
f, g: [0, 1]n
→ [0, 1] are Φ-conjugate, if there exists a ϕ ∈ Φ
such that g = fϕ , where
fϕ (x1, . . . , xn ) := ϕ−1
(f(ϕ(x1), . . . , ϕ(xn )))
for all x1, . . . , xn ∈ [0, 1]. If F is an associative binary operation
on [a, b] with neutral element e, then the power notation x
[n]
F ,
where n ∈ N0, is defined by
x
[n]
F :=



e, if n = 0
x, if n = 1
F x, x
[n−1]
F , if n > 1.
Definition 1 (see [20], [29]):
1) An associative, commutative, and increasing operation
T: [0, 1]2
→ [0, 1] is called a t-norm if it has the neutral
element 1.
2) An associative, commutative, and increasing operation
S: [0, 1]2
→ [0, 1] is called a t-conorm if it has the neutral
element 0.
Definition 2 [20, Definitions 2.9 and 2.13]: A t-norm T
(t-conorm S, respectively) is said to be
1) Archimedean, if for every x, y ∈ (0, 1), there is an n ∈ N
such that x
[n]
T < y (x
[n]
S > y, respectively);
2) strict, if T (S, respectively) is continuous and strictly
monotone, i.e., T(x, y) < T(x, z) (S(x, y) < S(x, z), re-
spectively) whenever x > 0 (x < 1, respectively) and
y < z;
3) nilpotent, if T (S, respectively) is continuous and if for
each x ∈ (0, 1), there exists n ∈ N such that x
[n]
T = 0
(x
[n]
S = 1, respectively).
Remark 1:
1) For a continuous t-conorm S, the Archimedean property
is given by the simpler condition (cf. [19, Prop. 5.1.2])
S(x, x) > x, x ∈ (0, 1).
2) If a t-conorm S is continuous and Archimedean, then
S is nilpotent if and only if there exists some nilpotent
element of S, which is equivalent to the existence of some
x, y ∈ (0, 1) such that S(x, y) = 1 (see [20, Th. 2.18]).
3) If a t-conorm S is strict or nilpotent, then it
is Archimedean. Conversely, every continuous and
Archimedean t-conorm is either strict or nilpotent (cf. [20,
Th. 2.18]).
We shall use the following characterizations of continuous
Archimedean t-conorms.
Theorem 1 ([23], cf. [20, Corollary 5.5]): For a function
S: [0, 1]2
→ [0, 1], the following statements are equivalent.
1) S is a continuous Archimedean t-conorm.
2) S has a continuous additive generator, i.e., there exists a
continuous, strictly increasing function s: [0, 1] → [0, ∞]
with s(0) = 0, which is uniquely determined up to a pos-
itive multiplicative constant, such that
S(x, y) = s(−1)
(s(x) + s(y)), x, y ∈ [0, 1] (5)
where s(−1)
is the pseudoinverse of s given by
s(−1)
(x) =
s−1
(x), if x ∈ [0, s(1)]
1, if x ∈ (s(1), ∞].
Remark 2:
1) A representation of a t-conorm S as earlier can be writ-
ten without explicitly using of the pseudo-inverse in the
following way:
S(x, y) = s−1
(min(s(x) + s(y), s(1))) (6)
for x, y ∈ [0, 1].
2) S is a strict t-conorm if and only if each continuous addi-
tive generator s of S satisfies s(1) = ∞.
3) S is a nilpotent t-conorm if and only if each continuous
additive generator s of S satisfies s(1) < ∞.
Next, two characterizations of strict t-norms and nilpotent t-
conorms are well known in literature and can be easily obtained
from the general characterizations of continuous Archimedean
t-norms and t-conorms (see [20, Sec. 5.2]).
Theorem 2: For a function T: [0, 1]2
→ [0, 1], the following
statements are equivalent.
1) T is a strict t-norm.
2) T is Φ-conjugate with the product t-norm, i.e., there exists
ϕ ∈ Φ, which is uniquely determined up to a positive
constant exponent, such that
T(x, y) = ϕ−1
(ϕ(x) · ϕ(y)), x, y ∈ [0, 1]. (7)
Theorem 3: For a function S: [0, 1]2
→ [0, 1], the following

1) S is a nilpotent t-conorm.
2) S is Φ-conjugate with the Łukasiewicz t-conorm, i.e., there
exists ϕ ∈ Φ, which is uniquely determined, such that for
all x, y ∈ [0, 1], we have
S(x, y) = ϕ−1
(min(ϕ(x) + ϕ(y), 1)). (8)
In the literature, we can find several diverse definitions of
fuzzy implications [12], [18]. In this paper, we will use the
following one, which is equivalent to the definition introduced
by Fodor and Roubens (see [18, Def. 1.15]).
Definition 3: A function I: [0, 1]2
→ [0, 1] is called a fuzzy
implication if it satisfies the following conditions:
I is decreasing in the first variable. (I1)
I is increasing in the second variable. (I2)
I(0, 0) = 1, I(1, 1) = 1, I(1, 0) = 0. (I3)
From the previous definition, we can deduce that, for each fuzzy
implication, I(0, x) = I(x, 1) = 1 for x ∈ [0, 1]. Moreover, I
also satisfies the normality condition
I(0, 1) = 1 (NC)
and consequently, every fuzzy implication restricted to the set
{0, 1}2
coincides with the classical implication.
There are many important methods for generating fuzzy im-
plications (see [17]–[19]). In this paper, we need only one
family—R-implications.
Definition 4: A function I: [0, 1]2
→ [0, 1] is called an R-
implication if there exist a t-norm T such that
I(x, y) = sup{t ∈ [0, 1] | T(x, t) ≤ y}, x, y ∈ [0, 1]. (9)
If I is generated from a t-norm T by (9), then we will sometimes
write IT .
It is very important to note that the name “R-implication” is
a short version of “residual implication,” and IT is also called
as “the residuum of T.” This class of implications is related to
a residuation concept from the intuitionistic logic. In fact, it has
been shown that in this context, this definition is proper only for
left-continuous t-norms.
Proposition 1 (cf. [19, Proposition 5.4.2 and Corollary
5.4.1]): For a t-norm T, the following statements are equiva-
lent.
1) T is left-continuous.
2) T and IT form an adjoint pair, i.e., they satisfy
T(x, t) ≤ y ⇐⇒ IT (x, y) ≥ t, x, y, t ∈ [0, 1].
3) The supremum in (9) is the maximum, i.e.,
IT (x, y) = max{t ∈ [0, 1] | T(x, t) ≤ y}
where the right side exists for all x, y ∈ [0, 1].
The following characterization of R-implications gener-
ated from left-continuous t-norms is also well known in the
literature.
Theorem 4 ([25], cf. [18, Th. 1.14]): For a function
I: [0, 1]2
1) I is an R-implication generated from a left-continuous
t-norm.
2) I is right-continuous with respect to the second variable,
and it satisfies (I2), the exchange principle
I(x, I(y, z)) = I(y, I(x, z)), x, y, z ∈ [0, 1] (EP)
and the ordering property
x ≤ y ⇐⇒ I(x, y) = 1, x, y ∈ [0, 1]. (OP)
It should be noted that each R-implication I satisfies the left
neutrality property
I(1, y) = y, y ∈ [0, 1]. (NP)
Further, if T is a strict t-norm, then we have the following
representation from [26, Th. 6.1.2] (see also [4, Th. 19]).
Theorem 5: If I is an R-implication generated from a strict
t-norm T, then I is Φ-conjugate to the Goguen implication,
i.e., there exists ϕ ∈ Φ, which is uniquely determined up to a
positive constant exponent, such that
I(x, y) =



1, if x ≤ y
ϕ−1 ϕ(y)
ϕ(x)
, otherwise
(10)
for all x, y ∈ [0, 1].
We would like to underline that the increasing bijection ϕ
earlier can be seen as a multiplicative generator of T in (7). The
proof that ϕ is uniquely determined up to a positive constant
exponent has been presented by Baczyński and Drewniak (see
[6, Th. 6]).
B. Some Results Pertaining to Functional Equations
Here, we present some results related to the additive and
multiplicative Cauchy functional equations:
f(x + y) = f(x) + f(y) (11)
f(xy) = f(x)f(y) (12)
which are crucial in the proofs of the main theorems.
Theorem 6 ([1], cf. [22, Th. 5.2.1]): For a continuous function
f: R → R, the following statements are equivalent.
1) f satisfies the additive Cauchy functional equation (11)
for all x, y ∈ R.
2) There exists a unique constant c ∈ R such that
f(x) = cx (13)
for all x ∈ R.
Theorem 7 ([22, Th. 13.5.3]): Let A ⊂ R be an interval such
that 0 ∈ clA, where clA denotes the closure of the set A and
let B = A + A = {a1 + a2 | a1 ∈ A, a2 ∈ A}. If a function
f: B → R satisfies the additive Cauchy functional equation (11)
for all x, y ∈ A, then f can be uniquely extended onto R to an
additive function g such that g(x) = f(x) for all x ∈ B.
By virtue of the previous theorems, we get the following new
results.
Proposition 2: For a function f: [0, ∞] → [0, ∞], the follow-
ing statements are equivalent.
for all x, y ∈ [0, ∞].

2) Either f = ∞, or f = 0, or
f(x) =
0, if x = 0,
∞, if x > 0,
x ∈ [0, ∞] (14)
or
f(x) =
0, if x < ∞,
∞, if x = ∞,
x ∈ [0, ∞] (15)
or there exists a unique constant c ∈ (0, ∞) such that f
admits the representation (13) for all x ∈ [0, ∞].
Proof: 2) =⇒ 1) It is a direct calculation that all the previous
functions satisfy (11).
1) =⇒ 2) Let f: [0, ∞] → [0, ∞] satisfy (11). Setting x =
y = 0 in (11), we get f(0) = f(0) + f(0), so f(0) = 0, or
f(0) = ∞. If f(0) = ∞, then for any x ∈ [0, ∞], we get
f(x) = f(x + 0) = f(x) + f(0) = f(x) + ∞ = ∞
and thus, we obtain the first possible solution f = ∞.
Now, setting x = y = ∞ in (11), we get f(∞) = f(∞) +
f(∞). Therefore, f(∞) = 0 or f(∞) = ∞. If f(∞) = 0, then
for any x ∈ [0, ∞], we have
0 = f(∞) = f(x + ∞) = f(x) + f(∞) = f(x) + 0 = f(x)
and thus, we obtain the second possible solution f = 0.
Let us assume that f = ∞ and f = 0. Considering the alter-
nate cases earlier, we get f(0) = 0 and f(∞) = ∞. Define a
set
Z = {x ∈ (0, ∞) | f(x) = ∞}.
If Z = ∅, then Z = (0, ∞). Indeed, let us fix some real x0 ∈ Z
and take any x ∈ (0, ∞). If x ≥ x0, then we get
f(x) = f((x − x0) + x0) = f(x − x0) + f(x0)
= f(x − x0) + ∞ = ∞.
If x < x0, then there exists a natural k ≥ 2 such that kx > x0.
From the previous point, we get that f(kx) = ∞. Further, by
the induction, we obtain
∞ = f(kx) = f(x + (k − 1)x) = f(x) + f((k − 1)x)
=
k-times
f(x) + · · · + f(x) = kf(x)
and thus, f(x) = ∞. This implies, with the assumptions f(0) =
0 and f(∞) = ∞, that if Z = ∅, then we obtain the third pos-
sible solution (14).
On the other hand, if Z = ∅, then f(x) ∈ [0, ∞) for x ∈
[0, ∞). By Theorem 7, for A = B = [0, ∞), f can be uniquely
extended to an additive function g: R → R, such that g(x) =
f(x) for all x ∈ [0, ∞). Consequently, g is bounded below on
the set [0, ∞). Further, −g is an additive function bounded above
on [0, ∞). Since any additive function is convex, by virtue of
theorem of Bernstein–Doetsch (see [11] or [22, Coro. 6.4.1]),
−g is continuous, and hence, g is continuous. Now, by Theorem
6, there exists a unique constant c ∈ R such that g(x) = cx for
every x ∈ R, i.e., f(x) = cx for every x ∈ [0, ∞). Since the
domain and the range of f are nonnegative, we see that c ≥ 0.
If c = 0, then we get the fourth possible solution (15), because
of our assumption f(∞) = ∞. If c > 0, then we obtain the last
possible solution (13), since c · ∞ = ∞ = f(∞).
Corollary 1: For a continuous function f: [0, ∞] → [0, ∞],
the following statements are equivalent.
for all x, y ∈ [0, ∞].
2) Either f = ∞, or f = 0, or there exists a unique constant
c ∈ (0, ∞) such that f admits the representation (13) for
all x ∈ [0, ∞].
Theorem 8 ([22, Th. 13.6.2]): Fix a real a > 0. Let A = [0, a],
and let H ⊂ R2
be the set
H = {(x, y) ∈ R2
| x ∈ A, y ∈ A, and x + y ∈ A}. (16)
If f: A → R is a function satisfying (11) on H, then there exists
a unique additive function g: R → R such that g(x) = f(x) for
all x ∈ A. Moreover, the closed interval [0, a] may be replaced
by any one of these intervals (0, a), [0, a), and (0, a].
Proposition 3: 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) (17)
for all x, y ∈ [0, a].
2) Either f = b, or f = 0, or
f(x) =
0, if x = 0,
b, if x > 0,
x ∈ [0, a] (18)
or there exists a unique constant c ∈ [b/a, ∞) such that
f(x) = min(cx, b), x ∈ [0, a]. (19)
Proof: 2) =⇒ 1) It is obvious that f = 0 and f = b satisfy
(17). Let f have the form (18). If x = y = 0, then the left side of
(17) is equal to f(min(0 + 0, a)) = f(0) = 0 and the right side
of (17) is min(f(0) + f(0), b) = min(0 + 0, b) = 0. If x = 0
or y = 0, then the both sides of (17) are equal to b.
Finally, if f has the form (19) with some c ∈ [b/a, ∞), then
the left side of (17) is equal to
f(min(x + y, a)) = min(c · min(x + y, a), b)
= min(c(x + y), ca, b)
= min(c(x + y), b), x, y ∈ [0, a]
since ca ≥ b. Now, the right side of (17) is equal to
min(f(x) + f(y), b) = min(min(cx, b) + min(cy, b), b)
= min(cx + cy, cx + b, cy + b, b + b, b)
= min(c(x + y), b), x, y ∈ [0, a]
which ends the proof in this direction.
1) =⇒ 2) Let f satisfy (17). Setting x = y = 0 in (17), we
get
f(0) = min(f(0) + f(0), b). (20)

If f(0) = b, then for any x ∈ [0, a], we have
f(x) = f(min(x + 0, a)) = min(f(x) + f(0), b)
= min(f(x) + b, b) = b
and thus, we obtain the first possible solution f = b.
Now, let us substitute x = y = a in (17). We have
f(a) = min(f(a) + f(a), b). (21)
If f(a) = f(a) + f(a), then f(a) = 0 and for every x ∈ [0, a],
we get
0 = f(a) = f(min(a + x, a)) = min(f(a) + f(x), b)
= min(f(x), b)
and therefore, f(x) = 0, and we obtain the second possible
solution f = 0.
Let us assume that f = 0 and f = b. Considering the alternate
cases in (20) and (21), we get that f(0) = 0 and f(a) = b. Let
us define
x0 = inf{x ∈ [0, a] | f(x) = b}.
First, we will show that if x0 < a, then
f(x) = b, x ∈ (x0, a]. (22)
Indeed, let us take any x ∈ (x0, a]. From the definition of
the element x0, there exists x1 ∈ (x0, x] such that f(x1) = b.
Further
f(x) = f(x1 + (x − x1)) = f(min(x1 + (x − x1), a))
= min(f(x1) + f(x − x1), b) = min(b + f(x − x1), b)
= b.
Thus, if x0 = 0, then we get the third possible solution (18).
Let us assume, that x0 > 0. Now we show, that for any
x, y ∈ [0, x0) such that x + y ∈ [0, x0), the function f is ad-
ditive, and therefore, it satisfies (11). Suppose that this does not
hold, i.e., there exist x1, y1 ∈ [0, x0) such that x1 + y1 ∈ [0, x0)
and f(x1 + y1) = f(x1) + f(y1). Setting x = x1 and y = y1
in (17), we get
f(x1 + y1) = f(min(x1 + y1, a))
= min(f(x1) + f(y1), b) = b.
However, x1 + y1 < x0, which is a contradiction
to the definition of x0. We proved that f satis-
fies the additive Cauchy functional equation (11)
on the set H defined by (16) for A = [0, x0). By
Theorem 8, the function f can be uniquely extended to
an additive function g: R → R, such that g(x) = f(x) for all
x ∈ [0, x0). Consequently, g is bounded on [0, x0), and by
virtue of theorem of Bernstein–Doetsch (see [11] or [22, Th.
6.4.2]), g is continuous. Because of Theorem 6, there exists a
unique constant c ∈ R such that g(x) = cx for every x ∈ R,
i.e., f(x) = cx for every x ∈ [0, x0). Since the domain and the
range of f are nonnegative, we get that c ≥ 0. Moreover
lim
x→x−
0
f(x) = lim
x→x−
0
cx = cx0 ≤ b
and consequently, c ∈ [0, b/x0]. If we assume that c ∈ [0, b/x0),
then we get
f(x0) = f
x0
2
+
x0
2
= f min
x0
2
+
x0
2
, a
= min f
x0
2
+ f
x0
2
, b
= min c
x0
2
+ c
x0
2
, b = min(cx0, b)
= cx0 < b
since c < b/x0. Hence, if x0 = a, then we get a contradiction to
our assumption f(a) = b. If x0 ∈ (0, a), then there exists x1 ∈
(0, x0) such that cx0 + cx1 < b. Setting x = x0 and y = x1 in
(17), we get, by (22), that
b = f(min(x0 + x1, a)) = min(f(x0) + f(x1), b)
= min(cx0 + cx1, b) = cx0 + cx1
which contradicts the previous assumption.
Consequently, we showed that if f = 0 and f = b and x0 ∈
(0, a], then there exists a unique c = b/x0 ≥ b/a such that
f(x) =
cx, if x ≤ x0,
b, if x > x0,
x ∈ [0, a].
Easy calculations show that for x ∈ [0, a], we have
f(x) =
cx, if b
x0
x ≤ b
b, if b
x0
x > b
=
cx, if cx ≤ b
b, if cx > b
= min(cx, b)
i.e., f has the last possible representation (19).
Corollary 2: Fix real a, b > 0. For a continuous function
f: [0, a] → [0, b], the following statements are equivalent.
1) f satisfies the functional equation (17) for all x, y ∈ [0, a].
2) Either f = 0, or f = b, or there exists a unique constant
c ∈ [b/a, ∞) such that f has the form (19).
Theorem 9 ([22, Th. 13.1.6]): Let D be one of the sets (0, 1),
[0, 1), [0, ∞), (0, ∞), R. For a continuous function f: D → R,
the following statements are equivalent.
1) f satisfies the multiplicative Cauchy functional equation
(12) for all x, y ∈ D.
2) Either f = 0, or f = 1, or f has one of the following
forms:
f(x) = |x|c
, x ∈ D
f(x) = |x|c
sgn(x), x ∈ D
with a certain c ∈ R. If 0 ∈ D, then c > 0.
Corollary 3: For an increasing bijection h: [0, 1] → [0, 1], the
following statements are equivalent.
1) h satisfies the multiplicative Cauchy functional equation
(12) for all x, y ∈ [0, 1].
2) There exists a unique constant c ∈ (0, ∞), such that
h(x) = xc
for all x ∈ [0, 1].
Corollary 4: For an increasing bijection h: [0, 1] → [0, 1], the

1) h satisfies the functional equation
h
y
x
=
h(y)
h(x)
, x, y ∈ [0, 1], x > y. (23)
2) There exists a unique constant c ∈ (0, ∞), such that
h(x) = xc
for all x ∈ [0, 1].
Proof: 1) =⇒ 2) Let x, y ∈ [0, 1] and x > y be fixed. Define
z = y/x. We see that z ∈ [0, 1). Setting y = zx in (23), we get
the multiplicative Cauchy functional equation
h(zx) = h(z)h(x), x, z ∈ [0, 1], x > 0, z < 1.
For x = 0 or z = 1, the function h also satisfies previous equa-
tion, since h(0) = 0 and h(1) = 1. By virtue of Corollary 3, we
get the thesis.
2) =⇒ 1) This implication is obvious.
III. ON THE EQUATION (4) WHEN S1, S2
ARE STRICT T-CONORMS
Our main goal in this section is to present the representa-
tions of some classes of fuzzy implications that satisfy (4) when
S1, S2 are strict t-conorms. Within this context, we firstly de-
scribe the general solutions of (4) when S1, S2 are strict t-
conorms. It should be noted that the general solutions of the
distributive equation
F(x, G(y, z)) = G(F(x, z), F(y, z))
where F is continuous and G is assumed to be continuous,
strictly increasing and associative were presented by Aczél (see
[1, Th. 6, p. 319]). Our results can be seen as a generalization
of the previous result without any assumptions on the function
F and less assumptions on the function G.
Theorem 10: Let S1, S2 be strict t-conorms. For a function
I: [0, 1]2
1) The triple of functions S1, S2, I satisfies the functional
equation (4) for all x, y, z ∈ [0, 1].
2) There exist continuous and strictly increasing func-
tions s1, s2: [0, 1] → [0, ∞] with s1(0) = s2(0) = 0 and
s1(1) = s2(1) = ∞, which are uniquely determined up
to positive multiplicative constants, such that S1, S2 ad-
mit the representation (5) with s1, s2, respectively, and for
every fixed x ∈ [0, 1], the vertical section I(x, ·) has one
of the following forms:
I(x, y) = 0, y ∈ [0, 1] (24)
I(x, y) = 1, y ∈ [0, 1] (25)
I(x, y) =
0, if y = 0,
1, if y > 0,
y ∈ [0, 1] (26)
I(x, y) =
0, if y < 1,
1, if y = 1,
y ∈ [0, 1] (27)
I(x, y) = s−1
2 (cx · s1(y)) , y ∈ [0, 1] (28)
with a certain cx ∈ (0, ∞) that is uniquely determined up
to a positive multiplicative constant, depending on con-
stants for s1 and s2.
Proof: 2) =⇒ 1) Let t-conorms S1, S2 have the represen-
tation (5) with some continuous and strictly increasing func-
tions s1, s2: [0, 1] → [0, ∞] such that s1(0) = s2(0) = 0 and
s1(1) = s2(1) = ∞. By Theorem 1 and part 2) of Remark 2,
the functions S1, S2 are strict t-conorms. Let us fix arbitrarily
x ∈ [0, 1]. We consider five cases.
If I(x, y) = 0 for all y ∈ [0, 1], then the left side of
(4) is I(x, S1(y, z)) = 0 and the right side of (4) is
S2(I(x, y), I(x, z)) = S2(0, 0) = 0 for all y, z ∈ [0, 1].
If I(x, y) = 1 for all y ∈ [0, 1], then the left side of
(4) is I(x, S1(y, z)) = 1, and the right side of (4) is
Let I(x, y) have the form (26) for all y ∈ [0, 1]. Fix arbitrarily
y, z ∈ [0, 1]. If y = 0, then the left side of (4) is I(x, S1(0, z)) =
I(x, z) and the right side of (4) is S2(I(x, 0), I(x, z)) =
S2(0, I(x, z)) = I(x, z). Analogously, if z = 0, then both
sides of (4) are equal to I(x, y). If y > 0 and z > 0, then
S1(y, z) > S1(0, 0) = 0 since S1 is strict. Now, the left side
of (4) is I(x, S1(y, z)) = 1, and the right side of (4) is
S2(I(x, y), I(x, z)) = S2(1, 1) = 1.
Let I have the form (27) for all y ∈ [0, 1]. Fix arbitrarily
y, z ∈ [0, 1]. If y = 1, then the left side of (4) is I(x, S1(1, z)) =
I(x, 1) = 1, and the right side of (4) is S2(I(x, 1), I(x, z)) =
S2(1, I(x, z)) = 1. Analogously, if z = 1, then both sides of (4)
are equal to 1. If y < 1 and z < 1, then S1(y, z) < S1(1, 1) = 1
since S1 is strict. Now, the left side of (4) is I(x, S1(y, z)) = 0,
and the right side of (4) is S2(I(x, y), I(x, z)) = S2(0, 0) = 0.
Let I have the form (28) for all y ∈ [0, 1]. Fix arbitrarily
y, z ∈ [0, 1]. If y, z ∈ [0, 1), then we have
I(x, S1(y, z)) = I(x, s−1
1 (s1(y) + s1(z)))
= s−1
2 (cx(s1(y) + s1(z)))
S2(I(x, y), I(x, z)) = S2 s−1
2 (cxs1(y)) , s−1
2 (cxs1(z))
= s−1
2 s2 ◦ s−1
2 (cxs1(y))
+ s2 ◦ s−1
2 (cxs1(z))
= s−1
2 (cxs1(y) + cxs1(z))
= s−1
2 (cx(s1(y) + s1(z)))
= I(x, S1(y, z))
since s1(y) < ∞ and s1(z) < ∞. If y = 1 or z = 1, then
I(x, S1(y, z)) = S2(I(x, y), I(x, z)) = 1.
Finally, let us assume that, for some x ∈ [0, 1], the vertical sec-
tion is given by (28). We know, by Theorem 1, that s1, s2 are
uniquely determined up to positive multiplicative constants. We
show that, in this case, the constant cx in (28) depends on pre-
vious constants. To prove this, let a, b ∈ (0, ∞) be fixed and as-
sume that s1(x) = as1(x) and s2(x) = bs2(x) for x ∈ [0, 1]. By
Theorem 1, functions s1 and s2 are also continuous additive gen-
erators of S1 and S2, respectively. Let us define cx := (b/a)cx.

For all y ∈ [0, 1], we get
s −1
2 (cx · s1(y)) = s −1
2
b
a
cxas1(y)
= s −1
2 (bcxs1(y))
= s−1
2
bcxs1(y)
b
= s−1
2 (cxs1(y))
i.e., the vertical section for s1, s2, and cx is the same as that of
s1, s2, and cx.
1) =⇒ 2) Let us assume that functions S1, S2, and I are
the solutions of the functional equation (4) satisfying the re-
quired properties. From Theorem 1 and part 2) of Remark 2,
the t-conorms S1 and S2 admit the representation (5) for some
continuous additive generators s1, s2: [0, 1] → [0, ∞] such that
s1(0) = s2(0) = 0 and s1(1) = s2(1) = ∞. Moreover, both
generators are uniquely determined up to positive multiplica-
tive constants. Now, (4) becomes
I x, s−1
1 (s1(y) + s1(z)) = s−1
2 (s2(I(x, y)) + s2(I(x, z)))
(29)
for all x, y, z ∈ [0, 1]. Let x ∈ [0, 1] be arbitrary but fixed. Define
a function Ix: [0, 1] → [0, 1] by the formula
Ix(y) = I(x, y), y ∈ [0, 1].
By routine substitutions, hx = s2 ◦ Ix ◦ s−1
1 , u = s1(y), v =
s1(z), for y, z ∈ [0, 1], from (29), we obtain the additive Cauchy
functional equation
hx(u + v) = hx(u) + hx(v), u, v ∈ [0, ∞]
where hx: [0, ∞] → [0, ∞]. By Proposition 2, we get either
hx = ∞, hx = 0, or
hx(u) =
0, if u = 0
∞, if u > 0
for u ∈ [0, ∞], or
hx(u) =
0, if u < ∞
∞, if u = ∞
for u ∈ [0, ∞], or there exists a constant cx ∈ (0, ∞) such that
hx(u) = cx · u, for u ∈ [0, ∞].
Because of the definition of the function hx, we get either
Ix = 1, or Ix = 0, or
Ix(y) =
0, if y = 0
1, if y > 0
for y ∈ [0, 1], or
Ix(y) =
0, if y < 1
1, if y = 1
for y ∈ [0, 1], or Ix(y) = s−1
2 (cx · s1(y)) for y ∈ [0, 1] and with
cx ∈ (0, ∞).
We show that in the last case, the constant cx is uniquely
determined up to a positive multiplicative constant depending on
constants for s1 and s2. Let s1(x) = as1(x) and s2(x) = bs2(x)
for all x ∈ [0, 1] and some a, b ∈ (0, ∞). Further, let cx be a
constant in (28) for s1, s2. If we assume that
s−1
2 (cx · s1(y)) = s −1
2 (cx · s1(y))
then we get
s−1
2 (cx · s1(y)) = s−1
2
cx · s1(y)
b
and therefore,
cx · s1(y) =
cx · a · s1(y)
b
and thus, when y = 0, we get
cx =
b
a
cx.
Remark 3: From the previous proof, it follows that if we
assume that S1 = S2 and for some x ∈ [0, 1], the vertical sec-
tion I(x, ·) has the form (28), then the constant cx is uniquely
determined.
Since we are interested in finding solutions of (4) in the
fuzzy logic context, we can easily obtain an infinite number of
solutions that are fuzzy implications. It should be noted that,
with this assumption, the vertical section (24) is not possible,
while for x = 0, the vertical section should be (25). Also, a
fuzzy implication is decreasing in the first variable while it is
increasing in the second one.
Example 1: If S1, S2 are both strict t-conorms, then the great-
est solution that is a fuzzy implication is the greatest fuzzy
implication [5]:
I1(x, y) =
0, if x = 1 and y = 0
1, otherwise.
The vertical sections are the following: For x ∈ [0, 1), this is
(25), and for x = 1, this is (26).
Example 2: If S1, S2 are both strict t-conorms, then the least
solution that is a fuzzy implication is the least fuzzy implication
[5]:
I0(x, y) =
1, if x = 0 or y = 1
0, otherwise.
The vertical sections are the following: For x = 0, this is (25),
and for x ∈ (0, 1], this is (27).
A. Continuous Solutions for I in (4) With Strict T-Conorms
From the previous result, we are in a position to describe the
continuous solutions I of (4).
Theorem 11 (cf. [1]): Let S1, S2 be strict t-conorms. For a
continuous function I: [0, 1]2
→ [0, 1], the following statements
are equivalent.

mit the representation (5) with s1, s2, respectively, and ei-
ther I = 0, or I = 1, or there exists a continuous function
c: [0, 1] → (0, ∞), uniquely determined up to a positive
multiplicative constant depending on constants for s1 and
s2, such that I has the form
I(x, y) = s−1
2 (c(x) · s1(y)) , x, y ∈ [0, 1]. (30)
Proof: 2) =⇒ 1) It is obvious that all functions I described in
2) are continuous. By previous general solution, they satisfy our
functional equation (4) with strict t-conorms S1, S2 generated
from s1 and s2, respectively.
1) =⇒ 2) From Theorem 10, we know what are the possible
vertical sections for the fixed x ∈ [0, 1]. Since I is continuous,
for every x ∈ [0, 1], the vertical sections are also continuous,
and consequently, the vertical sections (26) and (27) are not
possible.
Let us assume that there exists some x0 ∈ [0, 1] such that
I(x0, y) = 0 for all y ∈ [0, 1], i.e., the vertical section for x0
is (24). In particular, I(x0, 1) = 0, but for the other possible
vertical sections, we always have I(x, 1) = 1; therefore, the
only possibility in this case is I = 0.
Analogously, let us assume that there exists some x0 ∈ [0, 1]
such that I(x0, y) = 1 for all y ∈ [0, 1] i.e., the vertical section
for x0 is (25). In particular, I(x0, 0) = 1, but for the other pos-
sible vertical sections, we always have I(x, 0) = 0; therefore,
the only possibility in this case is I = 1.
Finally, assume that, for all x ∈ [0, 1], the vertical sections
Ix = 0 and Ix = 1. This implies that the vertical section is (28).
Therefore, there exists a function c: [0, 1] → (0, ∞) such that I
has the form (30). This function c is continuous since for any
fixed y ∈ (0, 1), it is a composition of continuous functions
c(x) =
s2(I(x, y))
s1(y)
, x ∈ [0, 1].
From the previous formula, one can immediately obtain, that the
function c is uniquely determined up to a positive multiplicative
constant, depending on constants for s1 and s2.
Example 3: If we assume that S1 = S2 and the function c = 1
in (30), then our solution is trivial
I(x, y) = s−1
(c(x) · s(y)) = s−1
(s(y)) = y, x, y ∈ [0, 1].
This function is not a fuzzy implication.
Since (4) is the generalization of a tautology from the classical
logic involving Boolean implication, it is reasonable to expect
that the solution I of (4) is also a fuzzy implication, but from
Theorem 11, we obtain the following result.
Corollary 5: If S1, S2 are strict t-conorms, then there are no
continuous solutions I of (4) that satisfy (I3).
Proof: Let a continuous function I satisfy (I3) and (4) with
some strict t-conorms S1, S2 with continuous additive genera-
tors s1, s2, respectively. Then, I has the form (30) with a con-
tinuous function c: [0, 1] → (0, ∞), but in this case, we get
I(0, 0) = s−1
2 (c(0) · s1(0)) = s−1
2 (c(0) · 0) = s−1
2 (0) = 0
and therefore, I does not satisfy the first condition in (I3).
B. Noncontinuous Solutions of (4) With Strict T-Conorms
From Corollary 5, it is obvious that we need to look for
solutions that are not continuous at the point (0, 0), and we
explore this case now.
Theorem 12: Let S1, S2 be strict t-conorms and let a func-
tion I: [0, 1]2
→ [0, 1] be continuous except at the point (0, 0),
which satisfies (I3) and (NC). Then, the following statements
are equivalent.
mit the representation (5) with s1, s2, respectively, and a
continuous function c: [0, 1] → (0, ∞] with c(x) < ∞ for
x ∈ (0, 1], c(0) = ∞ that is uniquely determined up to a
positive multiplicative constant, depending on constants
for s1 and s2, such that I has the form
I(x, y) =
1, if x = y = 0
s−1
2 (c(x) · s1(y)) , otherwise
(31)
for all x, y ∈ [0, 1].
Proof: 2) =⇒ 1) It is obvious that S1, S2 are strict t-conorms.
Moreover, the function I defined by (31) is continuous except
at the point (0, 0) and satisfies (I3) and (NC), since
I(0, 0) = 1
I(x, 1) = s−1
2 (c(x) · s1(1)) = s−1
2 (∞) = 1, x ∈ [0, 1]
I(0, x) = s−1
2 (c(0) · s1(x)) = s−1
2 (∞) = 1, x ∈ (0, 1].
By our previous general solution, they satisfy our functional
equation (4) with strict t-conorms S1, S2 generated from s1 and
s2, respectively.
1) =⇒ 2) Let us assume that the functions I and S1, S2 are
the solutions of (4) satisfying the required properties. From
Theorem 10, there exist continuous and strictly increasing
functions s1, s2: [0, 1] → [0, ∞] with s1(0) = s2(0) = 0 and
s1(1) = s2(1) = ∞, which are uniquely determined up to pos-
itive multiplicative constants, such that S1, S2 admit the repre-
sentation (5) with s1, s2, respectively.
Let x ∈ (0, 1] be arbitrary but fixed. Again from Theorem 10,
we get either Ix = 1, or Ix = 0, or Ix(y) = s−1
2 (c(x) · s1(y))
for all y ∈ [0, 1] and with c(x) ∈ (0, ∞).
From the continuity of the function I and the assumptions of
I, as shown in the proof of [2, Th. 5], the first two cases are
not possible. Indeed, if we take y = 1, then there are only two
possibilities, for any x ∈ (0, 1], either Ix(1) = 0, or Ix(1) = 1.
However, I1(1) = I(1, 1) = 1 and from the continuity of I on
the first variable (for x > 0 and y = 1), we get Ix(1) = 1 for
every x ∈ (0, 1], so Ix = 0 for every x ∈ (0, 1]. On the other
hand, taking y = 0, we also obtain two possibilities, for any x ∈
(0, 1], either Ix(0) = 0, or Ix(0) = 1, but I1(0) = I(1, 0) = 0,
and from the continuity of I on the first variable (for x > 0 and
y = 0), we get Ix(1) = 0 for every x ∈ (0, 1]; therefore, Ix =
1 for every x ∈ (0, 1]. We proved that there exists a function

c: (0, 1] → (0, ∞) such that I has the form (30). This function c
is continuous, since for any fixed y ∈ (0, 1), it is a composition
of continuous functions:
c(x) =
s2(I(x, y))
s1(y)
, x ∈ (0, 1]. (32)
If x = 0, then using similar steps as in the proof of Theorem
10, we obtain the additive Cauchy functional equation
h0(u + v) = h0(u) + h0(v), u, v ∈ (0, 1] (33)
where the function h0: (0, ∞] → [0, ∞] defined by the formula
h0 = s2 ◦ I0 ◦ s−1
1 is continuous (here, I0(y) = I(0, y) for all
y ∈ [0, 1]). Corollary 2 implies that h0 = ∞, or h0 = 0, or there
exists c(0) ∈ (0, ∞) such that h0(u) = c(0) · u for u ∈ (0, 1].
However
h0(1) = s2(I0(s−1
1 (1))) = s2(I(0, 1)) = s2(1) = ∞
and therefore, h0 = 0, and the solution h0 = ∞ implies c(0) =
∞. Therefore, we get
I(0, y) = I0(y) = s−1
2 (c(0) · s1(y)) , y ∈ (0, 1] (34)
and some c(0) ∈ (0, ∞]. We show that c(0) = ∞. From (29),
substituting x = 0 and z = 0, it follows that, for all y ∈ [0, 1],
we have
I 0, s−1
1 (s1(y) + s1(0)) = s−1
2 (s2(I(0, y)) + s2(I(0, 0))).
Since s1(0) = 0, s1(1) = ∞, and I(0, 0) = 1, we get
I(0, y) = 1, y ∈ [0, 1].
Let y ∈ (0, 1) be fixed. By (34), we get
1 = s−1
2 (c(0) · s1(y))
thus, c(0) · s1(y) = s2(1) = ∞. Since y ∈ (0, 1), we obtain
that c(0) = ∞.
Finally, we must prove the existence of the following limit
limx→0+ c(x) = c(0). To this end, we fix arbitrarily y ∈ (0, 1).
From the continuity, limx→0+ I(x, y) = I(0, y). Moreover, s2
is continuous, and therefore
lim
x→0+
c(x) = lim
x→0+
s2(I(x, y))
s1(y)
=
s2(limx→0+ I(x, y))
s1(y)
=
s2(I(0, y))
s1(y)
=
s2(1)
s1(y)
=
∞
s1(y)
= ∞ = c(0)
and c is a continuous function.
Remark 4: The function I in the previous theorem can also
be written in the form
I(x, y) = s−1
2 (c(x) · s1(y)) , x, y ∈ [0, 1]
with the convention that 0 × ∞ = ∞ × 0 = ∞.
From the previous proof, we see that a function I given by
(31) with a continuous function c satisfies conditions (I3). Ad-
ditionally, by the increasing nature of continuous generators s1
and s2, we get that I is increasing with respect to the second
variable. Unfortunately, we can say nothing about its mono-
tonicity with respect to the first one. The next result solves this
by showing some necessary and sufficient conditions.
Corollary 6: If S1, S2 are strict t-conorms and I is a fuzzy
implication that is continuous except at the point (0, 0), then the
mit the representation (5) with s1, s2, respectively, and
a continuous, decreasing function c: [0, 1] → (0, ∞] with
c(x) < ∞ for x ∈ (0, 1], c(0) = ∞, uniquely determined
up to a positive multiplicative constant depending on con-
stants for s1 and s2, such that I has the form (31) for all
x, y ∈ [0, 1].
We would like to underline the main difference between
Theorem 12 and the previous result. In Corollary 6, we have
the assumption that I is a fuzzy implication in the sense of
Definition 3.
Example 4: One specific example is the function c(x) = 1/x
for all x ∈ [0, 1], with the assumption that 1/0 = ∞. In this
case, the solution I is the following:
I(x, y) =



1, if x = y = 0
s−1
2
1
x
· s1(y) , otherwise
for all x, y ∈ [0, 1]. In the special case when s1 = s2, i.e., S1 =
S2, we obtain the function from the Yager’s class of g-generated
fuzzy implications (see [34, p. 202]).
IV. ON THE EQUATION (4) WHEN S1, S2 ARE
NILPOTENT T-CONORMS
In this section, our main goal is to present the characteriza-
tions of the classes of fuzzy implications that satisfy (4) when
S1, S2 are both nilpotent t-conorms, but we first describe the
general solutions of (4) when S1, S2 are nilpotent t-conorms.
From this result, we again show that there are no continuous
fuzzy implications I that are solutions for (4) for nilpotent t-
conorms and, hence, proceed to investigate noncontinuous so-
lutions for I obeying (4).
Theorem 13: Let S1, S2 be nilpotent t-conorms. For a function
I: [0, 1]2
2) There exist continuous and strictly increasing functions
s1, s2: [0, 1] → [0, ∞] with s1(0) = s2(0) = 0, s1(1) <
∞ and s2(1) < ∞, which are uniquely determined up to
positive multiplicative constants, such that S1, S2 admit
the representation (5) with s1, s2, respectively, and for
every fixed x ∈ [0, 1], the vertical section I(x, ·) has one
of the following forms:
I(x, y) = 0, y ∈ [0, 1] (35)
I(x, y) = 1, y ∈ [0, 1] (36)
I(x, y) =
0, if y = 0,
1, if y > 0,
y ∈ [0, 1] (37)

I(x, y) = s−1
2 (min(cx · s1(y), s2(1))) , y ∈ [0, 1]
(38)
with a certain cx ∈ [s2(1)/s1(1), ∞) uniquely determined
stants for s1 and s2.
Proof: 2) =⇒ 1) Let t-conorms S1, S2 have the representa-
tion (5) with some continuous and strictly increasing functions
s1, s2: [0, 1] → [0, ∞] with s1(0) = s2(0) = 0, s1(1) < ∞, and
s2(1) < ∞. By Theorem 1 and part 3) of Remark 2, the
functions S1, S2 are nilpotent t-conorms. Let us fix arbitrarily
x ∈ [0, 1]. We consider four cases.
If I(x, y) = 0 for all y ∈ [0, 1], then the left side of (4)
is equal to I(x, S1(y, z)) = 0, and the right is equal to
If I(x, y) = 1 for all y ∈ [0, 1], then the left side of (4)
is equal to I(x, S1(y, z)) = 1, and the right is equal to
Let I(x, y) have the form (37) for all y ∈ [0, 1]. Fix ar-
bitrarily y, z ∈ [0, 1]. If y = 0, then the left side of (4) is
equal to I(x, S1(0, z)) = I(x, z), and the right is equal to
S2(I(x, 0), I(x, z)) = S2(0, I(x, z)) = I(x, z). Analogously,
if z = 0, then both sides of (4) are equal to I(x, y). If y > 0 and
z > 0, then S1(y, z) ≥ min(S1(y, y), S1(z, z)) > min(y, z) >
0 since S1 is nilpotent, i.e., continuous and Archimedean. Now,
the left side of (4) is equal to I(x, S1(y, z)) = 1, and the right
is equal to S2(I(x, y), I(x, z)) = S2(1, 1) = 1.
If I has the form (38) for all y ∈ [0, 1] with some cx ∈
[s2(1)/s1(1), ∞), then one can check, similar to the proof of
Theorem 10, that the triple of functions S1, S2, I satisfies the
functional equation (4).
Finally, let us assume that, for some x ∈ [0, 1], the vertical
section is given by (38). We know, by Theorem 1, that continuous
additive generators s1, s2 are unique up to a positive multiplica-
tive constant. We show that, in this case, the constant cx in (38)
depends on previous constant. To prove this, let a, b ∈ (0, ∞) be
fixed and assume that s1(x) = as1(x) and s2(x) = bs2(x) for
all x ∈ [0, 1]. By Theorem 1, functions s1 and s2 are also contin-
uous additive generators of t-conorms S1 and S2, respectively.
Let us define cx := (b/a)cx. For all y ∈ [0, 1], we get
s −1
2 (min(cx · s1(y), s2(1)))
= s −1
2 min
b
a
cx · as1(y), bs2(1)
= s −1
2 (b · min(cx · s1(y), s2(1)))
= s−1
2
b · min(cx · s1(y), s2(1))
b
= s−1
2 (min(cx · s1(y), s2(1)))
i.e., the vertical section for s1, s2, and cx is the same as for
s1, s2, and cx.
1) =⇒ 2) Let us assume that functions S1, S2, and I are the
solutions of the functional equation (4) satisfying the required
properties. Then, from Theorem 1 and part 3) of Remark 2,
the t-conorms S1 and S2 admit the representation (5) for some
continuous additive generators s1, s2: [0, 1] → [0, ∞] such that
s1(0) = s2(0) = 0, s1(1) < ∞, and s2(1) < ∞. Moreover,
both generators are uniquely determined up to positive mul-
tiplicative constants. Now, (4) becomes, for all x, y, z ∈ [0, 1]
I x, s−1
1 (min(s1(y) + s1(z), s1(1)))
= s−1
2 (min(s2(I(x, y)) + s2(I(x, z)), s2(1))). (39)
Fix arbitrarily x ∈ [0, 1] and define a function Ix: [0, 1] → [0, 1]
by the formula
Ix(y) = I(x, y), y ∈ [0, 1].
By routine substitutions, hx = s2 ◦ Ix ◦ s−1
1 , u = s1(y), v =
s1(z) for y, z ∈ [0, 1], from (39), we obtain the following func-
tional equation, for u, v ∈ [0, s1(1)]:
hx(min(u + v, s1(1))) = min(hx(u) + hx(v), s2(1))
where the function hx: [0, s1(1)] → [0, s2(1)]. By Proposition
3, we get either hx = s2(1), hx = 0, or
hx(u) =
0, if u = 0
s2(1), if u > 0
for u ∈ [0, s1(1)], or there exists a constant cx ∈
[s2(1)/s1(1), ∞) such that hx(u) = min(cx · u, s2(1)), for
u ∈ [0, s1(1)].
Because of the definition of the function hx we get either
Ix = 1, Ix = 0, or
Ix(y) =
0, if y = 0
1, if y > 0
for y ∈ [0, 1], or Ix(y) = s−1
2 (min(cx · s1(y), s2(1))) for y ∈
[0, 1] and with cx ∈ [s2(1)/s1(1), ∞).
We show that, in the last case, the constant cx is uniquely
determined up to a positive multiplicative constant depending on
constants for s1 and s2. Let s1(x) = as1(x) and s2(x) = bs2(x)
for all x ∈ [0, 1] and some a, b ∈ (0, ∞). Further, let cx be a
constant in (38) for s1, s2. If we assume that
s−1
2 (min(cx · s1(y), s2(1))) = s −1
2 (min(cx · s1(y), s2(1)))
then we get
s−1
2 (min(cx · s1(y), s2(1)))
= s−1
2
min(cx · as1(y), bs2(1))
b
and therefore
min(cx · s1(y), s2(1)) = min
cx · as1(y)
b
, s2(1)
and thus, whenever cx · s1(y) < s2(1), we have
cx · s1(y) =
cx · a · s1(y)
b
.
Therefore, if y = 0, then we get cx = (b/a)cx.
Remark 5: From the previous proof, it follows that if we
assume that S1 = S2 and for some x ∈ [0, 1], the vertical sec-
tion I(x, ·) has the form (38), then the constant cx is uniquely
determined.

We can easily obtain an infinite number of solutions that are
fuzzy implications. It should be noted that, with this assumption,
the vertical section for x = 0 should be (36).
Example 5: If S1, S2 are both nilpotent t-conorms, then the
greatest solution of (4), which is a fuzzy implication, is the
greatest fuzzy implication I1.
Example 6: If S1, S2 are both nilpotent t-conorms, then
the least solution of (4), which is a fuzzy implication, is the
following:
I(x, y) =
1, if x = 1
s−1
2 min s2 (1)
s1 (1) · s1(y), s2(1) , if x < 1.
In the special case, when s1 = s2, i.e., S1 = S2, then we obtain
the following fuzzy implication:
I(x, y) =
1, if x = 1
y, if x < 1
which is also the least (S, N)-implication (see [7, Ex. 1.5]).
A. Continuous Solutions of (4) With Nilpotent T-Conorms
Similar to the proofs of Theorems 11 and 13, we can deduce
the following result.
Theorem 14: Let S1, S2 be nilpotent t-conorms. For a con-
tinuous function I: [0, 1]2
→ [0, 1], the following statements are
equivalent.
s1, s2: [0, 1] → [0, ∞] with s1(0) = s2(0) = 0, s1(1) <
the representation (5) with s1, s2, respectively, and either
I = 0, or I = 1, or there exists a continuous function
c: [0, 1] → [s2(1)/s1(1), ∞), uniquely determined up to
a positive multiplicative constant depending on constants
for s1 and s2, such that I has the form
I(x, y) = s−1
2 (min(c(x) · s1(y), s2(1))) (40)
for all x, y ∈ [0, 1].
Corollary 7: If S1, S2 are nilpotent t-conorms, then there are
no continuous solutions I of (4) that satisfy (I3).
Proof: Let a continuous function I satisfy (I3) and (4) with
some nilpotent t-conorms S1, S2 with continuous additive gen-
erators s1, s2, respectively. Then, I has the form (40) with a con-
tinuous function c: [0, 1] → [s2(1)/s1(1), ∞), but in this case,
we get
I(0, 0) = s−1
2 (min (c(0) · s1(0), s2(1)))
= s−1
2 (min(0, s2(1))) = 0
and therefore, I does not satisfy the first condition in (I3).
B. Noncontinuous Solutions of (4) With Nilpotent T-Conorms
From Corollary 7, it is obvious that we need to look for
solutions that are not continuous at the point (0, 0). Using similar
methods as earlier, we can prove the following fact.
Theorem 15: Let S1, S2 be nilpotent t-conorms and let a func-
tion I: [0, 1]2
→ [0, 1] be continuous except at the point (0, 0),
which satisfies (I3) and (NC). Then, the following statements
are equivalent.
s1, s2: [0, 1] → [0, ∞] with s1(0) = s2(0) = 0, s1(1) <
∞ and s2(1) < ∞, which are uniquely determined up
mit the representation (5) with s1, s2, respectively, and
a continuous function c: [0, 1] → [s2(1)/s1(1), ∞] with
c(x) < ∞ for x ∈ (0, 1], c(0) = ∞, uniquely determined
stants for s1 and s2, such that I has the form
I(x, y) =
1, if x=y=0
s−1
2 (min (c(x)s1(y), s2(1))) , otherwise
(41)
for x, y ∈ [0, 1].
Remark 6: The function I in the previous theorem can also
be written in the form
I(x, y) = s−1
2 (min (c(x)s1(y), s2(1))) , x, y ∈ [0, 1]
with the convention that 0 × ∞ = ∞ × 0 = ∞.
It can easily be verified that a function I given by the formula
(41) with a continuous function c satisfies conditions (I3). Ad-
ditionally, by the increasing nature of continuous generators s1
and s2, we get that I is increasing with respect to the second
variable, but we can say nothing about its monotonicity with
respect to the first one. The next result solves this by showing
some necessary and sufficient conditions.
Corollary 8: Let S1, S2 be nilpotent t-conorms and I be a
fuzzy implication that is continuous except at the point (0, 0).
Then, the following statements are equivalent.
s1, s2: [0, 1] → [0, ∞] with s1(0) = s2(0) = 0, s1(1) <
the representation (5) with s1, s2, respectively, and a con-
tinuous decreasing function c: [0, 1] → [s2(1)/s1(1), ∞]
with c(x) < ∞ for x ∈ (0, 1], c(0) = ∞, uniquely deter-
mined up to a positive multiplicative constant depending
on constants for s1 and s2, such that I has the form (41)
for all x, y ∈ [0, 1].
Here again, we would like to underline the main difference
between Theorem 15 and the previous result. In Corollary 8, we
have the assumption that I is a fuzzy implication in the sense of
Definition 3.
Example 7: One specific example, when s2(1) ≤ s1(1),
is again the function c(x) = 1/x for all x ∈ [0, 1], with the

assumption that 1/0 = ∞. In this case, the solution I is the
following:
I(x, y) =



1, if x = y = 0
s−1
2 min
1
x
· s1(y), s2(1) , otherwise
= s
(−1)
2
1
x
· s1(y)
for all x, y ∈ [0, 1]. In the special case when s1 = s2, i.e., S1 =
S2, we obtain the function from the Yager’s class of g-generated
fuzzy implications (see [34, p. 202]). Quite evidently, there are
other candidates for the function c, viz., c(x) = 1 + 1/x or
c(x) = 1/x2
for x ∈ [0, 1].
V. ON (4) WHEN I IS AN R-IMPLICATION
In this section, we discuss the distributive equation (4) when
I is an R-implication obtained from a continuous Archimedean
t-norm T. The case when T is a nilpotent t-norm has been
investigated by Balasubramaniam and Rao in [10]. The result
from this paper can be written in the following form.
Theorem 16 [10, Th. 4]: Let S1, S2 be t-conorms. For an R-
implication I obtained from a nilpotent t-norm, the following
2) S1 = S2 = max.
In [10], we can find the following sentence “. . . the authors
have a strong feeling that it holds for the case when the R-
implication is obtained from a strict t-norm . . ..” We will show
in this section that this is not true, i.e., for an R-implication
generated from a strict t-norm, there exist other solutions than
maximum.
We start our presentation with some connections between
solutions S1, S2 of (4) and the properties of R-implications.
Lemma 1: Let S1, S2 be t-conorms and I: [0, 1]2
→ [0, 1] be
a function that satisfies the left neutrality property (NP). If a
triple of functions S1, S2, I satisfies the functional equation (4),
then S1 = S2.
Proof: Let I satisfy (NP). Putting x = 1 in (4), we get
I(1, S1(y, z)) = S2(I(1, y), I(1, z)), y, z ∈ [0, 1]
and thus
S1(y, z) = S2(y, z), y, z ∈ [0, 1].
Hence, S1 = S2.
Note that the previous result is true for any binary operations
S1 and S2.
Lemma 2: Let S1, S2 be continuous Archimedean t-conorms,
and let I: [0, 1]2
→ [0, 1] be a function that satisfies the ordering
property (OP). If S1, S2, I satisfy the functional equation (4),
then S2 is a nilpotent t-conorm.
Proof: Since S1 is a continuous Archimedean t-conorm, from
part 1) of Remark 1, for every y ∈ (0, 1), we have S1(y, y) > y.
Let us fix any y ∈ (0, 1) and take some x ∈ (y, S1(y, y)). By
(OP), we get I(x, y) = y0 < 1, whereas from (4), we obtain
1 = I(x, S1(y, y)) = S2(I(x, y), I(x, y)) = S2(y0, y0).
Hence, by part 2) of Remark 1, the t-conorm S2 is nilpotent.
Since an R-implication I generated from left-continuous t-
norm satisfies (NP) and (OP), from previous Lemma 1, we have
S1 = S2 in (4), and hence, it suffices to consider the following
functional equation:
I(x, S(y, z)) = S(I(x, y), I(x, z)), x, y, z ∈ [0, 1].
(42)
Further, from Lemma 2, we get the following.
Corollary 9: Let S be a continuous Archimedean t-conorm,
and let I be an R-implication generated from some left-
continuous t-norm. If the couple of functions S, I satisfies the
functional equation (42), then S is nilpotent.
Corollary 10: An R-implication I obtained from a left-
continuous t-norm does not satisfy (4), when S1 and S2 are
both strict t-conorms.
From previous investigations, it follows that we should con-
sider the situation when S is a nilpotent t-conorm. As a result,
we obtain the following theorem.
Theorem 17: For a nilpotent t-conorm S and an R-implication
I generated from a strict t-norm, the following statements are
equivalent.
1) The couple of functions S, I satisfies the functional equa-
tion (42) for all x, y, z ∈ [0, 1].
2) There exist ϕ ∈ Φ, which is uniquely determined, such
that S admits the representation (8) with ϕ and I admits
the representation (10) with ϕ.
Proof: 2) =⇒ 1) Assume that there exists a ϕ ∈ Φ, such
that S admits the representation (8) with ϕ and I admits the
representation (10) with ϕ, i.e.,
S(x, y) = ϕ−1
(min(ϕ(x) + ϕ(y), 1)), x, y ∈ [0, 1]
and
I(x, y) =



1, if x ≤ y,
ϕ−1 ϕ(y)
ϕ(x)
, if x > y, x, y ∈ [0, 1].
We will show that I and S satisfy (42). Let us take any x, y, z ∈
[0, 1]. The left side of (42) is equal to
I(x, S(y, z))
= I(x, ϕ−1
(min(ϕ(y) + ϕ(z), 1)))
=



1, if x≤ϕ−1
(min(ϕ(y)+ϕ(z), 1))
ϕ−1 min(ϕ(y) + ϕ(z), 1)
ϕ(x)
, otherwise
=



1, if ϕ(x) ≤ min(ϕ(y) + ϕ(z), 1)
ϕ−1
min
ϕ(y) + ϕ(z)
ϕ(x)
,
1
ϕ(x)
, otherwise
=



1, if ϕ(x) ≤ ϕ(y) + ϕ(z)
ϕ−1 ϕ(y) + ϕ(z)
ϕ(x)
, otherwise.

On the other hand, the right side of (42) is equal to
S(I(x, y), I(x, z))
=



S(1, 1), if x ≤ y and x ≤ z
S ϕ−1 ϕ(y)
ϕ(x)
, 1 , if x > y and x ≤ z
S 1, ϕ−1 ϕ(z)
ϕ(x)
, if x ≤ y and x > z
S ϕ−1 ϕ(y)
ϕ(x)
, ϕ−1 ϕ(z)
ϕ(x)
, if x>y and x>z
=



1, if x≤y or x≤z
S ϕ−1 ϕ(y)
ϕ(x)
, ϕ−1 ϕ(z)
ϕ(x)
, if x>y and x>z
=



1, if x ≤ y or x ≤ z
ϕ−1
min
ϕ(y) + ϕ(z)
ϕ(x)
, 1 , if x > y and x > z
=



1, if x ≤ y or x ≤ z
1, if x > y and x > z and ϕ(x) ≤ ϕ(y) + ϕ(z)
ϕ−1 ϕ(y) + ϕ(z)
ϕ(x)
, otherwise
=



1, if ϕ(x) ≤ ϕ(y) + ϕ(z)
ϕ−1 ϕ(y) + ϕ(z)
ϕ(x)
, otherwise
which ends the proof in this direction.
1) =⇒ 2) Let S be a nilpotent t-conorm and I be an R-
implication generated from a strict t-norm. Because of Theorem
3, there exists a unique ϕ ∈ Φ such that S has the form (8).
In fact, the increasing bijection ϕ can be seen as a continuous
generator of S. Further, by virtue of Theorem 5, there exists
ψ ∈ Φ, uniquely determined up to a positive constant exponent,
such that
I(x, y) =



1, if x ≤ y
ψ−1 ψ(y)
ψ(x)
, if x > y x, y ∈ [0, 1].
(43)
It is obvious that I is a fuzzy implication that is continuous
except at the point (0, 0). Therefore, if functions S, I satisfy the
functional equation (42), then from Corollary 8, there exists a
continuous decreasing function c: [0, 1] → [1, ∞], with c(x) <
∞ for x ∈ (0, 1] and c(0) = ∞, such that
I(x, y) =
1, if x = y = 0
ϕ−1
(min (c(x) · ϕ(y), 1)) , otherwise
(44)
for x, y ∈ [0, 1].
Let us take any x ∈ (0, 1]. From (43), for any y < x, we
have I(x, y) < 1, whereas from (44), we have that I(x, y) =
ϕ−1
(c(x) · ϕ(y)). Hence, from the continuity of I (for x > 0),
we get
1 = I(x, x) = lim
y→x−
I(x, y) = lim
y→x−
ϕ−1
(c(x) · ϕ(y))
and therefore, c(x) · ϕ(x) = 1 for x ∈ (0, 1]. This implies that
c(x) =
1
ϕ(x)
< ∞ (45)
for x ∈ (0, 1]. Observe that c is a continuous decreasing function
from (0, 1] to [1, ∞). Moreover, this formula can be considered
also for x = 0, since ∞ = c(0) = 1/ϕ(0) = 1/0, i.e., c is well
defined. Now, comparing (43) with (44) and setting (45) in (44),
we obtain the functional equation
ϕ−1 1
ϕ(x)
· ϕ(y) = ψ−1 ψ(y)
ψ(x)
for x, y ∈ [0, 1] and x > y. By substitutions, hx = ϕ ◦ ψ−1
, u =
ψ(x), and v = ψ(y), we obtain the functional equation
h
v
u
=
h(v)
h(u)
, u, v ∈ [0, 1], u > v.
From Corollary 4, we get that there exists a unique constant
c ∈ (0, ∞) such that h(u) = uc
. By the definition of h, we get
ϕ ◦ ψ−1
(u) = uc
, thus ϕ(x) = (ψ(x))c
for all x ∈ [0, 1]. Since
the increasing bijection ψ is uniquely determined up to a positive
constant exponent, we get that I admits the representation (10)
also with ϕ.
Example 8: Taking ϕ = id[0,1], we obtain the interesting
example that the Łukasiewicz t-conorm and the Goguen impli-
cation satisfy the distributive equation (42).
VI. CONCLUSION
Recently, in [10] and [32], the authors have studied the dis-
tributivity of R- and S-implications over t-norms and t-conorms.
But the distributive equation (4) for R-implications obtained
from strict t-norms was not solved. In this paper, we have char-
acterized a function I that satisfies the functional equation (4),
when S1, S2 are either both strict or nilpotent t-conorms. Us-
ing the previous characterizations, we have shown that for an
R-implication I generated from a strict t-norm T, we have the
following.
1) Equation (4) does not hold when t-conorms S1, S2 are
strict.
2) Equation (4) holds if and only if t-conorms S1 = S2 are
Φ-conjugate with the Łukasiewicz t-conorm for some in-
creasing bijection ϕ, which is a multiplicative generator
of the strict t-norm T.
It is established that in the cases when I is an S-implication or
an R-implication, most of the equations (1)–(3) hold only when
the t-norms and t-conorms are either min or max. That the gen-
eralization (4) has more solutions in the case of R-implications
obtained from strict t-norms is bound to have positive impli-
cations in applications, especially in the new research area of
inference invariant rule reduction.
Also as part of characterizing (4), we have obtained a more
general class of fuzzy implications [see (31)] that contains the
Yager’s class [34] as a special case.
In our future works, we will try to concentrate on some cases
that are not considered in this paper, for example, when S1 is a
strict t-conorm and S2 is a nilpotent t-conorm, and vice versa.

Also, the situation when S1 and S2 are continuous, t-conorms
is still unsolved.
REFERENCES
[1] J. Aczél, Lectures on Functional Equations and Their Applications.
New York: Academic, 1966.
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Michał Baczyński was born in Katowice, Poland,
in 1971. He received the M.Sc. and Ph.D. degrees
in mathematics from the Department of Mathemat-
ics, Physics and Chemistry, University of Silesia,
Katowice, Poland, in 1995 and 2000, respectively.
He is currently with the Institute of Mathematics,
University of Silesia. His current research interests
include fuzzy logic connectives, fuzzy systems, func-
tional equations, algorithms, and data structures. He
is the author or coauthor of more than 15 published
papers in refereed international journals and confer-
ences and is a regular reviewer for many international journals and conferences.
Balasubramaniam Jayaram (S’02–A’03–M’04) re-
ceived the M.Sc. and Ph.D. degrees in mathemat-
ics from Sri Sathya Sai University (SSSU), Anan-
tapur, Andhra Pradesh, India, in 1999 and 2004,
respectively.
He is currently a Lecturer with the Department
of Mathematics and Computer Sciences, SSSU. His
current research interests include fuzzy aggregation
operations, chiefly fuzzy implications, and approxi-
mate reasoning. He is the author or coauthor of more
than 15 published papers in refereed international
journals and conferences and is a regular reviewer for many respected interna-
tional journals and conferences.
Dr. Jayaram is a member of many scientific societies.

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