Intuitionistic Fuzzy Predicate Logic

Abstract

The idea for evaluation of the propositions was extended for predicates (see Barwise (Handbook of mathematical logic, 1989, [1]), Crossley et al. (What is mathematical logic? 1972, [2]), van Dalen (Logic and structure, 2013, [3]), Ebbinghaus et al. (Mathematical logic, 1994, [4]), Mendelson (Introduction to mathematical logic, 1964, [5]), Shoenfield (Mathematical logic, 2001, [6]) as follows (see, e.g., Atanassov (Intuitionistic fuzzy sets, 1999, [7]), Atanassov (Modern approaches in fuzzy sets, intuitionistic fuzzy sets, generalized nets and related topics, 2014, [8]), Atanassov and Gargov (Elements of intuitionistic fuzzy logic I, 1998, [9]), Gargov and Atanassov (Two results in intuitionistic fuzzy logic, 1992, [10]).

2.1 Short Remarks on Intuitionistic Fuzzy Predicate Logic

The idea for evaluation of the propositions was extended for predicates (see [1,2,3,4,5,6]) as follows (see, e.g., [7,8,9,10]).

Let x be a variable, obtaining values in set E and let P(x) be a predicate with a variable x. Let

$$V(P(x)) = \langle \mu (P(x)), \nu (P(x)) \rangle .$$

The IF-interpretations of the (intuitionistic fuzzy) quantifiers for all ($\forall$) and there exists ($\exists$) are introduced in [7, 9, 10] by

$$\begin{aligned} V(\exists x P(x)) = \langle \sup _{y \in E} \mu (P(y)), \inf _{y \in E} \nu (P(y)) \rangle , \end{aligned}$$

(2.1.1)

$$\begin{aligned} V(\forall x P(x)) = \langle \inf _{y \in E} \mu (P(y)), \sup _{y \in E} \nu (P(y)) \rangle . \end{aligned}$$

(2.1.2)

If E is a finite set, then we can use the denotations

$$\begin{aligned} V(\exists x P(x)) = \langle \max _{y \in E} \mu (P(y)), \min _{y \in E} \nu (P(y)) \rangle , \end{aligned}$$

(2.1.3)

$$\begin{aligned} V(\forall x P(x)) = \langle \min _{y \in E} \mu (P(y)), \max _{y \in E} \nu (P(y)) \rangle . \end{aligned}$$

(2.1.4)

In general, below, we use the first forms of both quantifiers.

Their geometrical interpretations are illustrated in Figs. 2.1 and 2.2, respectively, where $x_1,\ldots , x_5$ are possible values of variable x and $V(x_1),\ldots ,$ $V(x_5)$ are their IF-evaluations.

The most important property of the two quantifiers is that each of them juxtaposes to predicate P a point (exactly one per quantifier) in the IF-interpretational triangle.

In [9, 10], for implication $\rightarrow _4$ the following two theorems are proved, where we used $\rightarrow$ instead of $\rightarrow _4$.

Fig. 2.1

Second geometrical interpretation of quantifier $\forall$

Fig. 2.2

Second geometrical interpretation of quantifier $\exists$

Theorem 2.1.1

The logical axioms of the $\mathcal K$ -theory (see [5]):

• (a) $A \rightarrow (B \rightarrow A),$

• (b) $(A \rightarrow (B \rightarrow C)) \rightarrow ((A \rightarrow B) \rightarrow (A \rightarrow C)),$

• (c) $( \lnot A \rightarrow \lnot B) \rightarrow (( \lnot A \rightarrow B) \rightarrow A),$

• (d) $\forall xA(x) \rightarrow A(t),\text{ for } \text{ the } \text{ fixed } \text{ variable } t,$

• (e) $\forall x(A \rightarrow B) \rightarrow (A \rightarrow \forall xB)$ are IFTs.

Proof

Assertions (a)–(c) coincide with those in Theorem 1.5.17 (IL2), (IL5) and Theorem 1.5.26, respectively. We will here prove only assertions (d) and (e).

(d) Let the variable t be fixed.

Then,

$$V( \forall xA(x) \rightarrow A(t))$$

$$= \langle \inf _{x} \mu (A(x)), \sup _{x} \nu (A(x))\rangle \rightarrow \langle \mu (A(t)), \nu (A(t))\rangle $$

$$= \langle \max (\sup _{x} \nu (A(x)), \mu (A(t))), \min (\inf _{x} \mu (A(x)), \nu (A(t)))\rangle $$

and

$$\max (\sup _{x} \nu (A(x)), \mu (A(t))) - \min (\inf _{x} \mu (A(x)), \nu (A(t)))$$

$$\ge \mu (A(t)) -\inf _{x} \mu (A(x)) \ge 0,$$

i.e., (d) is an IFT.

For (e) we sequentially obtain:

$$ V( \forall x(A \rightarrow B) \rightarrow (A \rightarrow \forall xB)) = V( \forall x(A \rightarrow B)) \rightarrow V(A \rightarrow \forall xB)$$

$$= \langle \inf _{x} \max ( \mu (B), \nu (A)), \sup _{x} \min ( \mu (A), \nu (B))\rangle $$

$$ \rightarrow \langle \max ( \nu (A), \inf _{x} \mu (B)), \min ( \mu (A), \sup _{x} \nu (B))\rangle $$

$$= \langle \max ( \nu (A), \inf _{x} \mu (B), \sup _{x} \min ( \mu (A), \nu (B))),$$

$$\min ( \mu (A), \sup _{x} \nu (B), \inf _{x} \max ( \mu (B), \nu (A)))\rangle $$

and

$$ \max ( \nu (A), \inf _{x} \mu (B), \sup _{x} \min ( \mu (A), \nu (B))) \ge \max ( \nu (A), \inf _{x} \mu (B))$$

$$= \inf _{x} \max ( \mu (B), \nu (A)) \ge \min ( \mu (A), \sup _{x} \nu (B), \inf _{x} \max ( \mu (B), \nu (A))),$$

i.e., (e) also is an IFT. $\square$

Below, we list some assertions, which are theorems of the classical first order logic (see, e.g. [5]).

Theorem 2.1.2

The following formulas are IFTs:

• (a) $( \forall x A(x) \rightarrow B) \equiv \exists x (A(x) \rightarrow B),$

• (b) $\exists x A(x) \rightarrow B \equiv \forall x (A(x) \rightarrow B),$

• (c) $B \rightarrow \forall x A(x) \equiv \forall x (B \rightarrow A(x)),$

• (d) $B \rightarrow \exists x A(x) \equiv \exists x (B \rightarrow A(x)),$

• (e) $( \forall x A \wedge \forall x B) \equiv \forall x (A \wedge B),$

• (f) $( \forall x A \vee \forall x B) \rightarrow \forall x (A \vee B),$

• (g) $\lnot \forall x A \equiv \exists x \lnot A,$

• (h) $\lnot \exists x A \equiv \forall x \lnot A,$

• (i) $\forall x \forall y A \equiv \forall y \forall x A,$

• (j) $\exists x \exists y A \equiv \exists y \exists x A,$

• (k) $\exists x \forall y A \rightarrow \forall y \exists x A,$

• (l) $\forall x (A \rightarrow B) \rightarrow ( \forall x A \rightarrow \forall x B).$

Proof

We shall use Lemma 1.5.1.

(a)

$$V( \forall x A(x) \rightarrow B) $$

$$= \langle \max ( \sup _{x} \nu (A(x)), \mu (B)), \min ( \inf _{x} \mu (A(x)), \nu (B))\rangle $$

$$= \langle \sup _{x} (\max ( \nu (A(x)), \mu (B))), \inf _{x} (\min ( \mu (A(x)), \nu (B)))\rangle $$

$$= V( \exists x (A(x) \rightarrow B));$$

(b)

$$V( \exists x A(x) \rightarrow B) $$

$$= \langle \max ( \inf _{x} \nu (A(x)), \mu (B)), \min ( \sup _{x} \mu (A(x)), \nu (B))\rangle $$

$$= \langle \inf _{x} (\max ( \nu (A(x)), \mu (B))), \sup _{x} (\min ( \mu (A(x)), \nu (B)))\rangle $$

$$= V( \forall x (A(x) \rightarrow B));$$

(c)

$$V(B \rightarrow \forall x A(x)) $$

$$= \langle \max (\inf _{x} \mu (A(x)), \nu (B)), \min (\sup _{x} \nu (A(x)), \mu (B))\rangle $$

$$= \langle \inf _{x} (\max ( \mu (A(x)), \nu (B))), \sup _{x} (\min ( \nu (A(x)), \mu (B)))\rangle $$

$$= V( \forall x (B \rightarrow A(x)));$$

(d) is proved analogically;

(e)

$$V( \forall x A \wedge \forall x B)$$

$$= \langle \min (\inf _{x} \mu (A), \min \mu (B)), \max (\sup _{x} \nu (A), \max \nu (B))\rangle $$

$$= \langle \inf _{x} (\min ( \mu (A), \mu (B))), \sup _{x} (\max ( \nu (A), \nu (B)))\rangle $$

$$= V( \forall x (A \wedge B));$$

(f) is proved analogically;

(g)

$$V( \lnot \forall x A) = \langle \sup _{x} \nu (A), \inf _{x} \mu (A) \rangle = V( \exists x \lnot A);$$

(h) is proved analogically;

(i)

$$V( \forall x \forall y A)$$

$$= \langle \inf _{x} \inf _{y} \mu (A), \sup _{x} \sup _{y} \nu (A) \rangle $$

$$= \langle \inf _{y} \inf _{x} \mu (A), \sup _{y} \sup _{x} \nu (A) \rangle $$

$$= V( \forall y \forall x A);$$

(j) is proved analogically;

(k)

$$V( \exists x \forall y A \rightarrow \forall y \exists x A)$$

$$= \langle \sup _{x} \inf _{y} \mu (A), \inf _{x} \sup _{y} \nu (A) \rangle \rightarrow \langle \inf _{y} \sup _{x} \mu (A), \sup _{y} \inf _{x} \nu (A)\rangle $$

$$= \langle \max ( \inf _{y} \sup _{x} \nu (A), \inf _{x} \sup _{y} \mu (A)), \min ( \sup _{x} \inf _{y} \mu (A), \sup _{y} \inf _{x} \nu (A)) \rangle $$

and

$$\max ( \inf _{y} \sup _{x} \nu (A), \inf _{x} \sup _{y} \mu (A)) - \min ( \sup _{x} \inf _{y} \mu (A), \sup _{y} \inf _{x} \nu (A))$$

$$\ge \sup _{y} \inf _{x} \nu (A) - \sup _{y} \inf _{x} \nu (A) = 0,$$

i.e., $\exists x \forall y A \rightarrow \forall y \exists x A$ is an IFT;

(l) is proved analogically. $\square$

Theorem 2.1.3

For a predicate P and for negation $\lnot _i, \ \forall x P(x) \vee \exists x \lnot _i P(x)$ is an IFT for $i = 1, 3, 4, 8, 9, 11, 12, 14, 15, 18,\ldots , 23, 25,\ldots , 32, 45, 52, 53.$

Proof

Let for the variable x,

$$V(\forall x P(x)) = \langle M, n \rangle ,$$

$$V(\exists x P(x)) = \langle m, N \rangle ,$$

where the pairs $\langle M, n \rangle$ and $\langle m, N \rangle$ are given by either (2.1.1) and (2.1.2), or (2.1.3) and (2.1.4). Below, we will discuss the proof for three of the negations: $\lnot _1$, $\lnot _{12}$ and $\lnot _{52}$.

For $\lnot _1$ we obtain:

$$\forall x P(x) \vee \exists x \lnot _1 P(x) = \langle m, N \rangle \vee V(\exists x \langle \nu (P(x)), \mu (P(x)) \rangle )$$

$$= \langle m, N \rangle \vee \langle N, m \rangle = \langle \max (m, N), \min (m,N) \rangle ,$$

that is an IFT, because $\max (m, N) - \min (m,N) \ge 0$.

For $\lnot _{12}$ we obtain:

$$\forall x P(x) \vee \exists x \lnot _{12} P(x)$$

$$ = \langle m, N \rangle \vee \langle \sup _x (\nu (P(x))(\mu (P(x)) + \nu (P(x)))),$$

$$\inf _x (\mu (P(x))(\mu (P(x)) + \mu (P(x))\nu (P(x)) + \nu (P(x))^2))\rangle $$

$$ = \langle \max (m, \sup _x (\nu (P(x))(\mu (P(x)) + \nu (P(x))))),$$

$$\min (N, \inf _x (\mu (P(x))(\mu (P(x)) + \mu (P(x))\nu (P(x)) + \nu (P(x))^2)))\rangle .$$

Then,

$$\max (m, \sup _x (\nu (P(x))(\mu (P(x)) + \nu (P(x)))))$$

$$- \min (N, \inf _x (\mu (P(x))(\mu (P(x)) + \mu (P(x))\nu (P(x)) + \nu (P(x))^2)))$$

$$\ge m - \inf _x (\mu (P(x))(\mu (P(x)) + \mu (P(x))\nu (P(x)) + \nu (P(x))^2)) \ge 0,$$

because for every two numbers $a, b \in [0, 1]$, such that $a + b \le 1$: $a + ab + b^2 = a + b(a + b) \le a + b \le 1$, i.e., the expression is an IFT.

For $\lnot _{52}$ we obtain:

$$\forall x P(x) \vee \exists x \lnot _{52} P(x)$$

$$ = \langle m, N \rangle \vee \langle \sup _x (\text {sg}(\nu (P(x)))+ \text {sg}(\mu (P(x))\nu (P(x)))), m \rangle $$

$$ = \langle \max (m, \sup _x (\text {sg}(\nu (P(x)))+ \text {sg}(\mu (P(x))\nu (P(x))))), \min (N, m) \rangle ,$$

which obviously is an IFT. The ($\sup - \inf$)-case is analogous.

All other checks are similar. $\square$

The link between the interpretations of quantifiers and the topological operators C (closure) and I (interior) defined over IFSs see [7] is obvious.

Open Problem 9. The basic problem which remains unsolved is related to the characterization of predicate IFTs by means of a calculus.

Following [9, 10], we mention that a partial solution of the problem of giving a calculus which generates all predicate IFTs is presented in the next theorem.

Theorem 2.1.4

A prenex normal form A is an IFT if and only if it is a classical predicate tautology and its quantifier free matrix is a propositional IFT.

Here, a prenex form means (see [9, 10]) a predicate formula in which all quantifiers are moved to the left. The proof is based on the fact that all predicate transformations leading to prenex forms in the classical logic are valid for the intuitionistic fuzzy case, too.

2.2 Extended Intuitionistic Fuzzy Quantifiers

In [8], we introduced the following six quantifiers and studied some of their properties.

$$V(\forall _{\mu } x P(x)) = \{ \langle x, \inf _{y \in E} \mu (P(y)), \nu (P(x)) \rangle | x \in E\rbrace ,$$

$$V(\forall _{\nu } x P(x)) = \{ \langle x, \min (1 - \sup _{y \in E} \nu (P(y)), \mu (P(x))), \sup _{y \in E} \nu (P(y)) \rangle | x \in E\rbrace ,$$

$$V(\exists _{\mu } x P(x)) = \{ \langle x, \sup _{y \in E} \mu (P(y)), \min (1 - \sup _{y \in E} \mu (P(y)), \nu (P(x))) \rangle | x \in E\rbrace ,$$

$$V(\exists _{\nu } x P(x)) = \{ \langle x, \mu (P(x)), \inf _{y \in E} \nu (P(y)) \rangle | x \in E\rbrace ,$$

$$V(\forall _{\nu }^* x P(x))$$

$$= \{ \langle x, \min (1 - \sup _{y \in E} \nu (P(y)), \mu (P(x))), \min (\sup _{y \in E} \nu (P(y)), 1 - \mu (P(x))) \rangle | x \in E\rbrace ,$$

$$V(\exists _{\mu }^* x P(x))$$

$$= \{ \langle x, \min (\sup _{y \in E} \mu (P(y)), 1 - \nu (P(x))), \min (1 - \sup _{y \in E} \mu (P(y)), \nu (P(x))) \rangle | x \in E\rbrace .$$

Let the possible values of the variable x be a, b, c and let their IF-evaluations V(a), V(b), V(c) be shown on Fig. 2.3. The geometrical interpretations of the new quantifiers are shown in Figs. 2.4, 2.5, 2.6, 2.7, 2.8 and 2.9.

Fig. 2.3

Example of a second geometrical interpretation

Fig. 2.4

Second geometrical interpretation of quantifier $\exists _{\mu }$

Fig. 2.5

Second geometrical interpretation of quantifier $\exists _{\nu }$

Fig. 2.6

Second geometrical interpretation of quantifier $\forall _{\mu }$

Fig. 2.7

Second geometrical interpretation of quantifier $\forall _{\nu }$

Fig. 2.8

Second geometrical interpretation of quantifier $\exists _{\mu }^*$

Fig. 2.9

Second geometrical interpretation of quantifier $\forall _{\nu }^*$

Now, we see that we can change the forms of the first two quantifiers to the forms

$$V(\forall x P(x)) = \{ \langle x, \inf _{y \in E} \mu (P(y)), \sup _{y \in E} \nu (P(y)) \rangle | x \in E\rbrace ,$$

$$V(\exists x P(x)) = \{ \langle x, \sup _{y \in E} \mu (P(y)), \inf _{y \in E} \nu (P(y)) \rangle | x \in E\rbrace .$$

Obviously, for every predicate P,

$$V(\forall x P(x)) \subseteq V(\forall _{\mu } x P(x)) \subseteq V(\forall _{\nu } x P(x)) \subseteq V(\exists _{\nu } x P(x))$$

$$\subseteq V(\exists _{\mu } x P(x)) \subseteq V(\exists x P(x))$$

and

$$V(\forall x P(x)) \subseteq V(\forall _{\nu } x P(x)) \subseteq V(\forall _{\nu }^* x P(x)) $$

$$\subseteq V(\exists _{\mu }^* x P(x)) \subseteq V(\exists _{\mu } x P(x)) \subseteq V(\exists x P(x)).$$

Open Problem 10. Which implications satisfy Theorem 2.1.1(d) and (e) in Sect. 2.1?

Now, we can modify the new six operators, so as to change their set form to the form of the first two operators.

Let a be one of the possible values for variable x. Then,

$$V((\forall _{\mu } x P(x)), a) = \langle \inf _{y \in E} \mu (P(y)), \nu (P(a)) \rangle ,$$

$$V((\forall _{\nu } x P(x)), a) = \langle \min (1 - \sup _{y \in E} \nu (P(y)), \mu (P(a))), \sup _{y \in E} \nu (P(y)) \rangle ,$$

$$V((\exists _{\mu } x P(x)), a) = \langle \sup _{y \in E} \mu (P(y)), \min (1 - \sup _{y \in E} \mu (P(y)), \nu (P(a))) \rangle ,$$

$$V((\exists _{\nu } x P(x)), a) = \langle \mu (P(a)), \inf _{y \in E} \nu (P(y)) \rangle ,$$

$$V((\forall _{\nu }^* x P(x)), a)$$

$$= \langle \min (1 - \sup _{y \in E} \nu (P(y)), \mu (P(a))), \min (\sup _{y \in E} \nu (P(y)), 1 - \mu (P(a))) \rangle ,$$

$$V((\exists _{\mu }^* x P(x)), a)$$

$$= \langle \min (\sup _{y \in E} \mu (P(y)), 1 - \nu (P(a))), \min (1 - \sup _{y \in E} \mu (P(y)), \nu (P(a))) \rangle .$$

We finish this section with an example.

Let the universe comprise the members of the European Union and let for each country the degree of government approval and disapproval be known. Let the predicate P(x) be “The government of country x is widely approved by the people of country x”. The first quantifier $\forall$ will give the minimal degree of approval which exists in the countries of the EU, and the maximal degree of disapproval in the countries (not necessarily the same). Conversely, the second operator $\exists$ will give us the maximal degree of approval in one of these countries and the minimal degree of disapproval.

Let us assume that for some reason we do not have complete information about either the approval or disapproval for a fixed country a from the EU (but we have such information about the rest). If we are missing information about the degree of approval for a, then, the third operator $\forall _\mu$ will give us a lower bound for this degree of approval for a. The fifth operator $\exists _\mu$ will give us an upper bound for the degree of approval for a.

Conversely, if we are missing information about the degree of disapproval, the fourth operator will give us $\forall _\nu$ will give us the upper bound and the sixth $\exists _\nu$ will give us the lower bound for the degree of disapproval for a.

The seventh and eighth operators act exactly like the fourth and the fifth operators, respectively, but provide a more precise evaluation for the respective degree.

2.3 Ideas for New Types of Quantifiers

It is well known from the classical logic that for each predicate P with argument x having a finite number of interpretations $a_1, a_2,\ldots , a_n$:

$$V(\forall x P(x)) = V(P(a_1) \wedge P(a_2) \wedge \cdots \wedge P(a_n)), $$

$$V(\exists x P(x)) = V(P(a_1) \vee P(a_2) \vee \cdots \vee P(a_n)). $$

Now, following [11] and having in mind the ideas from Sect. 1.7, we can construct a lot of new quantifiers. For each new pair of conjunction and disjunction, we obtain a pair of quantifiers that have the forms

$$V(\forall _{i,j} P(x)) = V(P(a_1) \wedge _{i,j} P(a_2) \wedge _{i,j} \cdots \wedge _{i,j} P(a_n),$$

$$V(\exists _{i,j} P(x)) = V(P(a_1) \vee _{i,j} P(a_2) \vee _{i,j} \cdots \vee _{i,j} P(a_n),$$

where $i \ (1\le i \le 185)$ and $j \ (1 \le j \le 3)$ are the indices of the respective pair of conjunction and disjunction that generates the new pair of quantifiers.

Obviously, $\forall _{4,1}$ coincides with the standard quantifier $\forall$, and $\exists _{4,1}$ coincides with the standard quantifier $\exists$.

One special case is the following: using implication $\rightarrow _{139}$ and negation $\lnot _1$ we obtain for $a, b, c, d \in [0, 1]$ and $a + b, c + d \le 1$:

$$V(\langle a, b \rangle \vee _{139,3} \langle c, d \rangle ) = \left\langle \frac{a + c}{2}, \frac{b + d}{2} \right\rangle = \langle a, b \rangle \wedge _{139,3} \langle c, d \rangle .$$

Therefore, if for each i: $V(P(x_i)) = \langle a_i, b_i \rangle$, then,

$$V(\forall _{139,3} x P(x)) = \left\langle \frac{\sum \limits _{i=1}^n a_i}{n}, \frac{\sum \limits _{i=1}^n b_i}{n} \right\rangle = V(\exists _{139,3} P(x)).$$

In this case, we check directly, that

$$\lnot _1 \forall _{139,3} x \lnot _1 P(x) = \forall _{139,3} x P(x).$$

Hence, there exists a quantifier’s interpretation for which both quantifiers “$\forall$” and “$\exists$” coincide.

It is very interesting that the topological weight-center operator W (see, e.g. [12]) is an exact analogue of quantifier $\forall _{139,3}$. So, we can denote it as WxP(x)).

The so defined quantifiers give us the possibility to clasify all of them in two groups.

• Global quantifiers: $\forall , \exists , W$,

• Local quantifiers: $\forall _{\mu }, \forall _{\mu }, \forall _{\nu }^*, \exists _{\mu }, \exists _{\nu }, \exists _{\mu }^*, U$.

Open Problem 11. Study in details the behaviour of these quantifiers.