Generalized valuations on MTL-algebras

Abstract

In this paper, we study some kinds of generalized valuations on MTL-algebras, discuss the relationship between the cokernel of generalized valuations and types of filters on MTL-algebras. Then, we give some equivalent characterizations of positive implicative generalized valuations on MTL-algebras. Finally, we characterize the structure theory of quotient MTL algebras based on the congruence relation, which is constructed by generalized valuations. The results of this paper not only generalize related theories of generalized valuations, but also enrich the algebraic conclusion of probability measure, on algebras of triangular norm based fuzzy logic.

Keywords

Fuzzy logic / MTL-algebra / generalized valuation / filter

1 Introduction

In classical logical reasoning, when the concepts and information are accurate, the results obtained from them are also accurate. This precise and strict logical reasoning has been widely used in artificial intelligence science and related topics and has been frequently applied in the fields of logical program design, automatic theorem proving, knowledge reasoning and so on. There is a growing evidence that classical logical reasoning not only provides the main idea for computer programming language, but also has some certain applications in computer hardware design, which is an important part of the theoretical basis of computers. However, the traditional computer can only recognize the classical logic, and is powerless to deal with the fuzzy concepts. In order to overcome these shortcomings, Zadeh [20] introduced the concept of fuzzy logic, which makes computers not only process fuzzy concepts, but also provide the accurate answers in the case of limited information. Fuzzy logic mainly adopts the research method of algebraic logic that uses logical algebra as a tool to study, so logical algebra plays a very important role in the research of fuzzy logic [22,23]. In the study of the algebraic system corresponding to fuzzy logic, scholars at home and abroad pay particular attention to MTL-algebras, which was introduced by Esteva and Godo [5] as the algebraic semantics of the logical system MTL, and extended BL-algebras by dropping the condition of divisibility. In the recent years, MTL-algebras have gradually become the most active object of all fuzzy logical algebras. The main reasons are: (1) MTL-algebras are the most fundamental residuated structures that contain all algebras induced by left continuous t-norms and their residua. (2) The logic MTL is indeed a kind of triangular norm based fuzzy logic with the broadest range of standard completeness provable. (3) MTL-algebras establish a connection between the different classes of fuzzy logic algebras, and provide a way to study them systematically and comprehensively. For a detailed consideration of MTL-algebras and their corresponding logics, we refer to [7,11-14].

In order to induce a generalized metric space on Hilbert algebras, Busneag [2] introduced the notion of generalized valuations on Hilbert algebras and studied some of their basic algebraic properties. Afterwards, Busneag [3] also obtained some new results of generalized valuations and proved that some extension theorems of on Hilbert algebras. Inspired by this, Busnea et al. [1] applied the notion of generalized valuations from Hilbert algebras to residuated lattices and established generalized metric spaces of residuated lattices. In recent years, some scholars have successively studied the generalized valuations on other logical algebras and obtained some important conclusions, for example, Doh [4] introduced the notion of generalized valuations on BCK/BCI-algebras, and constructed a generalized metric space of BCK/BCI-algebras by using generalized valuations. In the same year, Ghorbani [6] further studied generalized valuations on BCI-algebra and induced a congruence relation based on them, they also proved that the quotient algebra defined by this congruence relation is a BCI-algebra. Subsequently, Zhan [21] studied generalized valuations on R_0-algebras and gave some characterizations of several kinds of generalized valuations. In order to get a more general results of generalized valuations on the algebras of triangular norm based fuzzy logic, Yang [17] introduced the notion of generalized valuations on hoops and discussed relations between generalized valuations and filters. Tracking of Yang's work, Wang [15] further studied the quotient algebraic structure that induced by generalized valuations on hoops. Recently, Yang also have successively studied generalized valuations on EQ-algebras and MV-algebras in [18,19], and obtained some new results, which greatly broadens the theory of generalized valuations on triangular norm based fuzzy logical algebras.

As we have mentioned before that the main focus of existing research about generalized valuations is on MV-algebras, BL-algebras, R₀-algebras and residuated lattices, etc. All the above-mentioned algebraic structures satisfy the divisibility condition. In this case, the conjunction on the unit interval corresponds to a continuous t-norm. However, there is no research about generalized valuations on residuated structures without the divisibility condition so far. In fact, MTL-algebras are the widest possible residuated algebraic structure of triangular norm based fuzzy logic, which does not satisfy the divisibility condition. Therefore, it is interesting to study the generalized valuations on MTL-algebras for obtaining a more general algebraic results with respect to this topics on algebras of triangular norm based fuzzy logic. On the other hand, with the intent of measuring the average truth value of propositions in Łukasiewicz logic, the notion of states on MV-algebras was introduced by Mundici [9], which is a generalization of probability measures on Boolean algebras and can be interpreted as the probability of fuzzy events. However, fuzzy logical algebras with states are not universal algebras, and therefore, it is difficult for us to study the algebraic properties of states on fuzzy logical algebra deeply. Considering the close relationship between the generalized valuations and states on fuzzy logical algebras, and so studying the former is helpful to enrich the algebraic properties of the latter. This is the main motivation behind introducing the generalized valuations on MTL-algebras in the present paper.

The rest of the paper is structured as follows: in order to make the paper as self-contained as possible, we recapitulate in Section 2 the basic notions and results related to MTL-algebras that will be used in the paper. In Section 3 we introduce the concept of generalized valuations on MTL-algebras, and discuss the relationship among the generalized valuations, states and filters. In Section 4 we introduce the notion of (positive) implicative generalized valuations on MTL-algebras and obtain some equivalent conditions under which a generalized valuation to be a (positive) implicative generalized valuation. In Section 5 we study the quotient MTL-algebras that induced by generalized valuations and obtain some isomorphism theorem based on them.

2 Preliminaries

Definition 2.1 [5]　An algebraic structure L=(L,\wedge ,\vee ,\odot ,\to ,0,1) of type (2,2,2,2,0,0) is called an MTL-algebra if it satisfies the following conditions:

(1) (L,\wedge ,\vee ,0,1) is a bounded lattice,

(2) (L,\odot ,1) is a commutative monoid,

(3) x\odot y\le z if and only if x\le y\to z,

(4) (x\to y)\vee (y\to x)=1, for any x,y,z\in L.

In what follows, by L we denote the universe of an MTL-algebra (L,\wedge ,\vee \odot ,\to ,0,1). In any MTL-algebra L, we define

\mathrm{\neg }x=x\to 0,\mathrm{\neg }\mathrm{\neg }x=\mathrm{\neg }(\mathrm{\neg }x),x^0=1\text{and}{x}^n=x^{n-1}\odot x\text{for}n\ge 1.

Proposition 2.2 [10,24]　 Let L be an MTL-algebra. Then the following properties are valid: for all x,y,z\in L,

(1) x\le y if and only if x\to y=1,

(2) x\to x=1, x\to 1=1, 1\to x=x,

(3) x\le y implies x\odot z\le y\odot z,z\to x\le z\to y and y\to z\le x\to z,

(4) x\odot y\to z=x\to (y\to z),

(5) x\wedge y=((x\to y)\to y)\wedge ((y\to x)\to x),

(6) x\le y\to x,

(7) x\odot (x\to y)\le x,y,

(8) x\to y\le (y\to z)\to (x\to z).

Definition 2.3 [22]　An MTL-algebra L is called an IMTL-algebra if it satisfies \mathrm{\neg }\mathrm{\neg }x=x for any x\in L.

Definition 2.4 [25]　A non-empty subset F of an MTL-algebra L is called a filter if it satisfies the following conditions:

(1) x,y\in F implies x\odot y\in F,

(2) x\in F, y\in L and x\le y imply y\in F.

Definition 2.5 [25]　A filter F of an MTL-algebra L is called:

(1) proper if F\ne L,

(2) implicative if x\to (y\to z)\in F and x\to y\in F imply x\to z\in F,

(3) positive implicative if x\to ((y\to z)\to y)\in F and x\in F imply y\in F,

(4) obstinate if y\to x\in F implies ((x\to y)\to y)\to x\in F,

(5) maximal if it is not strictly contained in any proper filter of L.

Definition 2.6 [8]　A Bosbach state on an MTL-algebra L is a function s:L\to [0,1] such that the following conditions hold:

(1) s(0)=0,s(1)=1,

(2) s(x)+s(x\to y)=s(y)+s(y\to x).

Definition 2.7 [9]　Let L be an MTL-algebra and [0,1{]}_{MV} be a standard MV-algebra. A function \phi :L\to [0,1{]}_{MV} is called a valuation state if it satisfies the following conditions:

(1) \phi (x\to y)=\phi (x)\to \phi (y),

(2) \phi (x\wedge y)=min\{\phi (x),\phi (y)\},

(3) \phi (1)=1, \phi (0)=0.

Proposition 2.8 [8]　 Every valuation state is a Bosbach state on any MTL-algebra.

In order to discuss the relationship between generalized valuations and valuation states, we introduce the notion of ideals of MTL-algebras.

Definition 2.9 [16]　A non-empty subset I of an MTL-algebra L is called an ideal if it satisfies the following conditions:

(1) x,y\in I implies x\oslash y\in I, where x\oslash y=\mathrm{\neg }x\odot y,

(2) y\in I, x\in L and x\le y imply x\in I.

Proposition 2.10 [16]　 Let I be an ideal of an MTL-algebra L. Then the binary relation

{\sim }_I:x{\sim }_{I}yifandonlyif\mathrm{\neg }(x\to y),\mathrm{\neg }(y\to x)\in I

is a congruence, the set L/{\sim }_I=\{[x{]}_I|x\in L\} can form an MTL-algebra with respect to the binary operations of an MTL-algebra L.

Definition 2.11 [16]　An ideal I of an MTL-algebra L is called divisible if it satisfies the condition

\mathrm{\neg }((x\wedge y)\to (x\odot (x\to y)))\in I.

Proposition 2.12 [16]　 An ideal I of an MTL-algebra L is divisible if and only if L/{\sim }_I is a BL-algebra.

3 Generalized valuations on MTL-algebras

In this section, we introduce the concept of generalized valuations on MTL-algebras, obtaining some conditions under which a real-valued function can be a generalized valuation, and then discuss the relationship among the generalized valuations, states and filters of MTL-algebras.

Definition 3.1　Let L be an MTL-algebra. A real-valued function \nu :L\to \mathbb{R} is said to be a generalized valuation on L if it satisfies the following conditions:

(3.1)\nu (1)=0,

(3.2)\begin{array}{r}\nu (y)\le \nu (x\to y)+\nu (x).\end{array}

Moreover, a generalized valuation \nu is said to be a valuation if

(3.3)\begin{array}{r}\nu (x)=0\iff x=1.\end{array}

Example 3.2　Let L=\{0,a,b,1\} with 0\le a\le b\le 1. Define the binary operations \odot and \to on L as follows:

\begin{array}{ccccc}\odot & 0& a& b& 1\\ \\ 0& 0& 0& 0& 0\\ a& 0& a& 0& a\\ b& 0& 0& b& b\\ 1& 0& a& b& 1\end{array}\begin{array}{ccccc}\to & 0& a& b& 1\\ \\ 0& 1& 1& 1& 1\\ a& b& 1& b& 1\\ b& a& a& 1& 1\\ 1& 0& a& b& 1\end{array}

Then (L,\wedge ,\vee ,\odot ,\to ,0,1) is an MTL-algebra. Moreover, we define two real-valued functions {\nu }_1 and {\nu }_2 as

{\nu }_1=\left(\begin{array}{cccc}0& a& b& 1\\ 1& 1& 1& 0\end{array}\right),{\nu }_2=\left(\begin{array}{cccc}0& a& b& 1\\ 2& 0& 0& 0\end{array}\right).

It is easily checked that {\nu }_1 is a generalized valuation on L. However, {\nu }_2 is not a generalized valuation on L, since Eq. (3.2) does not,

2={\nu }_2(0)⪇{\nu }_2(a\to 0)+{\nu }_2(a)={\nu }_2(b)+{\nu }_2(a)=0.

Proposition 3.3　 If s is a Basbach state on an MTL-algebra L, then

\nu (x)=1-s(x)

is a generalized valuation on an MTL-algebra L.

Proof　Since s is a Basbach state, by Definition 2.6(1), we have

\nu (1)=1-s(1)=0.

Moreover, from Definition 2.6(2), we have

\begin{array}{rl}\nu (y)& =1-s(y)\\ & =1-(s(x)+s(x\to y)-s(y\to x))\\ & =1-s(x)-s(x\to y)+s(y\to x)\\ & =(1-s(x))+(1-s(x\to y))-(1-s(y\to x))\\ & \le (1-s(x))+(1-s(x\to y))\\ & =\nu (x)+\nu (x\to y).\end{array}

Therefore \nu is a generalized valuation on an MTL-algebra L.□

Remark 3.4　It is noted that if \nu is a generalized valuation on an MTL-algebra L, then

s(x)=1-\nu (x)

is not a Basbach state, since the following equation does not hold generally:

\nu (a)+\nu (a\to b)=\nu (b)+\nu (b\to a).

In fact, in Example 3.2,

{\nu }_1(0)+{\nu }_1(0\to a)=1\ne 2={\nu }_1(a)+{\nu }_1(a\to 0).

The relationship between the valuation states and generalized valuations on MTL-algebras is discussed.

Remark 3.5　If s is a valuation state on an MTL-algebra L, then 1-s is a generalized valuation on L.

The existence of generalized valuations on MTL-algebras is also studied.

Proposition 3.6　 Let L be an MTL-algebra. If L has a divisible ideal, then L has a generalized valuation.

Proof　If I is a divisible ideal of an MTL-algebra L, then by Proposition 2.12, L/I is a BL-algebra. Then there exists a Basbach state \tilde{\text{s}} on the BL-algebra L/I from [16]. Define a function s:L\to [0,1] such that

s(x)=\tilde{\text{s}}\left([x{]}_I\right),

and check that s is indeed a Basbach state on L, further by Proposition 3.3, \nu =1-s is also a generalized valuation on an MTL-algebra L.□

Remark 3.7　The converse of Proposition 3.6 does not hold generally, the reason can be seen in Remark 3.4.

Proposition 3.8　 Let \nu be a generalized valuation on an MTL-algebra L. Then the following hold: for any x,y,\in L,

(1) if x\le y, then \nu (y)\le \nu (x),

(2) 0\le \nu (x),

(3) \nu (x\to y)\le \nu (y),

(4) x\to (y\to z)=1 implies \nu (z)\le \nu (x)+\nu (y).

Proof　(1) If x\le y, that is, x\to y=1, then by Eq. (3.1) and Eq. (3.2), we have

\nu (y)\le \nu (x\to y)+\nu (x)=\nu (1)+\nu (x)=0+\nu (x)=\nu (x).

(2) Taking y=1 in Eq. (3.2), we have

0=\nu (1)\le \nu (x\to 1)+\nu (x)=\nu (1)+\nu (x)=0+\nu (x)=\nu (x).

(3) It follows from (1) and Proposition 2.2(6) that \nu (x\to y)\le \nu (y).

(4) It follows from Eq. (3.2) that

\nu (y\to z)\le \nu (x\to (y\to z))+\nu (x),

and further by x\to (y\to z)=1, we also have

\nu (y\to z)\le \nu (1)+\nu (x)=0+\nu (x)=\nu (x),

which implies \nu (z)\le \nu (y)+\nu (y\to z)\le \nu (y)+\nu (x).□

Theorem 3.9　 Let L be an MTL-algebra and \nu :L\to \mathbb{R} be a real-valued function. Then the following statements are equivalent:

(1) \nu is a generalized valuation on an MTL-algebra L,

(2) \nu satisfies Eq. (3.2) and Proposition 3.8(4),

(3) \nu satisfies Eq. (3.1) and

(3.4)\begin{array}{r}x\odot y\le zimplies\nu (z)\le \nu (x)+\nu (y).\end{array}

Proof　 (1)\Rightarrow (2). The proof is obvious and hence we omit it.

(2)\Rightarrow (3). Since x\to (y\to z)=1 if and only if x\le y\to z, if and only if x\odot y\le z, by Proposition 3.8(4), Eq. (3.4) holds.

(3)\Rightarrow (1). If (x\to y)\to (x\to y)=1, that is, x\to y\le x\to y, and hence (x\to y)\odot x\le y, then by Eq. (3.4), we have \nu (y)\le \nu (x\to y)+\nu (x).□

Proposition 3.10　 If \nu is a generalized valuation on an MTL-algebra L, then the following hold: for any x,y,z\in L,

(1) \nu (x\odot y)\le \nu (x)+\nu (y),

(2) \nu (x\wedge y)\le \nu (x)+\nu (y),

(3) \nu (x\to (y\to z))\le \nu ((x\to y)\to z),

(4) \nu (x\to z)\le \nu (x\to y)+\nu (y\to z).

Proof　(1) Taking z=x\odot y in Eq. (3.4), we have \nu (x\odot y)\le \nu (x)+\nu (y).

(2) It follows from x\odot y\le x\wedge y and Proposition 3.8(1) that \nu (x\wedge y)\le \nu (x\odot y), and further by (1), we have \nu (x\wedge y)\le \nu (x\odot y)\le \nu (x)+\nu (y).

(3) Since 1=y\to (x\to y)\le ((x\to y)\to z))\to (y\to z), we have (x\to y)\to z\le y\to z, further by Proposition 3.8(1), we also have

\nu (y\to z)\le \nu ((x\to y)\to z).

On the other hand, by y\to z\le x\to (y\to z), we have

\nu (x\to (y\to z))\le \nu (y\to z)\le \nu ((x\to y)\to z).

(4) It follows from (x\to y)\odot (y\to z)\le x\to z and Proposition 3.8(1) that

\nu (x\to z)\le \nu ((x\to y)\odot (y\to z)),

and further by (1), we also have

\nu (x\to z)\le \nu ((x\to y)\odot (y\to z))\le \nu (x\to y)+\nu (y\to z).◻

Proposition 3.11　 If \nu is a generalized valuation on an MTL-algebra L, then

\mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu =\{x\in L|\nu (x)=0\},

which is called the \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r} of \nu, is a filter of an MTL-algebra L.

Proof　Notice that Eq. (3.1) implies 1\incoker\nu. If x,x\to y\incoker\nu, then

\nu (x)=\nu (x\to y)=0.

Moreover, it follows from Eq. (3.2) that

\nu (y)\le \nu (x\to y)+\nu (x)=0,

and hence \nu (y)=0, which implies that y\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu.□

However, the converse of Proposition 3.11 is not true in general.

Example 3.12　Let L=\{0,a,b,c,1\} with 0\le a\le b\le c\le 1. Define the binary operations \odot and \to as follows:

\begin{array}{rlr}& \begin{array}{cccccc}\odot & 0& a& b& c& 1\\ \\ 0& 0& 0& 0& 0& 0\\ a& 0& 0& 0& 0& a\\ b& 0& 0& 0& b& b\\ c& 0& 0& b& c& c\\ 1& 0& a& b& c& 1\end{array}& \begin{array}{cccccc}\to & 0& a& b& c& 1\\ \\ 0& 1& 1& 1& 1& 1\\ a& c& 1& 1& 1& 1\\ b& b& b& 1& 1& 1\\ c& a& a& b& 1& 1\\ 1& 0& a& b& c& 1\end{array}\end{array}

Then (L,\wedge ,\vee ,\odot ,\to ,0,1) is an MTL-algebra. Also, define a real-valued function \nu as

\nu =\left(\begin{array}{ccccc}0& a& b& c& 1\\ 2& 1& 1& 0& 0\end{array}\right).

It is easily checked that \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu =\{1,c\} is a filter of L. But \nu is not a generalized valuation on L, since

\phi (0)=2⪇\phi (a\to 0)+\phi (a)=\phi (c)+\phi (a)=1.

4 (Positive) implicative generalized valuations on MTL-algebras

In this section, we introduce the notion of (positive) implicative generalized valuations on MTL-algebras and give some characterizations of them.

Definition 4.1　Let L be an MTL-algebra. A real-valued function \nu :L\to \mathbb{R} is said to be an implicative generalized valuation on L if it satisfies Eq. (3.1) and

(4.1)\begin{array}{r}\nu (x\to z)\le \nu (x\to (y\to z))+\nu (x\to y).\end{array}

The example of implicative generalized valuations on MTL-algebras is given.

Example 4.2　Let L=\{0,a,b,c,d,1\} with 0\le a\le b\le c\le d\le 1. Define the binary operations \odot and \to as follows:

\begin{array}{rlr}& \begin{array}{ccccccc}\odot & 0& a& b& c& d& 1\\ \\ 0& 0& 0& 0& 0& 0& 0\\ a& 0& 0& 0& 0& 0& a\\ b& 0& 0& 0& 0& b& b\\ c& 0& 0& 0& c& c& c\\ d& 0& 0& b& c& d& d\\ 1& 0& a& b& c& d& 1\end{array}& \begin{array}{ccccccc}\to & 0& a& b& c& d& 1\\ \\ 0& 1& 1& 1& 1& 1& 1\\ a& d& 1& 1& 1& 1& 1\\ b& c& c& 1& 1& 1& 1\\ c& b& b& b& 1& 1& 1\\ d& a& a& b& c& 1& 1\\ 1& 0& a& b& c& d& 1\end{array}\end{array}

Then (L,\wedge ,\vee ,\odot ,\to ,0,1) is an MTL-algebra. Define a real-valued function \nu by

\nu =\left(\begin{array}{cccccc}0& a& b& c& d& 1\\ 1& 1& 1& 0& 0& 0\end{array}\right).

It is checked that \nu is an implicative generalized valuation on L.

Proposition 4.3　 An implicative generalized valuation is a generalized valuation on an MTL-algebra L.

Proof　Taking x=1 in Eq. (4.1), we have

\nu (1\to z)\le \nu (1\to (y\to z))+\nu (1\to y)=\nu (y\to z)+\nu (y),

that is,

\nu (z)\le \nu (y\to z)+\nu (y),

which implies that \nu is a generalized valuation by Definition 2.1.□

The following example shows that the converse of Proposition 4.3 is not true.

Example 4.4　Considering the MTL-algebra L in Example 3.2. Define a generalized valuation \nu by

\nu =\left(\begin{array}{ccccc}0& a& b& c& 1\\ 2& 2& 1& 1& 0\end{array}\right).

However, \nu is not an implicative generalized valuation on the MTL-algebra L, since

1=\nu (c)=\nu (a\to 0)⪇\nu (a\to (b\to 0))+\nu (a\to b)=\nu (1)+\nu (1)=0.

The equivalent conditions under which a generalized valuation can be an implicative generalized valuation on MTL-algebras are given.

Theorem 4.5　 Let \nu be a generalized valuation on an MTL-algebra L. Then the following statements are equivalent:

(1) \nu is an implicative generalized valuation on L,

(2) for any x,y\in L, \nu ((x\wedge (x\to y))\to y)=0,

(3) for any x,y\in L, \nu (x\to y)=\nu (x\to (x\to y)).

Proof　 (1)\Rightarrow (2). If \nu is implicative, then by

(x\wedge (x\to y))\to x=1,(x\wedge (x\to y))\to (x\to y)=1,

and Definition 3.1, we have

\begin{array}{rl}\nu ((x\wedge (x\to y))\to y)& \le \nu ((x\wedge (x\to y))\to (x\to y))+\nu ((x\wedge (x\to y))\to x)\\ & =\nu (1)+\nu (1)\\ & =0.\end{array}

(2)\Rightarrow (3). If \nu ((x\wedge (x\to y))\to y)=0, then by

y\le x\to y,x\to y\le x\to (x\to y),

and Proposition 3.4(1), we have

\nu (x\to (x\to y))\le \nu (x\to y).

On the other hand, since

1\to ((x\wedge (x\to y))\to y)\le x\to ((x\wedge (x\to y))\to y),

that is,

(x\wedge (x\to y))\to y\le x\to ((x\wedge (x\to y))\to y),

we have

\nu (x\to ((x\wedge (x\to y))\to y))\le \nu ((x\wedge (x\to y))\to y)\le \nu ((x\odot (x\to y))\to y)=\nu (1)=0.

So \nu (x\to ((x\wedge (x\to y))\to y))=0.Further by Definition 3.1,

\begin{array}{rl}\nu (x\to y)& \le \nu (x\to ((x\wedge (x\to y))\to y))+\nu (x\to (x\wedge (x\to y)))\\ & =\nu (x\to (x\wedge (x\to y)))\\ & =\nu (x\to (x\to y)).\end{array}

Therefore \nu (x\to y)=\nu (x\to (x\to y)).

(3)\Rightarrow (1). Assume that \nu (x\to y)=\nu (x\to (x\to y)). First, by the transitivity of \nu, we have

x\to (y\to z)\le ((y\to z)\to (x\to z))\to (x\to (x\to z)),x\to y\le (y\to z)\to (x\to z).

Then it follows from Proposition 3.4(1) that

\begin{array}{rl}\nu (((y\to z)\to (x\to z))\to & (x\to (x\to z)))\le \nu (x\to (y\to z)),\\ & \nu ((y\to z)\to (x\to z))\le \nu (x\to y).\end{array}

Finally, by Definition 3.1, we have

\begin{array}{rl}\nu (x\to z)& =\nu (x\to (x\to z))\\ & \le \nu (((y\to z)\to (x\to z))\to (x\to (x\to z)))+\nu ((y\to z)\to (x\to z))\\ & \le \nu (x\to (y\to z))+\nu (x\to y).\end{array}

This shows that

\nu (x\to z)\le \nu (x\to (y\to z))+\nu (x\to y).

Therefore, \nu is an implicative generalized valuation on L.□

The equivalent conditions under which a generalized valuation to be an implicative generalized valuation on IMTL-algebra are also given.

Theorem 4.6　 Let \nu be a generalized valuation on an IMTL-algebra L. Then the following statements are equivalent:

(1) \nu is an implicative generalized valuation on L,

(2) for any x,y,z\in L, \nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to y))+\nu (y\to z).

Proof　 (1)\Rightarrow (2). If L is an IMTL-algebra, then

x\to z=\mathrm{\neg }z\to \mathrm{\neg }x.

Assume that \nu is implicative. By Definition 3.1, we have

\begin{array}{rl}\nu (x\to z)& =\nu (\mathrm{\neg }z\to \mathrm{\neg }x)\\ & \le \nu (\mathrm{\neg }z\to (\mathrm{\neg }y\to \mathrm{\neg }x))+\nu (\mathrm{\neg }z\to \mathrm{\neg }y)\\ & =\nu (\mathrm{\neg }z\to (x\to y))+\nu (y\to z)\\ & =\nu (x\to (\mathrm{\neg }z\to y))+\nu (y\to z),\end{array}

which implies \nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to y))+\nu (y\to z).

(2)\Rightarrow (1). If the condition (2) holds, then

\begin{array}{rl}\nu (x\to z)& =\nu (\mathrm{\neg }z\to \mathrm{\neg }x)\\ & \le \nu (\mathrm{\neg }z\to (\mathrm{\neg }\mathrm{\neg }x\to \mathrm{\neg }y))+\nu (\mathrm{\neg }y\to \mathrm{\neg }x)\\ & =\nu (\mathrm{\neg }z\to (x\to \mathrm{\neg }y))+\nu (x\to y)\\ & =\nu (x\to (\mathrm{\neg }z\to \mathrm{\neg }y))+\nu (x\to y)\\ & =\nu (x\to (y\to z))+\nu (x\to y),\end{array}

which implies \nu (x\to z)\le \nu (x\to (y\to z))+\nu (x\to y).

Therefore \nu is an implicative generalized valuation on L.□

Theorem 4.7　 Let \nu be a generalized valuation on an IMTL-algebra L. Then the following statements are equivalent:

(1) \nu is an implicative generalized valuation on L,

(2) for any x,y,z\in L, \nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to z)),

(3) for any x,y,z\in L, \nu (x\to z)\le \nu (y\to (x\to (\mathrm{\neg }z\to z)))+\nu (y).

Proof　 (1)\Rightarrow (2). If \nu is implicative, then taking y=z in Theorem 4.6(2), we have

\begin{array}{rl}\nu (x\to z)& \le \nu (x\to (\mathrm{\neg }z\to z))+\nu (z\to z)\\ & =\nu (x\to (\mathrm{\neg }z\to z))+\nu (1)\\ & =\nu (x\to (\mathrm{\neg }z\to z)),\end{array}

which implies \nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to z)).

(2)\Rightarrow (3). If the condition (2) holds, by Definition 3.1, we have

\nu (x\to (\mathrm{\neg }z\to z))\le \nu (y\to (x\to (\mathrm{\neg }z\to z)))+\nu (y),

and hence

\nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to z))\le \nu (y\to (x\to (\mathrm{\neg }z\to z)))+\nu (y),

which shows that the condition (3) holds.

(3)\Rightarrow (1). If the condition (3) holds, then by Proposition 3.4(3), we have

\nu (x\to (\mathrm{\neg }z\to z))=\nu ((x\odot \mathrm{\neg }z)\to z)\le \nu ((x\odot \mathrm{\neg }z)\to y)+\nu (y\to z).

Taking y=1 in the condition (3),

\begin{array}{rl}\nu (x\to z)& \le \nu (1\to (x\to (\mathrm{\neg }z\to z)))+\nu (1)\\ & =\nu (x\to (\mathrm{\neg }z\to z))\\ & \le \nu ((x\odot \mathrm{\neg }z)\to y)+\nu (y\to z)\\ & =\nu (x\to (\mathrm{\neg }z\to y))+\nu (y\to z),\end{array}

which implies

\nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to y))+\nu (y\to z).

By Theorem 4.6, \nu is an implicative generalized valuation on L.□

Theorem 4.8　 Let \nu be a generalized valuation on an IMTL-algebra L. Then the following statements are equivalent:

(1) \nu is an implicative generalized valuation on L;

(2) for any x\in L, \nu (x)\le \nu (\mathrm{\neg }x\to x),

(3) for any x,y\in L, \nu (x)\le \nu ((x\to y)\to x),

(4) for any x,y,z\in L, \nu (x)\le \nu (z\to ((x\to y)\to x))+\nu (z).

Proof　 (1)\Rightarrow (2). If \nu is implicative, then by Theorem 4.7(2), we have

\nu (x)=\nu (1\to x)\le \nu (1\to (\mathrm{\neg }x\to x))=\nu (\mathrm{\neg }x\to x),

which implies \nu (x)\le \nu (\mathrm{\neg }x\to x).

(2)\Rightarrow (3). If the condition (2) holds, then by \mathrm{\neg }x\le x\to y, we have

(x\to y)\to x\le \mathrm{\neg }x\to x,

and hence

\nu (\mathrm{\neg }x\to x)\le \nu ((x\to y)\to x).

Further by (2), we also have

\nu (x)\le \nu (\mathrm{\neg }x\to x)\le \nu ((x\to y)\to x),

which shows \nu (x)\le \nu ((x\to y)\to x).

(3)\Rightarrow (4). If the condition (3) holds, then by Definition 3.1, we have

\nu (x)\le \nu ((x\to y)\to x)\le \nu (z\to ((x\to y)\to x))+\nu (z),

which implies \nu (x)\le \nu (z\to ((x\to y)\to x))+\nu (z).

(4)\Rightarrow (1). If the condition (4) holds, then by z\le x\to z, we have \mathrm{\neg }(x\to z)\le \mathrm{\neg }z, and hence

\mathrm{\neg }z\to (x\to z)\le \mathrm{\neg }(x\to z)\to (x\to z),

So

\nu (\mathrm{\neg }(x\to z)\to (x\to z))\le \nu (\mathrm{\neg }z\to (x\to z)).

Further by (4), we have

\begin{array}{rl}\nu (x\to z)& \le \nu (1\to (((x\to z)\to 0)\to (x\to z)))+\nu (1)\\ & =\nu (\mathrm{\neg }(x\to z)\to (x\to z))\\ & \le \nu (\mathrm{\neg }z\to (x\to z))\\ & =\nu (x\to (\mathrm{\neg }z\to z)),\end{array}

which implies \nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to z)).

Therefore \nu is an implicative generalized valuation on an IMTL-algebra L.□

Theorem 4.9　 Let \nu be a generalized valuation on an IMTL-algebra L. Then the following statements are equivalent:

(1) \nu is an implicative generalized valuation on L,

(2) for any x\in L, \nu (x\vee \mathrm{\neg }x)=0.

Proof　 (1)\Rightarrow (2). If \nu is implicative, then

\begin{array}{rl}\mathrm{\neg }x\to (((\mathrm{\neg }x\to x)\to x)\to (\mathrm{\neg }(\mathrm{\neg }x\to x)))& =((\mathrm{\neg }x\to x)\to x)\to (\mathrm{\neg }x\to (\mathrm{\neg }(\mathrm{\neg }x\to x)))\\ & =((\mathrm{\neg }x\to x)\to x)\to ((\mathrm{\neg }x\to x)\to x)\\ & =1,\end{array}

and

\mathrm{\neg }x\to ((\mathrm{\neg }x\to x)\to x)=(\mathrm{\neg }x\to x)\to (\mathrm{\neg }x\to x)=1,

and hence

\begin{array}{rl}\nu ((\mathrm{\neg }x\to x)\to x)& =\nu (\mathrm{\neg }x\to \mathrm{\neg }(\mathrm{\neg }x\to x))\\ & \le \nu (\mathrm{\neg }x\to (((\mathrm{\neg }x\to x)\to x)\to \mathrm{\neg }(\mathrm{\neg }x\to x)))\\ & +\nu (\mathrm{\neg }x\to ((\mathrm{\neg }x\to x)\to x))\\ & =\nu (1)+\nu (1)\\ & =0,\end{array}

which implies \nu ((\mathrm{\neg }x\to x)\to x)=0. In the similar way, one can prove that

\nu ((x\to \mathrm{\neg }x)\to \mathrm{\neg }x)=0.

Further by Proposition 3.4(4), we have

\begin{array}{rl}\nu (x\vee \mathrm{\neg }x)& =\nu (((\mathrm{\neg }x\to x)\to x)\wedge ((x\to \mathrm{\neg }x)\to \mathrm{\neg }x))\\ & \le \nu ((\mathrm{\neg }x\to x)\to x)+\nu ((x\to \mathrm{\neg }x)\to \mathrm{\neg }x)\\ & =0.\end{array}

(2)\Rightarrow (1). If \nu (x\vee \mathrm{\neg }x)=0, then by y\le x\to y, we have y\to (x\to y)=1, and hence

\nu (y\to (x\to y))=\nu (1)=0.

Moreover, since \nu is a generalized valuation, we have

\begin{array}{rl}\nu (x\to y)& \le \nu ((y\vee \mathrm{\neg }y)\to (x\to y))+\nu (y\vee \mathrm{\neg }y)\\ & =\nu ((y\vee \mathrm{\neg }y)\to (x\to y))\\ & =\nu ((y\to (x\to y))\wedge (\mathrm{\neg }y\to (x\to y))\\ & \le \nu (y\to (x\to y))+\nu (\mathrm{\neg }y\to (x\to y))\\ & =\nu (1)+\nu (\mathrm{\neg }y\to (x\to y))\\ & =\nu (\mathrm{\neg }y\to (x\to y))\\ & =\nu (x\to (\mathrm{\neg }y\to y)),\end{array}

which implies \nu (x\to y)\le \nu (x\to (\mathrm{\neg }y\to y)).

Therefore \nu is an implicative generalized valuation on L.□

The condition of an involution in Theorem 4.9 is indeed necessary.

Example 4.10　Let L=\{0,a,b,1\} with 0\le a\le b\le 1. Define the binary operations \odot and \to as follows:

\begin{array}{rlr}& \begin{array}{ccccc}\odot & 0& a& b& 1\\ \\ 0& 0& 0& 0& 0\\ a& 0& 0& a& a\\ b& 0& a& b& b\\ 1& 0& a& b& 1\end{array}& \begin{array}{ccccc}\to & 0& a& b& 1\\ \\ 0& 1& 1& 1& 1\\ a& 0& 1& 1& 1\\ b& 0& a& 1& 1\\ 1& 0& a& b& 1\end{array}\end{array}

Then (L,\wedge ,\vee ,\odot ,\to ,0,1) is an MTL-algebra. Define a real-valued function \nu by

\nu =\left(\begin{array}{cccc}0& a& b& 1\\ 2& 1& 0& 0\end{array}\right).

It is easily checked that \nu is an implicative generalized valuation on an MTL-algebra L. However,

\nu (a\vee \mathrm{\neg }a)=\nu (a)=1\ne 0.

Corollary 4.11　 Let \nu be a generalized valuation on an IMTL-algebra L. Then the following statements are equivalent:

(1) \nu is an implicative generalized valuation on L;

(2) for any x,y,z\in L, \nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to y))+\nu (y\to z);

(3) for any x,z\in L, \nu (x\to z)\le \nu (x\to (\mathrm{\neg }z\to z));

(4) for any x,y,z\in L, \nu (x\to z)\le \nu (y\to (x\to (\mathrm{\neg }z\to z)))+\nu (y);

(5) for any x\in L, \nu (x)\le \nu (\mathrm{\neg }x\to x);

(6) for any x,y\in L, \nu (x)\le \nu ((x\to y)\to x);

(7) for any x,y,z\in L, \nu (x)\le \nu (z\to ((x\to y)\to x))+\nu (z);

(8) for any x\in L, \nu (x\vee \mathrm{\neg }x)=0.

Proof　These results can be deduced from Theorems 4.3−4.9.□

Proposition 4.12　 If \nu is an implicative generalized valuation on an IMTL-algebra L, then \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu is an implicative filter of an MTL-algebra L.

Proof　(3.1) implies 1\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu. If x\to (y\to z),x\to y\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu, then

\nu (x\to (y\to z))=\nu (x\to y)=0.

Moreover, by Definition 4.1, we have

\nu (x\to z)\le \nu (x\to (y\to z))+\nu (x\to y)=0,

and hence \nu (x\to z)=0, which implies that x\to z\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu.

Therefore \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu is an implicative filter of an MTL-algebra L.□

The converse of Proposition 4.12 is not true in general.

Example 4.13　Consider the MTL-algebra in Example 4.2. Define a real-valued function \nu on an MTL-algebra L by

\nu =\left(\begin{array}{cccccc}0& a& b& c& d& 1\\ 2& 1& 1& 0& 0& 0\end{array}\right).

Then

\mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu =\{x\in L\mid \nu (x)=0\}=\{c,d,1\}

is not an implicative generalized valuation on an IMTL-algebra L, since

2=\nu (0)=\nu (1\to 0)⪇\nu (1\to (a\to 0))+\nu (1\to a)=\nu (d)+\nu (a)=1.

Here we introduce the notion of positive implicative generalized valuations on MTL-algebras and study some of their algebraic properties, and obtain some equivalent conditions under which an implicative generalized valuation becomes a positive implicative generalized valuation.

Definition 4.14　Let L be an MTL-algebra. A real-valued function \nu :L\to \mathbb{R} is said to be a positive implicative generalized valuation of L if it satisfies Eq. (3.1) and

(4.2)\begin{array}{r}\nu (x)\le \nu (z\to ((x\to y)\to x))+\nu (z).\end{array}

Example 4.15　Let L=\{0,a,b,c,d,1\} with 0\le b\le a\le 1, 0\le d\le a,c\le 1. Define the binary operations \odot and \to as follows:

\begin{array}{cllllll}\odot & 0& a& b& c& d& 1\\ \\ 0& 0& 0& 0& 0& 0& 0\\ a& 0& b& b& d& 0& a\\ b& 0& b& b& 0& 0& b\\ c& 0& d& 0& c& d& c\\ d& 0& 0& 0& d& 0& d\\ 1& 0& a& b& c& d& 1\end{array}\begin{array}{ccccccc}\to & 0& a& b& c& d& 1\\ \\ 0& 1& 1& 1& 1& 1& 1\\ a& d& 1& a& c& c& 1\\ b& c& 1& 1& c& c& 1\\ c& b& a& b& 1& a& 1\\ d& a& 1& a& 1& 1& 1\\ 1& 0& a& b& c& d& 1\end{array}

Then (L,\wedge ,\vee ,\odot ,\to ,0,1) is an MTL-algebra. Define a real-valued function \nu as

\nu =\left(\begin{array}{cccccc}0& a& b& c& d& 1\\ 1& 0& 0& 1& 1& 0\end{array}\right).

It is easily checked that \nu is a positive implicative generalized valuation on L.

Proposition 4.16　 Every positive implicative generalized valuation is implicative on an MTL-algebra L.

Proof　First, we show that every positive implicative generalized valuation \nu is an inverse order function. Indeed, if x\le y, that is, x\to y=1, then by y\le 1\to y, we have

x\to y\le x\to (1\to y),

and hence

1=x\to (1\to y)=x\to ((y\to y)\to y).

Further by Definition 4.14, one has

\nu (y)\le \nu (x\to ((y\to y)\to y))+\nu (x)=\nu (1)+\nu (x)=\nu (x),

that is, x\le y implies \nu (y)\le \nu (x).

Then we will prove that \nu is an implicative generalized valuation on L.

If \nu is a positive implicative generalized valuation on L, then for any x,y,z\in L,

\begin{array}{rl}\nu (x\to y)& \le \nu (1\to (((x\to y)\to y)\to (x\to y)))+\nu (1)\\ & =\nu (((x\to y)\to y)\to (x\to y)),\end{array}

which implies \nu (x\to y)\le \nu (((x\to y)\to y)\to (x\to y)).

Moreover, since x\to (x\to y)\le ((x\to y)\to y)\to (x\to y), \nu (((x\to y)\to y)\to (x\to y))\le \nu (x\to (x\to y)), we have \nu (x\to y)\le \nu (x\to (x\to y)).

On the other hand, since x\to y\le x\to (x\to y), one has \nu (x\to (x\to y))\le \nu (x\to y).

This implies \nu (x\to y)=\nu (x\to (x\to y)), and hence by Theorem 4.5, \nu is an implicative generalized valuation on an MTL-algebra L.□

The converse of Proposition 4.16 is not true in general.

Example 4.17　Consider the MTL-algebra L in Example 4.2. Define a real-valued function \nu as

\nu =\left(\begin{array}{ccccc}0& a& b& c& 1\\ 4& 3& 1& 2& 0\end{array}\right).

It is checked that \nu is an implicative generalized valuation on L. However, \nu is not a positive implication generalized valuation, since

2=\nu (c)⪇\nu (1\to ((c\to 0)\to c))+\nu (1)=\nu (1)+\nu (1)=0.

Proposition 4.18　 A positive implicative generalized valuation is a generalized valuation on an MTL-algebra L.

Proof　The result can be deduced directly from Propositions 4.3 and 4.16.□

Example 4.19　Consider the MTL-algebra L in Example 4.10. Define a real-valued function \nu by

\nu =\left(\begin{array}{cccc}0& a& b& 1\\ 7& 6& 3& 0\end{array}\right).

It is checked that \nu is a generalized valuation, but not a positive implicative generalized valuation, since

3=\nu (b)⪇\nu (1\to ((b\to a)\to b))+\nu (1)=\nu (1)+\nu (1)=0.

Here we give some equivalent conditions under which a generalized valuation becomes a positive implicative generalized valuation on an MTL-algebra.

Theorem 4.20　 Let \nu be a generalized valuation on an MTL-algebra L. Then the following statements are equivalent:

(1) \nu is a positive implicative generalized valuation on L,

(2) for any x,y\in L, \nu ((x\to y)\to x)=\nu (x).

Proof　 (1)\Rightarrow (2). If \nu is positive implicative, then

\nu (x)\le \nu (1\to ((x\to y)\to x))+\nu (1)=\nu ((x\to y)\to x),

which implies

\nu (x)\le \nu ((x\to y)\to x).

Further by Proposition 3.8(1) and x\le (x\to y)\to x, we have

\nu ((x\to y)\to x)\le \nu (x).

Therefore \nu ((x\to y)\to x)=\nu (x).

(2)\Rightarrow (1). Assume that for any x,y\in L,

\nu ((x\to y)\to x)=\nu (x).

Since \nu is a generalized valuation on an MTL-algebra L, by Definition 3.1,we have

\nu ((x\to y)\to x)\le \nu (z\to ((x\to y)\to x))+\nu (z),

and hence

\nu (x)\le \nu (z\to ((x\to y)\to x))+\nu (z),

which implies that \nu is a positive implicative generalized valuation on an MTL-algebra L.□

Theorem 4.21　 Let \nu be a generalized valuation on an MTL-algebra L. Then the following statements are equivalent:

(1) \nu is a positive implicative generalized valuation on L,

(2) for any x\in L, \nu (\mathrm{\neg }x\to x)=\nu (x).

Proof　 (1)\Rightarrow (2). If \nu is positive implicative, then by Corollary 4.11, we have

\nu (x)=\nu ((x\to 0)\to x)=\nu (\mathrm{\neg }x\to x).

(2)\Rightarrow (1). Assume that for any x,y\in L, \nu (\mathrm{\neg }x\to x)=\nu (x). Since \mathrm{\neg }x=x\to 0\le x\to y, we have (x\to y)\to x\le \mathrm{\neg }x\to x, and hence

\nu (\mathrm{\neg }x\to x)\le \nu ((x\to y)\to x).

Further by the hypothesis,

\nu (x)\le \nu ((x\to y)\to x),

and then by Proposition 3.8(1) and x\le (x\to y)\to x, we have

\nu ((x\to y)\to x)\le \nu (x),

and hence \nu ((x\to y)\to x)=\nu (x).

Therefore \nu is a positive implicative by Corollary 4.11.□

The next results provide some conditions under which an implicative generalized valuation becomes a positive implicative generalized valuation.

Theorem 4.22　 Let \nu be an implicative generalized valuation on an MTL-algebra L. Then the following statements are equivalent:

(1) \nu is a positive implicative generalized valuation on L,

(2) for any x\in L, \nu (x)=\nu (\mathrm{\neg }\mathrm{\neg }x).

Proof　 (1)\Rightarrow (2). If \nu is positive implicative, then by \mathrm{\neg }\mathrm{\neg }x=\mathrm{\neg }x\to 0\le \mathrm{\neg }x\to x, we have

\nu (\mathrm{\neg }x\to x)\le \nu (\mathrm{\neg }\mathrm{\neg }x).

Further by Theorem 4.10, we also have

\nu (x)=\nu (\mathrm{\neg }x\to x)\le \nu (\mathrm{\neg }\mathrm{\neg }x).

On the other hand, notice that x\le (x\to 0)\to 0=\mathrm{\neg }\mathrm{\neg }x, and hence

\nu (\mathrm{\neg }\mathrm{\neg }x)\le \nu (x).

Therefore we can conclude that \nu (x)=\nu (\mathrm{\neg }\mathrm{\neg }x).

(2)\Rightarrow (1). Assume that for any x\in L,

\nu (x)=\nu (\mathrm{\neg }\mathrm{\neg }x).

Since

\mathrm{\neg }x\to x\le (x\to 0)\to (\mathrm{\neg }x\to 0)=\mathrm{\neg }x\to (\mathrm{\neg }x\to 0),

we have

\nu (\mathrm{\neg }x\to (\mathrm{\neg }x\to 0))\le \nu (\mathrm{\neg }x\to x).

Further by Theorem 4.5(3),

\nu (\mathrm{\neg }x\to (\mathrm{\neg }x\to 0))=\nu (\mathrm{\neg }x\to 0)=\nu (\mathrm{\neg }\mathrm{\neg }x),

and hence

\nu (\mathrm{\neg }\mathrm{\neg }x)\le \nu (\mathrm{\neg }x\to x),

which shows \nu (x)\le \nu (\mathrm{\neg }x\to x).

On the other hand, since x\le (\mathrm{\neg }x)\to x, we have \nu (\mathrm{\neg }x\to x)\le \nu (x), and hence \nu (x)=\nu (\mathrm{\neg }x\to x), further by Theorem 4.20, we conclude that \nu is a positive implicative generalized valuation on an MTL-algebra L.□

Proposition 4.23　 If \nu is a positive implicative generalized valuation on an MTL-algebra L, then \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu is a positive implicative filter on L.

Proof　(3.1) implies 1\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu. If x\to ((y\to z)\to y),x\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu, then

\nu (x\to ((y\to z)\to y))=\nu (x)=0.

Moreover, by Eq. (4.2), we have

\nu (y)\le \nu (x\to ((y\to z)\to y))+\nu (x)=0,

and hence \nu (y)=0, which implies that y\incoker\nu.

Therefore \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu is a positive implicative filter of MTL-algebra L.□

The converse of Proposition 4.23 is not true in general.

Example 4.24　Let L=\{0,a,b,1\} with 0\le a,b\le 1. Define the binary operations \odot and \to as follows:

\begin{array}{ccccc}\odot & 0& a& b& 1\\ \\ 0& 0& 0& 0& 0\\ a& 0& a& 0& a\\ b& 0& 0& b& b\\ 1& 0& a& b& 1\end{array}\begin{array}{ccccc}\to & 0& a& b& 1\\ \\ 0& 1& 1& 1& 1\\ a& b& 1& b& 1\\ b& a& a& 1& 1\\ 1& 0& a& b& 1\end{array}

Then (L,\wedge ,\vee ,\odot ,\to ,0,1) is an MTL-algebra. Define a real-valued function \nu by

\nu =\left(\begin{array}{cccc}0& a& b& 1\\ 4& 0& 2& 0\end{array}\right).

It is checked that \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu =\{a,1\} is a positive implicative filter of L, but \nu is not a positive implicative generalized valuation on the MTL-algebra L, since

4=\nu (0)⪇\nu (b\to ((0\to a)\to 0))+\nu (b)=\nu (a)+\nu (b)=2.

5 Quotient structures induced by generalized valuations

In this section, we study the quotient MTL-algebras that induced by generalized valuations.

Definition 5.1　Let \nu be a generalized valuation on an MTL-algebra L. Define a binary relation {\theta }_{\nu } by

(5.1)\begin{array}{r}(x,y)\in {\theta }_{\nu }\iff \nu (x\to y)=\nu (y\to x)=0.\end{array}

Proposition 5.2　 If \nu is a generalized valuation on an MTL-algebra L, then {\theta }_{\nu } is a congruence on L, which is called the binary relation induced by the generalized valuation.

Proof　First we prove that {\theta }_{\nu } is an equivalence relation on L.

(1) Since \nu (x\to x)=\nu (1)=0, we have (x,x)\in {\theta }_{\nu }.

(2) If (x,y)\in {\theta }_{\nu }, then (y,x)\in {\theta }_{\nu }.

(3) If (x,y),(y,z)\in {\theta }_{\nu }, then by Eq. (5.1), we have

\nu (x\to y)=0,\nu (y\to z)=0,

and further by Proposition 3.8(4), we also have

\nu (x\to z)\le \nu (x\to y)+\nu (y\to z)=0,

which implies \nu (x\to z)=0. Similarly, \nu (z\to x)=0, and thus (x,z)\in {\theta }_{\nu }.

Then we prove that {\theta }_{\nu } is a congruence on L.

(4) If (x,y)\in {\theta }_{\nu }, then

\nu (x\to y)=\nu (y\to x)=0.

Notice that from Proposition 2.2, we have

x\le (x\to z)\to z\Rightarrow y\to x\le y\to ((x\to z)\to z)=(x\to z)\to (y\to z),

and hence

\begin{array}{c}y\to x\le (x\to z)\to (y\to z),\\ \nu ((x\to z)\to (y\to z))\le \nu (y\to x)=0,\end{array}

which implies

\nu ((x\to z)\to (y\to z))=0.

Similarly, \nu ((y\to z)\to (x\to z))=0.

Therefore (x\to z,y\to z)\in {\theta }_{\nu }.

(5) If (x,y)\in {\theta }_{\nu }, then

\nu (x\to y)=\nu (y\to x)=0.

Notice that from Proposition 2.2, we have y\le z\to (y\odot z)\Rightarrow x\to y\le x\to (z\to (y\odot z))=(x\odot z)\to (y\odot z), which implies x\to y\le (x\odot z)\to (y\odot z), and hence

\begin{array}{c}\nu ((x\odot z)\to (y\odot z))\le \nu (x\to y)=0,\\ \nu ((x\odot z)\to (y\odot z))=0.\end{array}

Similarly, \nu ((y\odot z)\to (x\odot z))=0.

Therefore (x\odot z,y\odot z)\in {\theta }_{\nu }.□

Example 5.3　Let L=\{0,a,b,c,1\} with 0\le a,b\le c\le 1. Define the binary operations \odot and \to as follows:

\begin{array}{rlr}& \begin{array}{llllll}\odot & 0& a& b& c& 1\\ \\ 0& 0& 0& 0& 0& 0\\ a& 0& a& 0& a& a\\ b& 0& 0& b& b& b\\ c& 0& a& b& c& c\\ 1& 0& a& b& c& 1\end{array}& \begin{array}{cccccc}\to & 0& a& b& c& 1\\ \\ 0& 1& 1& 1& 1& 1\\ a& b& 1& b& 1& 1\\ b& a& a& 1& 1& 1\\ c& 0& a& b& 1& 1\\ 1& 0& a& b& c& 1\end{array}\end{array}

Then (L,\wedge ,\vee ,\odot ,\to ,0,1) is an MTL-algebra. Define a real-valued function \nu as

\nu =\left(\begin{array}{ccccc}0& a& b& c& 1\\ 1& 1& 1& 0& 0\end{array}\right).

It is easily checked that \nu is a generalized valuation on L. Since

\nu (c\to 1)=\nu (1\to c)=0,\nu (c\to c)=0,\nu (1\to 1)=0,

we have \{(c,1),(1,c),(c,c),(1,1)\}\in {\theta }_{\nu }, which implies that

\{(c,1),(1,c),(c,c),(1,1)\}

is a congruence on MTL-algebra L.

Proposition 5.4　 Let \nu be a generalized valuation on an MTL-algebra L and {\theta }_{\nu } be the congruence induced by \nu. The set of all equivalence classes [x{]}_{\nu }=\{y\in L|(x,y)\in {\theta }_{\nu }\} is denoted by L/\nu. On the set, we define

\begin{array}{rl}& [x{]}_{\nu }\vee [y{]}_{\nu }=[x\vee y{]}_{\nu },[x{]}_{\nu }\wedge [y{]}_{\nu }=[x\wedge y{]}_{\nu },\\ & [x{]}_{\nu }\cdot [y{]}_{\nu }=[x\odot y{]}_{\nu },[x{]}_{\nu }\rightharpoonup [y{]}_{\nu }=[x\to y{]}_{\nu }.\end{array}

Then the resulting algebra is denoted by L/\nu and is called the quotient algebra of L induced by the generalized valuation \nu.

Proof　The proof is clear and hence we omit it.□

Theorem 5.5　 If \nu is a generalized valuation on an MTL-algebra L, then the following statements are equivalent:

(1) [x{]}_{\nu }\le [y{]}_{\nu },

(2) \nu (x\to y)=0.

Proof　 (1)\Rightarrow (2). If [x{]}_{\nu }\le [y{]}_{\nu }, then [x{]}_{\nu }\rightharpoonup [y{]}_{\nu }=[1{]}_{\nu }, that is, [x\to y{]}_{\nu }=[1{]}_{\nu }, which implies (x\to y,1)\in {\theta }_{\nu }, and hence \nu (1\to (x\to y))=\nu (x\to y)=0.

(2)\Rightarrow (1). Since

\nu (1\to (x\to y))=\nu (x\to y)=0,\nu ((x\to y)\to 1)=\nu (1)=0,

we have (x\to y,1)\in {\theta }_{\nu }, hence [x{]}_{\nu }\rightharpoonup [y{]}_{\nu }=[1{]}_{\nu }, which implies [x{]}_{\nu }\le [y{]}_{\nu }.□

Proposition 5.6　 If \nu is a generalized valuation on an MTL-algebra L and \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu is closed under the operation \to, then F_{\nu }=[1{]}_{\nu }.

Proof　If x\in [1{]}_{\nu }, that is, (x,1)\in {\theta }_{\nu }, then

\nu (x)=\nu (1\to x)=\nu (x\to 1)=\nu (1)=0,

which implies \nu (x)=0, and hence x\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu. Therefore [1{]}_{\nu }\subseteq \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu.

If x\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu, then \nu (1\to x)=\nu (x)=0. Since \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu is closed under \to with 1\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu, we have x\to 1\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu, and hence \nu (x\to 1)=0, which implies (x,1)\in {\theta }_{\nu }. So \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu \subseteq [1{]}_{\nu }.□

Proposition 5.7　 If \nu is a generalized valuation on an MTL-algebra L, then {\theta }_{\mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu }={\theta }_{\nu }.

Proof　If (x,y)\in {\theta }_{\mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu }, that is, x\to y,y\to x\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu, then

\nu (x\to y)=0,\nu (y\to x)=0,

and hence (x,y)\in {\theta }_{\nu }, which implies {\theta }_{\mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu }\subseteq {\theta }_{\nu }.

On the other hand, if (x,y)\in {\theta }_{\nu }, that is, \nu (x\to y)=\nu (y\to x)=0, then x\to y,y\to x\in \mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu, and hence (x,y)\in {\theta }_{\mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu }, which implies {\theta }_{\nu }\subseteq {\theta }_{\mathrm{c}\mathrm{o}\mathrm{k}\mathrm{e}\mathrm{r}\nu }.□

Proposition 5.8　 If {\nu }_1 and {\nu }_2 are two generalized valuations on an MTL-algebra L with [1{]}_{{\nu }_1}=[1{]}_{{\nu }_2}, then {\theta }_{{\nu }_1}={\theta }_{{\nu }_2}.

Proof　If for any x,y\in L, (x,y)\in {\theta }_{{\nu }_1}, then (x\to y,y\to y)=(x\to y,1)\in {\theta }_{{\nu }_1}, and hence x\to y\in [1{]}_{{\nu }_1}. Similarly, we have y\to x\in [1{]}_{{\nu }_1}. Since [1{]}_{{\nu }_1}=[1{]}_{{\nu }_2}, we have x\to y\in [1{]}_{{\nu }_2} and y\to x\in [1{]}_{{\nu }_2}, and so [x{]}_{{\nu }_2}\rightharpoonup [y{]}_{{\nu }_2}=[x\to y{]}_{{\nu }_2}=[1{]}_{{\nu }_2} and [y{]}_{{\nu }_2}\rightharpoonup [x{]}_{{\nu }_2}=[y\to x{]}_{{\nu }_2}=[1{]}_{{\nu }_2}. Further by Proposition 4.2, we obtain that L/{\nu }_2 is an MTL-algebra, and hence [x{]}_{{\nu }_2}=[y{]}_{{\nu }_2}, that is, (x,y)\in {\theta }_{{\nu }_2}. So {\theta }_{{\nu }_1}\subseteq {\theta }_{{\nu }_2}. Similarly, {\theta }_{{\nu }_2}\subseteq {\theta }_{{\nu }_1}.

Therefore {\theta }_{{\nu }_1}={\theta }_{{\nu }_2}.□

Proposition 5.9　 If \nu is a generalized valuation on an MTL-algebra L and F is a filter of L with F_{\nu }\subseteq F. Define the set F/\nu =\{[x{]}_{\nu }|x\in F\}. Then the following hold:

(1) x\in F if and only if [x{]}_{\nu }\in F/\nu,

(2) F/\nu is a filter of L.

Proof　 (1) Sufficiency is obviously established. Conversely, if [x{]}_{\nu }\in F/\nu, then there exists y\in F such that [x{]}_{\nu }=[y{]}_{\nu }, that is, \nu (y\to x)=0, which implies y\to x\in F_{\nu }\subseteq F. Since F is a filter of L with y\in F, we have x\in F.

(2) Since F is a filter of L, we have 1\in F. By (1), [1{]}_{\nu }\in F/\nu. If [x{]}_{\nu },[x{]}_{\nu }\rightharpoonup [y{]}_{\nu }\in F/\nu, then x\in F, x\to y\in F, and hence y\in F, which implies [y{]}_{\nu }\in F/\nu. Therefore F/\nu is also a filter of L/\nu.□

Proposition 5.10　 If L_1,L_2 are two MTL-algebras and f:L_1\to L_2 is an epimorphism with \nu is a generalized valuation on L_2, then L_1/(\nu \circ f)\cong L_2/\nu.

Proof　It follows from Proposition 5.4 that L_1/(\nu \circ f) and L_2/\nu are two MTL-algebras. For any x\in L_1, define a function \psi :L_1/(\nu \circ f)\to L_2/\nu as \psi ([x{]}_{(\nu \circ f)})=[f(x){]}_{\nu }. Here we will prove that \psi is isomorphism.

First, we show that \psi is well defined.

For any x,y\in L_1, [x{]}_{(\nu \circ f)}=[y{]}_{(\nu \circ f)}, that is, (x,y)\in {\theta }_{(\nu \circ f)}, and hence (\nu \circ f)(x\to y)=(\nu \circ f)(y\to x)=0. Since f is an epimorphism, we have \nu (f(x)\to f(y))=\nu (f(y)\to f(x))=0, which implies (f(x),f(y))\in {\theta }_{\nu }, that is, [f(x){]}_{\nu }=[f(y){]}_{\nu }. So \psi ([x{]}_{(\nu \circ f)})=\psi ([y{]}_{(\nu \circ f)}).

Then we prove that \psi is a homomorphism. Since f is an epimorphism, we have

(1) \psi ([1{]}_{(\nu \circ f)})=[f(1){]}_{\nu }=[1{]}_{\nu };

(2) \psi ([x{]}_{(\nu \circ f)}\wedge [y{]}_{(\nu \circ f)})=\psi ([x\wedge y{]}_{(\nu \circ f)})=[f(x\wedge y){]}_{\nu }=[f(x)\wedge f(y){]}_{\nu }=[f(x){]}_{\nu }\wedge [f(y){]}_{\nu }=\psi ([x{]}_{(\nu \circ f)})\wedge \psi ([y{]}_{(\nu \circ f)});

(3) \psi ([x{]}_{(\nu \circ f)}\vee [y{]}_{(\nu \circ f)})=\psi ([x\vee y{]}_{(\nu \circ f)})=[f(x\vee y){]}_{\nu }=[f(x)\vee f(y){]}_{\nu }=[f(x){]}_{\nu }\vee [f(y){]}_{\nu }=\psi ([x{]}_{(\nu \circ f)})\vee \psi ([y{]}_{(\nu \circ f)});

(4) \psi ([x{]}_{(\nu \circ f)}\cdot [y{]}_{(\nu \circ f)})=\psi ([x\odot y{]}_{(\nu \circ f)})=[f(x\odot y){]}_{\nu }=[f(x)\odot f(y){]}_{\nu }=[f(x){]}_{\nu }·[f(y){]}_{\nu }=\psi ([x{]}_{(\nu \circ f)})\cdot \psi ([y{]}_{(\nu \circ f)})\cdot \psi ([x{]}_{(\nu \circ f)}\cdot [y{]}_{(\nu \circ f)})=\psi ([x{]}_{(\nu \circ f)})\cdot \psi ([y{]}_{(\nu \circ f)});

(5) \psi ([x{]}_{(\nu \circ f)}\rightharpoonup [y{]}_{(\nu \circ f)})=\psi ([x\to y{]}_{(\nu \circ f)})=[f(x\to y){]}_{\nu }=[f(x)\to f(y){]}_{\nu }=[f(x){]}_{\nu }\rightharpoonup [f(y){]}_{\nu }=\psi ([x{]}_{(\nu \circ f)})\rightharpoonup \psi ([y{]}_{(\nu \circ f)}).\psi ([x{]}_{(\nu \circ f)}\rightharpoonup [y{]}_{(\nu \circ f)})=\psi ([x{]}_{(\nu \circ f)})\rightharpoonup \psi ([y{]}_{(\nu \circ f)}).

Finally, we show that \psi is a bijection.

(6) If [y{]}_{\nu }\in L_2/\nu, then there exists x\in L_1 such that f(x)=y, and hence \psi ([x{]}_{(\nu \circ f)})=[f(x){]}_{\nu }=[y{]}_{\nu }, which implies that \psi is surjection.

(7) If \psi ([x{]}_{(\nu \circ f)})=\psi ([y{]}_{(\nu \circ f)}), then [f(x){]}_{\nu }=[f(y){]}_{\nu }, that is, (f(x),f(y))\in {\theta }_{\nu }. So \nu (f(x)\to f(y))=\nu (f(y)\to f(x))=0. Moreover, since f is an epimorphism, we have (\nu \circ f)(x\to y)=(\nu \circ f)(y\to x)=0, and hence (x,y)\in {\theta }_{(\nu \circ f)}, which implies [x{]}_{(\nu \circ f)}=[y{]}_{(\nu \circ f)}. So \psi is injection.

Therefore, we can conclude that L_1/(\nu \circ f)\cong L_2/\nu.□

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