2.1. Negation Functions

Negation functions constitute a well-known tool allowing to represent fuzzy negations.

2.2. Conjunctors With a Neutral Element

Conjunctors with a neutral element constitute a special class of aggregation functions (see Refs. [10–13]). In this paper we will only be interested in the bivariate case (where aggregation functions are increasing functions A:[0,1]2→[0,1] verifying A(0,0)=0 and A(1,1)=1) and in the following properties.

Definition 2.

[10–13] Let A be an aggregation function and let N be a negation function.

• A is a conjunctor when it satisfies A(1,0)=A(0,1)=0.

• A has a neutral element e∈[0,1] (NE(e)) when A(x,e)=A(e,x)=x for all x∈[0,1].

• A has zero divisors (0Div) when there exist a,b∈[0,1] such that A(a,b)=0.

• A satisfies the Non-Contradiction principle with respect to N (NC(N)) when A(x,N(x))=0 for all x∈[0,1].

Note that when A is a conjunctor, due to its increasingness, A has zero as annihilator element.

Conjunctors with a neutral element may be classified into the two following categories:

1. Conjunctors with neutral element e=1, also known as semi-copulas.² Semi-copulas are conjunctive (C⩽min) and may or may not have zero divisors. Some distinguished families of semi-copulas are the following:

   • Triangular norms (t-norms for short) [21,22], which are associative and commutative semi-copulas.

   • Copulas [27], which are semi-copulas C satisfying C(0,x)=C(x,0)=0 for all x∈[0,1] and the so-called 2-increasing property: C(x1,y1)−C(x1,y2)−C(x2,y1) + C(x2,y2)⩾0 for all x1,y1,x2,y2∈[0,1] such that x1⩽x2 and y1⩽y2.

   The least and the greatest semi-copulas are, respectively, the drastic product t-norm TD (given by TD(x,y)=0 if (x,y)∈[0,1]2 and TD(x,y)=min(x,y) otherwise) and the minimum t-norm TM (TM(x,y)=min(x,y)). Other well-known families of t-norms are the so-called strict t-norms (or t-norms in the Product family), defined as (TP)φ(x,y)=φ−1(φ(x)φ(y)), and nilpotent t-norms (or t-norms in the Łukasiewicz family), (TLK)φ(x,y)=φ−1(max(0,φ(x) + φ(y) − 1)), where φ:[0,1]→[0,1] is an increasing bijection [21,22].

2. Conjunctors with a neutral element e∈[0,1], which are conjunctive (C⩽min) on the square [0,e]2, disjunctive (C⩾max) on [e,1]2 and averaging (min⩽C⩽max) elsewhere. These conjunctors are averaging on the whole unit square only when the limiting functions min and max are chosen in the regions [0,e]2 and [e,1]2, respectively. They have zero divisors if and only if C|[0,e]2 has zero divisors. One distinguished family of aggregation functions in this category is the class of conjunctive uninorms, which are associative and commutative aggregation functions U:[0,1]2→[0,1] with a neutral element e∈[0,1] satisfying U(0,1)=0 that behave as scaled t-norms/t-conorms on [0,e]2/[e,1]2 (see Ref. [28] for a recent survey on these functions).

   Figure 1 depicts the structure of Ce⊥ and Ce⊤, the least and the greatest conjunctors with a neutral element e≠1 (note that the first one is also the least uninorm and belongs to the class Umin).

Figure 1

The least (Ce⊥, left) and the greatest (Ce⊤, right) conjunctors with a neutral element e∈]0,1[.

2.3. Fuzzy Implication Functions

The most accepted definition of fuzzy implication function is the following.

2.4. The Modus Ponens Inequality

The Modus Ponens scheme, a well-known classical inference rule allowing to perform forward reasoning, has usually been translated to the fuzzy framework as follows:

Proposition 3.

[20] Let C be a conjunctor with a neutral element e∈]0,1], let I be an implication function satisfying the Modus Ponens inequality with respect to C and let NI be the natural negation of the fuzzy implication function (NI(x)=I(x,0)). Then:

1. C(x,y)=0 for all x,y∈[0,1] such that y⩽NI(x), which means that C (C|[0,e]2 when e≠1) must satisfy (0Div) unless it is NI=N⊥.

2. [∀x⩾e:NI(x)=0] and [∀x<e,x≠0:NI(x)<e] (note that this implies, in particular, that NI is non-filling and, whenever e≠1, NI is in addition vanishing and non-continuous at least on x=0).

3. If e=1, C=(〈aj,bj,Tj〉)j∈J is a non-trivial ordinal sum t-norm and NI≠N⊥, then:

   1. There exists i∈J such that ai=0 and bi≠1.

   2. [∀x⩾bi:NI(x)=0] and [∀x<bi,x≠0:NI(x)<bi] (note that this implies that NI must be vanishing and non-continuous at least on x=0).

   3. Ti(x∕bi,y∕bi)=0 for all x∈]0,bi],y∈[0,NI(x)].

In the following we present some additional necessary conditions for fuzzy implication functions satisfying the Modus Ponens inequality with respect to conjunctors with a neutral element, and we analyze their consequences.

Proposition 4.

Let C be a conjunctor with a neutral element e∈]0,1] and let I be an implication function satisfying the Modus Ponens inequality with respect to C. Then

1. I(x,y)<e∀x,y∈[0,1] such that x>y.

2. I(x,y)⩽y∀x,y∈[0,1] such that x⩾e.

Proof.

We prove the items step by step.

• Suppose there exist x0,y0∈[0,1] such that x0>y0 and I(x0,y0)⩾e. In this case, from the monotonicity of C and the fact that e is the neutral element of C, we deduce

  C(x0,I(x0,y0))⩾C(x0,e)=x0>y0,
  obtaining a contradiction.

• Note first that choosing x=e in (MP(C)) provides C(e,I(e,y))=I(e,y)⩽y for any y∈[0,1]. Thus, by monotonicity of I we have

  I(x,y)⩽I(e,y)⩽y
  for any x,y∈[0,1] such that x⩾e.

Remark 2

When e=1 the first item in the proposition above may be written as I(x,y)≠1 for all x>y and hence, it is a generalization of Proposition 7.4.2. in Ref. [1]. Note also that, if C=min, the stronger condition I(x,y)⩽y when x>y is obtained. Similarly, the second item when e=1 can be written as I(1,y)⩽y for all y∈[0,1] and hence, it is a generalization of the result given in Ref. [4] for continuous t-norms.

On the other hand, when e≠1, Proposition 4 recovers some weaker statements related to uninorms proved in Ref. [19].

Next result involves several additional properties of fuzzy implication functions.

Proposition 5.

Let C be a conjunctor with a neutral element e∈]0,1] and let I be an implication function satisfying the Modus Ponens inequality with respect to C. Then:

1. NI(x)<e for any x≠0 and NI(x)=0 for any x⩾e.

2. If I satisfies (IP), then it necessarily satisfies (OP).

3. If I satisfies (CB), then necessarily e=1 and I satisfies (NP).

4. If e≠1, then

   • I(x,y)<y for any x>y⩾e, in particular I(1,y)<y for any 1>y⩾e (i.e., I does not satisfy (CB) either (NP)).

   • I does not satisfy (IP) either (OP).

Proof.

Again we give the proof step by step.

1. It suffices to choose y=0 in Proposition 4 (note that this is also a consequence of item 2 in Proposition 3).

2. Directly comes from the fact that (OP) is equivalent to (IP) and [I(x,y)≠1 if x>y].

3. Suppose on the contrary that I satisfies (CB) and e≠1. Then choosing x>y⩾e we get I(x,y)<e⩽y by Proposition 4. This contradicts (CB) and hence e=1. Now (NP) follows directly by taking x=1 in the second item of Proposition 4.

4. The first part of this item is a direct consequence of the previous one. To prove the second part just take x=y⩾e with x≠1, then Proposition 4 provides I(x,x)⩽x≠1.

We finally deal with some sufficient conditions that apply to a broad class of conjunctors that includes, as a particular case, conjunctors with a neutral element.

Proposition 6.

Let C be a conjunctor, let e∈]0,1] and let I be a fuzzy implication. Then:

1. Suppose C(x,e)⩽x for any x∈[0,1] and I(x,y)⩽e for any x,y∈[0,1] such that x⩽y,x≠0,y≠1. If C(x,I(x,y))⩽y for all x,y∈[0,1] such that x>y then I satisfies (MP(C)).

2. Suppose C(e,y)⩽y for any y∈[0,1] and I(x,y)⩽y for any x,y∈[0,1] such that x⩽e,x≠0. If C(x,I(x,y))⩽y for all x,y∈[0,1] such that x>e then I satisfies (MP(C)).

Proof.

1. It suffices to check that (MP(C)) is true for any x,y∈[0,1] such that x⩽y. This is obvious if x=0 or y=1, and otherwise we have, by monotonicity of C, C(x,I(x,y))⩽C(x,e)⩽x⩽y.

2. We now have to prove that (MP(C)) is true for any x,y∈[0,1] such that x⩽e. If x=0, C(x,I(x,y))=0⩽y for any y∈[0,1]. Otherwise, using the monotonicity of C it is C(x,I(x,y))⩽C(e,I(x,y))⩽I(x,y)⩽y.

In the limit case where e=1 we get the following results, which are valid, in particular, for any conjunctive aggregation function and hence for any semi-copula:

Corollary 7.

Let C be a conjunctor and let I be a fuzzy implication function.

1. Suppose C(x,1)⩽x for any x∈[0,1]. If C(x,I(x,y))⩽y for all x,y∈[0,1] such that x>y then I satisfies (MP(C)) (i.e., the values of I when x⩽y are indifferent).

2. Suppose C(1,y)⩽y for any y∈[0,1] and I(x,y)⩽y for any x,y∈[0,1] such that x≠0. Then I satisfies (MP(C)).

3. Suppose C⩽min (i.e., C is conjunctive) and I satisfies I(x,y)⩽y for any x,y∈[0,1] such that x>y. Then I satisfies (MP(C)).

Proof.

The two first items come directly from Proposition 6, and the third one is a combination of them (indeed, because of monotonicity, the conditions C(x,1)⩽x and C(1,y)⩽y are equivalent to C⩽min).

Proposition 8.

Let C be a conjunctor with a neutral element e∈]0,1[ and let I be a fuzzy implication function. If I is an (S,N)-implication, an R-implication, a QL-implication, an f, g, h or hg-generated implication, a probabilistic implication or an S-implication, then I does not satisfy (MP(C)).

Proof.

All the abovementioned families of fuzzy implication functions satisfy (NP) (see e.g., Refs. [8,14,30,31]) and hence Proposition 5 item 4 can be applied.

Nevertheless, there are still many fuzzy implication functions that could satisfy the Modus Ponens inequality with respect to a conjunctor with a neutral element e∈]0,1[. Observe, as it was recalled in Section 2.2, that the behavior of these conjunctors depends on the region of the unit square which is considered, since they are below min on [0,e]2, above max on [e,1]2 and averaging otherwise. The next result analyzes the consequences of choosing the least possible values on each of these regions.

Proposition 9.

Let C be a conjunctor with neutral element e∈]0,1[ and let I be an implication function.

1. Suppose C|[0,e[2=0 and let R1={(x,y)∈[0,1]2:x>y, x<e}. Then I satisfies (MP(C)) on R1 if and only if I(x,y)<e for any (x,y)∈R1.

2. Suppose C|[e,1]2=max and let R2={(x,y)∈[0,1]2:x⩽y<1,x⩾e}. Then I satisfies (MP(C)) on R2 if and only if I(x,y)⩽y for any (x,y)∈R2.

3. Suppose C|[e,1]×[0,e[=min and let R3={(x,y)∈[0,1]2:x>y,x⩾e}. Then I satisfies (MP(C)) on R3 if and only if I(x,y)<e for any (x,y)∈R3 such that y⩾e and I(x,y)⩽y for any (x,y)∈R3 such that y<e.

4. Suppose C|[0,e[×[e,1]=min and let R4={(x,y)∈[0,1]2:0<x⩽y<1,x<e}. Then I satisfies (MP(C)) on R4.

Proof.

The necessity part of any of the three first items is given by Proposition 4. The sufficiency is proved below:

1. If (x,y)∈R1 and I(x,y)<e then we have C(x,I(x,y))=0⩽y.

2. Suppose that (x,y)∈R2 and I(x,y)⩽y. If it were I(x,y)⩽e then Proposition 6 proves the conclusion. Otherwise, if I(x,y)>e then C(x,I(x,y))=max(x,I(x,y))⩽y since both x and I(x,y) are smaller than or equal to y.

3. Suppose now that (x,y)∈R3 and we distinguish two cases. If y⩾e and I(x,y)<e we have (x,I(x,y))∈[e,1]×[0,e[ and consequently

   C(x,I(x,y))=min(x,I(x,y))=I(x,y)<e⩽y.

   On the other hand, if y<e and I(x,y)⩽y we again have (x,I(x,y))∈[e,1]×[0,e[ and

   C(x,I(x,y))=min(x,I(x,y))=I(x,y)⩽y.

4. Finally, if (x,y)∈R4 we again distinguish two cases. If it were I(x,y)<e this would obviously entail C(x,I(x,y))⩽C(x,e)=x⩽y, and otherwise it would be (x,I(x,y))∈[0,e[×[e,1] and then C(x,I(x,y))=min(x,I(x,y))=x⩽y.

Note that the combination of the results given in Proposition 9 allows us to characterize the capacity of the least conjunctor with a neutral element e≠1 (the least uninorm with neutral element e, depicted on the left part of Figure 1) for solving the Modus Ponens inequality.

Proposition 10.

Let Ce⊥ be the least conjunctor with a neutral element e∈]0,1[ and let I be a fuzzy implication function. Then I satisfies (MP(C)) with C=Ce⊥ if and only if the two following conditions hold:

• I(x,y)⩽y for any x,y∈[0,1] such that x⩾e and either y⩽e or x⩽y.

• I(x,y)<e for any x,y∈[0,1] such that x>y and either y⩾e or x<e.

Remark 3

The two following remarks regarding the above characterization are worth noting:

• I satisfies (MP(Ce⊥)) is not equivalent to I⩽ICe⊥, where ICe⊥ is the residuated implication built from Ce⊥ and given (see e.g., Ref. [37]) by

  ICe⊥(x,y)=1if x⩽y,x<e,eif x>y,(x<e or y⩾e), yotherwise.

  Indeed, as stated in Proposition 2, if I satisfies (MP(Ce⊥)) then clearly I⩽ICe⊥, but the converse is not necessarily true if C is not left-continuous, as it happens with Ce⊥. Note in particular that ICe⊥ itself does not satisfy (MP(Ce⊥)), since taking e.g., (x0,y0) such that x0>y0,x0<e provides Ce⊥(x0,ICe⊥(x0,y0))=Ce⊥(x0,e)=x0>y0.

• Proposition 10 is actually a characterization of fuzzy implication functions satisfying (MP(C)) with respect to at least some conjunctor with a neutral element e∈]0,1[, since Ce⊥ is the least of them.

The regions R1 to R4 considered in Proposition 9 are depicted in the top of Figure 3, whereas the possible values of fuzzy implication functions I satisfying (MP(C)) with C=Ce⊥ proved in Proposition 10 can be viewed in the bottom of the figure.

Figure 3

Regions R₁ to R₄ considered in Proposition 9 can be viewed in the top of the figure. Dashed segments do not belong to the corresponding region. The bottom of the figure corresponds to the possible values of the fuzzy implication functions satisfying (MP(Ce⊥)) characterized in Proposition 10.

Example 1.

Fixed some e∈]0,1[, let us show that there are many examples of fuzzy implication functions satisfying conditions in Proposition 10 and, consequently, satisfying (MP(Ce⊥)).

1. The parameterized family of fuzzy implication functions given by

   Ia,e(x,y)=1if x⩽y,x<e,aif a<y<x, yotherwise,
   satisfies (MP(Ce⊥)), for any a such that 0⩽a<e.

2. There are also examples among the residual implications IU derived from uninorms, or RU-implications for short. For instance, it can be easily viewed from Proposition 10 that all RU-implications derived from uninorms in some of the following classes³ satisfy (MP(Ce⊥)):

   1. Uninorms in Umin with neutral element e′<e whose underlying t-conorm is given by the maximum (this can be derived also from Proposition 7 in Ref. [19]).

   2. All representable uninorms with neutral element e (this can be also derived from Theorem 5 in Ref. [19]).

   3. All idempotent uninorms with g(e)=0 where g is the Id-symmetrical function associated to the uninorm (see also Proposition 10 in Ref. [19]).

The next Proposition deals with the cases where the greatest values are chosen in each of the four different regions R1 to R4 (note that in this case there are some small differences in the location of the border values to each region) of the unit square.

Proposition 11.

Let C be a conjunctor with neutral element e∈]0,1[ and let I be an implication function.

1. Suppose C|[0,e]2=min and let R1={(x,y)∈[0,1]2:x>y, x⩽e}. Then I satisfies (MP(C)) on R1 if and only if I(x,y)⩽y for any (x,y)∈R1.

2. Suppose C|]e,1]2=1 and let R2={(x,y)∈[0,1]2:x⩽y, x>e,y≠1}. Then I satisfies (MP(C)) on R2 if and only if I(x,y)⩽e for any (x,y)∈R2.

3. Suppose C|]e,1]×]0,e]=max and let R3={(x,y)∈[0,1]2: x>y,x>e}. Then I satisfies (MP(C)) on R3 if and only if I(x,y)=0 for any (x,y)∈R3.

4. Suppose C|]0,e]×]e,1]=max and let R4={(x,y)∈[0,1]2: x⩽y,x⩽e,x≠0,y≠1}. Then I satisfies (MP(C)) on R4 if and only if I(x,y)⩽max(e,y) for any (x,y)∈R4.

Proof.

The sufficiency part of items 1, 2, and 4 is given by Proposition 6, whereas the sufficiency of item 3 is obvious since I(x,y)=0 implies C(x,I(x,y))=0⩽y. The necessity is proved below:

1. Take (x,y)∈R1. Since x>y we have I(x,y)<e by Proposition 4, and consequently

   C(x,I(x,y))=min(x,I(x,y)).

   Now, if it were I(x,y)>y we would deduce C(x,I(x,y))>y obtaining a contradiction with the Modus Ponens, thus it must be I(x,y)⩽y.

2. Consider now (x,y)∈R2 and suppose that I(x,y)>e. Since x>e, y≠1 and C|]e,1]2=1 we obtain C(x,I(x,y))=1≰y contradicting again the Modus Ponens. Thus, we have I(x,y)⩽e in this case.

3. Let (x,y)∈R3 and suppose that I(x,y)≠0. In this case, Proposition 4 provides I(x,y)<e, and since x>e we get the contradiction

   C(x,I(x,y))=max(x,I(x,y))=x>y.

4. Finally, consider (x,y)∈R4 and let us distinguish two cases. If I(x,y)⩽e then obviously I(x,y)⩽max(e,y). On the other hand, if I(x,y)>e and taking into account that x⩽e we obtain by the Modus Ponens that

   I(x,y)=max(x,I(x,y))=C(x,I(x,y))⩽y,
   and hence I(x,y)⩽max(e,y).

Note that the first item in Proposition 11 forces implications functions to verify NI=N⊥ (see Proposition 3, item 1) and it may be applied, in particular, to the greatest conjunctor with a neutral element e≠1, Ce⊤. Moreover, the combination of the four items of this Proposition characterizes the family of implications functions that satisfy the Modus Ponens inequality with respect to Ce⊤.

Proposition 12.

Let Ce⊤ be the greatest conjunctor with a neutral element e∈]0,1[ and let I be a fuzzy implication function. Then I satisfies (MP(C)) with C=Ce⊤ if and only if the three following conditions hold:

• I(x,y)=0 for any x,y∈[0,1] such that x>y,x>e.

• I(x,y)⩽y for any x,y∈[0,1] such that (x>y,x⩽e) or (x⩽e,y⩾e,x≠0).

• I(x,y)⩽e for any x,y∈[0,1] such that (x⩽y,x>e,y≠1) or (x⩽y,y<e,x≠0).

The regions R1 to R4 slightly modified in Proposition 11 are depicted in the top of Figure 4, whereas the possible values of implications I satisfying (MP(C)) with C=Ce⊤ proved in Proposition 12 can be viewed in the bottom of the figure.

Figure 4

Regions R₁ to R₄ considered in Proposition 11 can be viewed in the top of the figure. Dashed segments do not belong to the corresponding region. The bottom of the figure corresponds to the possible values of the fuzzy implication functions satisfying (MP(Ce⊤)) characterized in Proposition 12.

Remark 4

Since Ce⊤ is left-continuous, Proposition 2 shows that the characterization of the satisfaction of (MP(Ce⊤)), given in the previous proposition, is equivalent to the inequality I⩽ICe⊤, where the latter is the residuated implication of Ce⊤, given by

Observe finally that the fact that Ce⊤ is the greatest conjunctor with neutral element e, along with Remarks 1 and 4 and Proposition 12, allows for the following result.

Proposition 13.

Let I be a fuzzy implication function and let Ce⊤ be the greatest conjunctor with a neutral element e∈]0,1[. The following items are equivalent:

1. I satisfies (MP(Ce⊤)).

2. I satisfies (MP(C)) with respect to any conjunctor C such that C⩽Ce⊤, in particular any conjunctor with a neutral element e∈]0,1].

3. I satisfies the three conditions given in Proposition 12 (bottom of Figure 4).

4. I⩽ICe⊤, where ICe⊤ is given in Remark 4.

Example 2.

Fixed some e∈]0,1[, let us give some examples of fuzzy implication functions satisfying conditions in Proposition 12, i.e., satisfying (MP(Ce⊤)), and consequently (MP(C)) with respect to any conjunctor C with neutral element e. Of course ICe⊤ itself is one of them, as it has been proved before. The following parameterized family of fuzzy implication functions presents other possibilities.

Results in Propositions 9 and 11 can be also used to characterize those fuzzy implication functions satisfying (MP(C)) with C=Ue⊥, where Ue⊥ is the least idempotent uninorm with neutral element e and with C=Ue⊤, where Ue⊤ is the greatest conjunctive idempotent uninorm with neutral element e. These uninorms are respectively given by (see for instance Ref. [28]):

Proposition 14.

Let Ue⊥ be the least idempotent uninorm with neutral element e, Ue⊤ the greatest conjunctive idempotent uninorm with neutral element e and let I be a fuzzy implication function. Then

1. I satisfies (MP(C)) with C=Ue⊥ if and only if the following two conditions hold:

   • I(x,y)⩽y for any x,y∈[0,1] such that [x>y,y<e] or [e⩽x⩽y<1].

   • I(x,y)<e for any x,y∈[0,1] such that e⩽y<x.

2. I satisfies (MP(C)) with C=Ue⊤ if and only if the following three conditions hold:

   • I(x,y)=0 for any x,y∈[0,1] such that e,y<x.

   • I(x,y)⩽y for any x,y∈[0,1] such that [y<x⩽e] or [0<x⩽y,y⩾e].

   • I(x,y)⩽e for any x,y∈[0,1] such that x⩽y<e.

Note that Ue⊥ is an example of a uninorm in Umin which is not left-continuous and whose residual implication IUe⊥ does not satisfy the Modus Ponens with Ue⊥. On the contrary, Ue⊤ is a left-continuous uninorm and consequently the characterization given in the second item of the previous proposition is equivalent to say I⩽IUe⊤.

Proposition 15.

Let I be a fuzzy implication function. The following statements are true:

• I satisfies (MP(TD)) if and only if I(1,y)⩽y for any y∈[0,1] and I(x,y)≠1 for all x,y∈[0,1] such that x>y.

• I satisfies (MP((TLK)φ)) if and only if I(x,y)⩽φ−1(φ(y) +1−φ(x)) for all x,y∈[0,1] such that x>y, i.e., I⩽ILKφ.

• I satisfies (MP((TP)φ)) if and only if I(x,y)⩽φ−1(φ(y)∕φ(x)) for all x,y∈[0,1] such that x>y, i.e., I⩽IGGφ.

• I satisfies (MP(TM)) if and only if I(x,y)⩽y for all x,y∈[0,1] such that x>y, i.e., I⩽IGD.

Proof.

The first item is a matter of calculation (note that TD is not left-continuous and the satisfaction of (MP(TD)) implies but is not equivalent to I⩽IWB-the Weber implication is the residuated implication associated to TD-). The following three items are obtained from Proposition 2 since it is well-known (see e.g., Ref. [1]) that ILK,IGG, and IGD are, respectively, the residuated implication functions associated to the continuous t-norms TLK,TP, and TM.