A note on the fuzzy Xor

Łukasik, Radosław

doi:10.2478/amsil-2025-0018

Full Article

1.

Introduction

The connective exclusive or plays an important role in computer programming especially in many encryption algorithms (see e.g. [9], [14], [15]), neural networks ([7], [13]), Quantum Computing ([10], [12]) or other fields ([4], [5], [16], [17]). The authors of [2] and [3] introduced an autonomous definition of the fuzzy Xor connective, which is independent of the other connectives. In the earlier literature we can find a fuzzy Xor operation as a composition of the fuzzy negation, and triangular conorms and norms (see [6], [8], [11], [13]).

In this paper we show what conditions must be met by the fuzzy Xor to be used for cryptographic purposes. We also provide new constructions of fuzzy Xor based on the composition of fuzzy implications and other fuzzy connectives.

2.

Fuzzy connectives

In this article we use the notation $\overset{(property)}{=}$ {\buildrel {\left( {property} \right)}\over =} , which means that this equality holds due to the indicated property.

First, we recall definitions of some fuzzy connectives.

Definition 2.1.

A t-norm is a function T : [0, 1]² → [0, 1] satisfying

(T1)
T(x, y) = T(y, x), x, y ∈ [0, 1];
(T2)
T(x, T(y, z)) = T(T(x, y), z), x, y, z ∈ [0, 1];
(T3)
T(x, z) ≤ T(y, z), x, y, z ∈ [0, 1], x ≤ y;
(T4)
T(x, 1) = x, x ∈ [0, 1].

Definition 2.2.

A t-conorm is a function S : [0, 1]² → [0, 1] satisfying

(S1)
S(x, y) = S(y, x), x, y ∈ [0, 1];
(S2)
S(x, S(y, z)) = S(S(x, y), z), x, y, z ∈ [0, 1];
(S3)
S(x, z) ≤ S(y, z), x, y, z ∈ [0, 1], x ≤ y;
(S4)
S(x, 0) = x, x ∈ [0, 1].

Definition 2.3.

A non-increasing function N : [0, 1] → [0, 1] is called a fuzzy negation if N(0) = 1, N(1) = 0. A fuzzy negation N is called

strict if it is strictly decreasing and continuous;
strong if it is an involution, i.e., N(N(x)) = x, x ∈ [0, 1].

Given a fuzzy negation N, a value e ∈ (0, 1) such that N(e) = e is said to be an equilibrium point of N. If the equilibrium point of N exists (e.g. when N is continuous), then it is unique.

Definition 2.4.

A function I : [0, 1]² → [0, 1] is called a fuzzy implication if it satisfies the following properties:

(I1)
I(x₂, y) ≤ I(x₁, y), x₁, x₂, y ∈ [0, 1], x₁ ≤ x₂;
(I2)
I(x, y₁) ≤ I(x, y₂), x, y₁, y₂ ∈ [0, 1], y₁ ≤ y₂;
(I3)
I(0, 0) = 1;
(I4)
I(1, 1) = 1;
(I5)
I(1, 0) = 0.

The natural negation of a fuzzy implication I is defined by N_I (x) := I(x, 0), x ∈ [0, 1].

A fuzzy implication can have many properties (see [1]) but we will remind here only those that we will need, so for a fuzzy implication I and a fuzzy negation N we have the following properties: (R-CP(N)) $I (x, N (y)) = I (y, N (x)), x, y \in 0, 1];$ I\left( {x,N\left( y \right)} \right) = I\left( {y,N\left( x \right)} \right), \;\;\;x,y \in \left[ {0,1} \right]; (L-CP(N)) $I (N (x), y) = I (N (y), x), x, y \in 0, 1];$ I\left( {N\left( x \right),y} \right) = I\left( {N\left( y \right),x} \right), \;\;\;x,y \in \left[ {0,1} \right]; (NP) $I (1, x) = x, x \in 0, 1];$ I\left( {1,x} \right) = x, \;\;\;x \in \left[ {0,1} \right]; (IP) $I (x, x) = 1, x \in 0, 1] .$ I\left( {x,x} \right) = 1, \;\;\;x \in \left[ {0,1} \right].

B. Bedregal, R.H.S. Reiser and G.P. Dimuro in their two works [2], [3] tried to introduce an intrinsic definition of the connective fuzzy exclusive or.

Definition 2.5.

A function E: [0, 1]² → [0, 1] is called a fuzzy Xor if it satisfies the following properties:

(E0)
E(0, 0) = E(1, 1) = 0, E(1, 0) = E(0, 1) = 1;
(E1)
E(x, y) = E(y, x), x, y ∈ [0, 1];
(E2a)
functions E(0, ・), E(・, 0) are non-decreasing;
(E2b)
functions E(1, ・), E(・, 1) are non-increasing.

Remark 2.6.

Let E: [0, 1]² → [0, 1] satisfy (E0) and (E2b). Then N_E : [0, 1] → [0, 1] given by $N_{E} (x) : = E (x, 1), x \in 0, 1]$ {N_E}\left( x \right): = E\left( {x,1} \right), \;\;\;x \in \left[ {0,1} \right] is a fuzzy negation. When E is a fuzzy Xor, then it is called the natural fuzzy negation of the fuzzy Xor E.

In [2], [3] we can find also the following properties connected with a fuzzy Xor function E:

(E3)
E(x, x) ≠ 1, x ∈ [0, 1];
(E4)
E(0, x) = x, x ∈ [0, 1];
(E5)
E(E(x, y), z) = E(x,E(y, z)), x, y, z ∈ [0, 1];
(E6)
E(x, x) = 0, x ∈ [0, 1];
(E7)
E(E(x, y), y) = x, x, y ∈ [0, 1];
(E9)
N_E is a strict fuzzy negation;
(E10)
N_E is a strong fuzzy negation;
(E13)
E(N_E(x), x) = 1, x ∈ [0, 1];
(E15)
N_E(E(x, y)) = E(x,N_E(y)), x, y ∈ [0, 1].

In this work we will pay special attention to the property (E7), which is related to encryption. The authors of [3] showed that some of the above properties can be obtained from others (e.g. obvious implications (E6) ⇒ (E3), (E10) ⇒ (E9), (E5) ⇒ (E15)) but they missed that some of them are incompatible.

Lemma 2.7.

Let E: [0, 1]² → [0, 1] satisfy (E13) and there exists an equilibrium point of N_E. Then (E3) does not hold.

Proof

Let e ∈ (0, 1) be such that N_E(e) = e. We have $E (e, e) = E (N_{E} (e), e) = 1.$ E\left( {e,e} \right) = E\left( {{N_E}\left( e \right),e} \right) = 1.

Theorem 2.8.

Let E: [0, 1]² → [0, 1] satisfy (E7). Then, the following relations hold:

(i)
(E2b) ⇒ ((E10)∧￢(E15)∧￢(E5));
(ii)
(E1) ⇒ (E13);
(iii)
(E4) ⇔ (E6);
(iv)
(E2a) ⇒ E(x, 0) = x for x ∈ [0, 1];
(v)
(E2b)∧(E13) ⇒ (￢(E3) ∧ ￢(E4) ∧ ￢(E6));
(vi)
(E1)∧(E2b) ⇒ (￢(E2a) ∧ ￢(E3) ∧ ￢(E4) ∧ ￢(E6));
(vii)
(E1)∧(E2a) ⇒ (￢(E2b) ∧ (E3) ∧ (E4) ∧ (E6));
(viii)
((E2a)∧(E2b)∧(E7)) ⇒ (E0).

Proof

(i)
We have $x \overset{(E 7)}{=} E (E (x, 1), 1) = N_{E}^{2} (x), x \in 0, 1] .$ x{\buildrel {\left( {{\rm{E}}7} \right)}\over =} E\left( {E\left( {x,1} \right),1} \right) = N_E^2\left( x \right), \;\;\;x \in \left[ {0,1} \right]. Hence $N_{E}^{2} (0) = 0$ N_E^2\left( 0 \right) = 0 and $N_{E}^{2} (1) = 1$ N_E^2\left( 1 \right) = 1 , so from (E2b) we get N_E(0) = 1 and N_E(1) = 0.

Suppose that (E15) holds. Since N_E is strong, there exists y₀ ∈ (0, 1) such that y₀ = N_E(y₀). Then for x₀ := E(1, y₀) we have $\begin{array}{l} 0 = N_{E} (1) \overset{(E 7)}{=} N_{E} (E (E (1, y_{0}), y_{0})) = N_{E} (E (x_{0}, y_{0})) \\ \overset{(E 15)}{=} E (x_{0}, N_{E} (y_{0})) = E (x_{0}, y_{0}) = E (E (1, y_{0}), y_{0}) \overset{(E 7)}{=} 1, \end{array}$ \matrix{ {0 = {N_E}\left( 1 \right){\buildrel {\left( {{\rm{E}}7} \right)}\over =} {N_E}\left( {E\left( {E\left( {1,{y_0}} \right),{y_0}} \right)} \right) = {N_E}\left( {E\left( {{x_0},{y_0}} \right)} \right)} \hfill \cr {\;\;\;\;{\buildrel {\left( {{\rm{E}}15} \right)}\over =} E\left( {{x_0},{N_E}\left( {{y_0}} \right)} \right) = E\left( {{x_0},{y_0}} \right) = E\left( {E\left( {1,{y_0}} \right),{y_0}} \right){\buildrel {\left( {E7} \right)}\over =} 1,} \hfill \cr } which gives us a contradiction. Hence (E15) and (E5) do not hold.
(ii)
We observe that $E (N_{E} (x), x) = E (E (x, 1), x) \overset{(E 1)}{=} E (E (1, x), x) \overset{(E 7)}{=} 1, x \in 0, 1] .$ E\left( {{N_E}\left( x \right),x} \right) = E\left( {E\left( {x,1} \right),x} \right){\buildrel {\left( {{\rm{E}}1} \right)}\over =} E\left( {E\left( {1,x} \right),x} \right){\buildrel {\left( {{\rm{E}}7} \right)}\over =} 1, \;\;\;x \in \left[ {0,1} \right].
(iii)
We have $E (x, x) \overset{(E 4)}{=} E (E (0, x), x) \overset{(E 7)}{=} 0, x \in 0, 1]$ E\left( {x,x} \right){\buildrel {\left( {{\rm{E}}4} \right)}\over =} E\left( {E\left( {0,x} \right),x} \right){\buildrel {\left( {{\rm{E}}7} \right)}\over =} 0, \;\;\;x \in \left[ {0,1} \right] and $E (0, x) \overset{(E 6)}{=} E (E (x, x), x) \overset{(E 7)}{=} x, x \in 0, 1] .$ E\left( {0,x} \right){\buildrel {\left( {{\rm{E}}6} \right)}\over =} E\left( {E\left( {x,x} \right),x} \right){\buildrel {\left( {{\rm{E}}7} \right)}\over =} x, \;\;\;x \in \left[ {0,1} \right].
(iv)
Let f(x) := E(x, 0), x ∈ [0, 1]. We notice that $f^{2} (x) = E (E (x, 0), 0) \overset{(E 7)}{=} x, x \in 0, 1],$ {f^2}\left( x \right) = E\left( {E\left( {x,0} \right),0} \right){\buildrel {\left( {{\rm{E}}7} \right)}\over =} x, \;\;\;x \in \left[ {0,1} \right], so since f is non-decreasing, f(x) = x, x ∈ [0, 1].
(v)
From (i) and Lemma 2.7 we know that (E3) does not hold, so (E6) does not hold. Hence from (iii) (E4) does not hold.
(vi)
From (ii) and (v) we obtain that (E3), (E4), (E6) do not hold.

Suppose that (E2a) holds. We have $E (0, x) \overset{(E 1)}{=} E (x, 0) \overset{(iv)}{=} x, x \in 0, 1]$ E\left( {0,x} \right){\buildrel {\left( {{\rm{E}}1} \right)}\over =} E\left( {x,0} \right){\buildrel {\left( {{\rm{iv}}} \right)}\over =} x, \;\;\;x \in \left[ {0,1} \right] , so (E4) holds and we obtain the contradiction.
(vii)
We have $E (0, x) \overset{(E 1)}{=} E (x, 0) \overset{(iv)}{=} x, x \in 0, 1]$ E\left( {0,x} \right){\buildrel {\left( {{\rm{E}}1} \right)}\over =} E\left( {x,0} \right){\buildrel {\left( {{\rm{iv}}} \right)}\over =} x, \;\;\;x \in \left[ {0,1} \right] , so (E4) holds. From (iii) (E6) (and (E3)) holds. From (vi) (E2b) does not hold.
(viii)
We have $\begin{array}{l} 0 \overset{(E 7)}{=} E (E (0, 0), 0) \overset{(E2a)}{\geq} E (0, 0), \\ 1 \overset{(E 7)}{=} E (E (1, 0), 0) \overset{(E2a)}{\leq} E (1, 0), \\ 0 \overset{(E 7)}{=} E (E (1, 1), 1) \overset{(E2b)}{\leq} E (0, 1), \\ 0 \overset{(E 7)}{=} E (E (0, 1), 1) \overset{(E2b)}{\geq} E (1, 1) . \end{array}$ \matrix{ {0{\buildrel {\left( {{\rm{E}}7} \right)}\over =} E\left( {E\left( {0,0} \right),0} \right) {{\buildrel {\left( {{\rm{E2a}}} \right)}\over \ge }} E\left( {0,0} \right),} \hfill \cr {1{\buildrel {\left( {{\rm{E}}7} \right)}\over =} E\left( {E\left( {1,0} \right),0} \right) {{\buildrel {\left( {{\rm{E2a}}} \right)}\over \le }} E\left( {1,0} \right),} \hfill \cr {0{\buildrel {\left( {{\rm{E}}7} \right)}\over =} E\left( {E\left( {1,1} \right),1} \right) {{\buildrel {\left( {{\rm{E2b}}} \right)}\over \le }} E\left( {0,1} \right),} \hfill \cr {0{\buildrel {\left( {{\rm{E}}7} \right)}\over = } E\left( {E\left( {0,1} \right),1} \right) {{\buildrel {\left( {{\rm{E2b}}} \right)}\over \ge }} E\left( {1,1} \right).} \hfill \cr }

Remark 2.9.

Let

(E1-0)
E(x, 0) = E(0, x), x ∈ [0, 1];
(E1-1)
E(x, 1) = E(1, x), x ∈ [0, 1].

In the previous theorem instead of (E1) we can assume in (ii) that (E1-1) holds, in (vi) and (vii) that (E1-0), (E1-1) hold. Moreover, we have (E2a)∧(E4) ⇒ (E1-0).

Remark 2.10.

There is no function that satisfies (E0), (E1), (E2a), (E2b) and (E7). Since conditions (E2a) and (E2b) are similar, then among functions satisfying (E7) it is better to look for these which do not satisfy (E1).

Remark 2.11.

Let E: [0, 1]² → [0, 1] satisfy (E2a), (E2b) and (E7). Then we have the following possibilities:

E satisfies (E3), (E4), (E6) and does not satisfy (E13).
E satisfies (E13) and does not satisfy (E3), (E4), (E6).
E does not satisfy (E3), (E4), (E6), (E13).

All of the above possibilities can occur, as illustrated by the following examples.

Example 2.12.

Let N₁, N₂ : [0, 1] → [0, 1] be fuzzy negations, N₁ be strong, E: [0, 1]² → [0, 1] be a function given by $E (x, y) = \begin{array}{l} N_{2} (y) + (1 - N_{2} (y)) N_{1} (\frac{x - N_{2} (y)}{1 - N_{2} (y)}), & x \geq N_{2} (y) \neq 1, \\ x, & x < N_{2} (y) \lor N_{2} (y) = 1. \end{array}$ E\left( {x,y} \right) = \left\{ {\matrix{ {{N_2}\left( y \right) + \left( {1 - {N_2}\left( y \right)} \right){N_1}\left( {{{x - {N_2}\left( y \right)} \over {1 - {N_2}\left( y \right)}}} \right),} \hfill & {x \ge {N_2}\left( y \right) \ne 1,} \hfill \cr {x,} \hfill & {x < {N_2}\left( y \right) \vee {N_2}\left( y \right) = 1.} \hfill \cr } } \right. We will show that E satisfies (E2a), (E2b), (E7) and does not satisfy (E3), (E4), (E6). Moreover (E13) holds iff N₁ = N₂. We have $E (x, 0) = x, x \in 0, 1], E (0, y) = \begin{array}{l} 1, & N_{2} (y) = 0, \\ 0, & N_{2} (y) \neq 0, \end{array}$ E\left( {x,0} \right) = x, \;\;\;x \in \left[ {0,1} \right], \;\;\;E\left( {0,y} \right) = \left\{ {\matrix{ {1,} \hfill & {{N_2}\left( y \right) = 0,} \hfill \cr {0,} \hfill & {{N_2}\left( y \right) \ne 0,} \hfill \cr } } \right. so (E2a) holds and (E4) does not hold, $\begin{array}{l} N_{E} (x) = E (x, 1) = N_{1} (x), & x \in 0, 1], \\ E (1, y) = N_{2} (y), & y \in 0, 1], \end{array}$ \matrix{ {{N_E}\left( x \right) = E\left( {x,1} \right) = {N_1}\left( x \right),} \hfill & {x \in \left[ {0,1} \right],} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;E\left( {1,y} \right) = {N_2}\left( y \right),} \hfill & {\;y \in \left[ {0,1} \right],} \hfill \cr } so (E2b) holds. Finally, for x, y ∈ [0, 1] we have

if x ≥ N₂(y) ≠ 1, then E(x, y) ≥ N₂(y) and $\begin{array}{l} E (E (x, y), y) & = N_{2} (y) + (1 - N_{2} (y)) N_{1} (\frac{E (x, y) - N_{2} (y)}{1 - N_{2} (y)}) \\ = N_{2} (y) + (1 - N_{2} (y)) N_{1} (\frac{N_{2} (y) + (1 - N_{2} (y)) N_{1} (\frac{x - N_{2} (y)}{1 - N_{2} (y)}) - N_{2} (y)}{1 - N_{2} (y)}) \\ = N_{2} (y) + (1 - N_{2} (y)) N_{1} (N_{1} (\frac{x - N_{2} (y)}{1 - N_{2} (y)})) \\ = N_{2} (y) + (1 - N_{2} (y)) \frac{x - N_{2} (y)}{1 - N_{2} (y)} = x, \end{array}$ \matrix{ {E\left( {E\left( {x,y} \right),y} \right)} \hfill & { = {N_2}\left( y \right) + \left( {1 - {N_2}\left( y \right)} \right){N_1}\left( {{{E\left( {x,y} \right) - {N_2}\left( y \right)} \over {1 - {N_2}\left( y \right)}}} \right)} \hfill \cr {} \hfill & { = {N_2}\left( y \right) + \left( {1 - {N_2}\left( y \right)} \right){N_1}\left( {{{{N_2}\left( y \right) + \left( {1 - {N_2}\left( y \right)} \right){N_1}\left( {{{x - {N_2}\left( y \right)} \over {1 - {N_2}\left( y \right)}}} \right) - {N_2}\left( y \right)} \over {1 - {N_2}\left( y \right)}}} \right)} \hfill \cr {} \hfill & { = {N_2}\left( y \right) + \left( {1 - {N_2}\left( y \right)} \right){N_1}\left( {{N_1}\left( {{{x - {N_2}\left( y \right)} \over {1 - {N_2}\left( y \right)}}} \right)} \right)} \hfill \cr {} \hfill & { = {N_2}\left( y \right) + \left( {1 - {N_2}\left( y \right)} \right){{x - {N_2}\left( y \right)} \over {1 - {N_2}\left( y \right)}} = x,} \hfill \cr }
if x < N₂(y) ∨ N₂(y) = 1, then E(x, y) = x and $E (E (x, y), y) = E (x, y) = x,$ E\left( {E\left( {x,y} \right),y} \right) = E\left( {x,y} \right) = x,

so (E7) holds. From Theorem 2.8 (iii) (E6) and (E3) do not hold.

Now, we notice that $E (N_{2} (y), y) = 1,$ E\left( {{N_2}\left( y \right),y} \right) = 1, so (E13) ⇔ N₁ = N₂.

Example 2.13.

Let N₁, N₂ : [0, 1] → [0, 1] be fuzzy negations, N₁ be strong, N₂ has not an equilibrium point. For y ∈ (0, 1) let A_y := (0, 1) \ {y, N₂(y)} and φ_y : A_y → A_y be a bijective involution. Let further E: [0, 1]² → [0, 1] be a function given by $E (x, y) = \begin{array}{l} φ_{y} (x), & y \in (0, 1), x \in A_{y}, \\ y, & x = 0, \\ N_{2} (y), & x = 1, \\ 0, & x = y \in (0, 1), \\ 1, & x = N_{2} (y) \in (0, 1), \\ x, & y = 0, x \in (0, 1), \\ N_{1} (x), & y = 1, x \in (0, 1) . \end{array}$ E\left( {x,y} \right) = \left\{ {\matrix{ {{\varphi _y}\left( x \right),} \hfill & {y \in \left( {0,1} \right), \;\;\;x \in {A_y},} \hfill \cr {y,} \hfill & {x = 0,} \hfill \cr {{N_2}\left( y \right),} \hfill & {x = 1,} \hfill \cr {0,} \hfill & {x = y \in \left( {0,1} \right),} \hfill \cr {1,} \hfill & {x = {N_2}\left( y \right) \in \left( {0,1} \right),} \hfill \cr {x,} \hfill & {y = 0, \;\;\;x \in \left( {0,1} \right),} \hfill \cr {{N_1}\left( x \right),} \hfill & {y = 1, \;\;\;x \in \left( {0,1} \right).} \hfill \cr } } \right. We will show that E satisfies (E2a), (E2b), (E3), (E4), (E6), (E7) and does not satisfy (E13).

First, we observe that $\begin{array}{l} E (E (x, y), y) & = \begin{array}{l} E (φ_{y} (x), y), & y \in (0, 1), x \in A_{y}, \\ E (y, y), & x = 0, \\ E (N_{2} (y), y), & x = 1, \\ E (0, y), & x = y \in (0, 1), \\ E (1, y), & x = N_{2} (y) \in (0, 1), \\ E (x, y), & y = 0, x \in (0, 1), \\ E (N_{1} (x), y), & y = 1, x \in (0, 1) . \end{array} \\ = \begin{array}{l} φ_{y} (φ_{y} (x)), & y \in (0, 1), x \in A_{y}, \\ 0, & x = 0, \\ 1, & x = 1, \\ y, & x = y \in (0, 1), = x, x, y \in 0, 1], \\ N_{2} (y), & x = N_{2} (y) \in (0, 1), \\ x, & y = 0, x \in (0, 1), \\ N_{1} (N_{1} (x)), & y = 1, x \in (0, 1) . \end{array} \end{array}$ \matrix{ {E\left( {E\left( {x,y} \right),y} \right)} \hfill & { = \left\{ {\matrix{ {E\left( {{\varphi _y}\left( x \right),y} \right),} \hfill & {y \in \left( {0,1} \right), \;\;\;x \in {A_y},} \hfill \cr {E\left( {y,y} \right),} \hfill & {x = 0,} \hfill \cr {E\left( {{N_2}\left( y \right),y} \right),} \hfill & {x = 1,} \hfill \cr {E\left( {0,y} \right),} \hfill & {x = y \in \left( {0,1} \right),} \hfill \cr {E\left( {1,y} \right),} \hfill & {x = {N_2}\left( y \right) \in \left( {0,1} \right),} \hfill \cr {E\left( {x,y} \right),} \hfill & {y = 0, \;\;\;x \in \left( {0,1} \right),} \hfill \cr {E\left( {{N_1}\left( x \right),y} \right),} \hfill & {y = 1, \;\;\;x \in \left( {0,1} \right).} \hfill \cr } } \right.} \hfill \cr {} \hfill & { = \left\{ {\matrix{ {{\varphi _y}\left( {{\varphi _y}\left( x \right)} \right),} \hfill & {y \in \left( {0,1} \right), \;\;\;x \in {A_y},} \hfill \cr {0,} \hfill & {x = 0,} \hfill \cr {1,} \hfill & {x = 1,} \hfill \cr {y,} \hfill & {x = y \in \left( {0,1} \right), \;\;\;\;\;\;\;\; = x, \;\;\;\;\;x,y \in \left[ {0,1} \right],} \hfill \cr {{N_2}\left( y \right),} \hfill & {x = {N_2}\left( y \right) \in \left( {0,1} \right),} \hfill \cr {x,} \hfill & {y = 0, \;\;\;x \in \left( {0,1} \right),} \hfill \cr {{N_1}\left( {{N_1}\left( x \right)} \right),} \hfill & {y = 1, \;\;\;x \in \left( {0,1} \right).} \hfill \cr } } \right.} \hfill \cr } so E satisfies (E7). We have also $\begin{array}{l} E (x, 0) = x, x \in 0, 1], \\ E (0, y) = y, y \in 0, 1], \\ E (x, 1) = N_{1} (x), x \in 0, 1], \\ E (1, y) = N_{2} (y), y \in 0, 1], \end{array}$ \matrix{ {E\left( {x,0} \right) = x, \;\;\;x \in \left[ {0,1} \right],} \hfill \cr {E\left( {0,y} \right) = y, \;\;\;y \in \left[ {0,1} \right],} \hfill \cr {E\left( {x,1} \right) = {N_1}\left( x \right), \;\;\;x \in \left[ {0,1} \right],} \hfill \cr {E\left( {1,y} \right) = {N_2}\left( y \right), \;\;\;y \in \left[ {0,1} \right],} \hfill \cr } whence (E2a), (E4), (E2b) hold. From Theorem 2.8 we obtain that (E6), (E3) hold and (E13) does not hold.

3.

Fuzzy Xor generated from the other fuzzy connectives

In [2] and [3] we can find some classes of functions: $\begin{array}{l} E_{S, T, N} (x, y) = S (T (x, N (y)), T (N (x), y)), x, y \in 0, 1]; \\ E_{T, S, N} (x, y) = T (S (x, y), S (N (x), N (y))), x, y \in 0, 1]; \\ E_{S, T} (x, y) = S (x, y) - T (x, y), x, y \in 0, 1]; \end{array}$ \matrix{ {{E_{S,T,N}}\left( {x,y} \right) = S\left( {T\left( {x,N\left( y \right)} \right),T\left( {N\left( x \right),y} \right)} \right), \;\;\;x,y \in \left[ {0,1} \right];} \hfill \cr {{E_{T,S,N}}\left( {x,y} \right) = T\left( {S\left( {x,y} \right),S\left( {N\left( x \right),N\left( y \right)} \right)} \right), \;\;\;x,y \in \left[ {0,1} \right];} \hfill \cr {{E_{S,T}}\left( {x,y} \right) = S\left( {x,y} \right) - T\left( {x,y} \right), \;\;\;x,y \in \left[ {0,1} \right];} \hfill \cr } where T : [0, 1]² → [0, 1] is a t-norm, S : [0, 1]² → [0, 1] is a t-conorm, and N : [0, 1] → [0, 1] is a fuzzy negation. Each of them is a fuzzy Xor satisfying (E4). We would like to present others classes which may not have this property.

3.1.

Fuzzy Xor generated from fuzzy implications and negation

In a classical logic we can obtain a Xor in the following way $α \oplus β = ((α \Rightarrow β) \Rightarrow \neg (β \Rightarrow α)) .$ \alpha \oplus \beta = \left( {\left( {\alpha \Rightarrow \beta } \right) \Rightarrow \neg \left( {\beta \Rightarrow \alpha } \right)} \right). Hence, we can define the following class of functions.

Definition 3.1.

Let I : [0, 1]² → [0, 1] be a fuzzy implication, N : [0, 1] → [0, 1] be a fuzzy negation. We define the function E_I,N : [0, 1]² → [0, 1] by the formula (1) $E_{I, N} (x, y) = I (I (x, y), N (I (y, x))), x, y \in 0, 1] .$ {E_{I,N}}\left( {x,y} \right) = I\left( {I\left( {x,y} \right),N\left( {I\left( {y,x} \right)} \right)} \right), \;\;\;x,y \in \left[ {0,1} \right].

We can generalize the above definition in the following way:

Definition 3.2.

Let I₁, I₂ : [0, 1]² → [0, 1] be fuzzy implications and N : [0, 1] → [0, 1] be a fuzzy negation. We define the function E_I₁,I₂,N : [0, 1]² → [0, 1] by the formula (2) $E_{I_{1}, I_{2}, N} (x, y) = I_{1} (I_{2} (x, y), N (I_{2} (y, x))), x, y \in 0, 1] .$ {E_{{I_1},{I_2},N}}\left( {x,y} \right) = {I_1}\left( {{I_2}\left( {x,y} \right),N\left( {{I_2}\left( {y,x} \right)} \right)} \right), \;\;\;x,y \in \left[ {0,1} \right].

Remark 3.3.

Let S : [0, 1]² → [0, 1] be a t-conorm, T : [0, 1]² → [0, 1] be a t-norm, N : [0, 1] → [0, 1] be a fuzzy negation. Let further $\tilde{N} : 0, 1 \to[0, 1]$ \tilde N:\left[ {0,1\left] \to \right[0,1} \right] be any strong negation. We define $\begin{array}{l} I_{1} (x, y) : = S (\tilde{N} (x), y), x, y \in 0, 1], \\ I_{2} (x, y) : = \tilde{N} (T (x, N (y))), x, y \in 0, 1] . \end{array}$ \matrix{ {{I_1}\left( {x,y} \right): = S\left( {\tilde N\left( x \right),y} \right), \;\;\;x,y \in \left[ {0,1} \right],} \hfill \cr {{I_2}\left( {x,y} \right): = \tilde N\left( {T\left( {x,N\left( y \right)} \right)} \right), \;\;\;x,y \in \left[ {0,1} \right].} \hfill \cr } Then $\begin{array}{l} E_{I_{1}, I_{2}, \tilde{N}} (x, y) & = I_{1} (I_{2} (x, y), \tilde{N} (I_{2} (y, x))) \\ = S (\tilde{N} (\tilde{N} (T (x, N (y)))), \tilde{N} (\tilde{N} (T (y, N (x))))) \\ = S (T (x, N (y)), T (N (x), y)) = E_{S, T, N} (x, y), x, y \in 0, 1] . \end{array}$ \matrix{ {{E_{{I_1},{I_2},\tilde N}}\left( {x,y} \right)} \hfill & { = {I_1}\left( {{I_2}\left( {x,y} \right),\tilde N\left( {{I_2}\left( {y,x} \right)} \right)} \right)} \hfill \cr {} \hfill & { = S\left( {\tilde N\left( {\tilde N\left( {T\left( {x,N\left( y \right)} \right)} \right)} \right),\tilde N\left( {\tilde N\left( {T\left( {y,N\left( x \right)} \right)} \right)} \right)} \right)} \hfill \cr {} \hfill & { = S\left( {T\left( {x,N\left( y \right)} \right),T\left( {N\left( x \right),y} \right)} \right) = {E_{S,T,N}}\left( {x,y} \right), \;\;\;\;x,y \in \left[ {0,1} \right].} \hfill \cr }

Theorem 3.4.

Let I₁, I₂ : [0, 1]² → [0, 1] be fuzzy implications, N : [0, 1] → [0, 1] be a fuzzy negation. Then for E_I₁,I₂,N given by (2) we have:

(i)
(E0), (E2a), (E2b) hold and (3) $E_{I_{1}, I_{2}, N} (x, 0) = (N_{I_{1}} \circ N_{I_{2}}) (x), x \in 0, 1],$ {E_{{I_1},{I_2},N}}\left( {x,0} \right) = \left( {{N_{{I_1}}} \circ {N_{{I_2}}}} \right)\left( x \right), \;\;\;x \in \left[ {0,1} \right], (4) $E_{I_{1}, I_{2}, N} (0, x) = (g_{I_{1}} \circ N \circ N_{I_{2}}) (x), x \in 0, 1],$ {E_{{I_1},{I_2},N}}\left( {0,x} \right) = \left( {{g_{{I_1}}} \circ N \circ {N_{{I_2}}}} \right)\left( x \right), \;\;\;x \in \left[ {0,1} \right], (5) $E_{I_{1}, I_{2}, N} (x, 1) = (g_{I_{1}} \circ N \circ g_{I_{2}}) (x), x \in 0, 1],$ {E_{{I_1},{I_2},N}}\left( {x,1} \right) = \left( {{g_{{I_1}}} \circ N \circ {g_{{I_2}}}} \right)\left( x \right), \;\;\;x \in \left[ {0,1} \right], (6) $E_{I_{1}, I_{2}, N} (1, x) = (N_{I_{1}} \circ g_{I_{2}}) (x), x \in 0, 1],$ {E_{{I_1},{I_2},N}}\left( {1,x} \right) = \left( {{N_{{I_1}}} \circ {g_{{I_2}}}} \right)\left( x \right), \;\;\;x \in \left[ {0,1} \right], where g_I₁(x) := I₁(1, x), g_I₂(x) := I₂(1, x) for x ∈ [0, 1];
(ii)
if I₁ satisfies (R-CP(N)), then (E1) holds, (E7) does not hold;
(iii)
(E4) holds iff (g_I₁ ◦ N ◦ N_I₂)(x) = x for x ∈ [0, 1];
(iv)
if I₁, I₂ satisfy (R-CP(N)), then N_{E_I₁,I₂,N} ◦ N = N_I₁ ◦ N_I₂;
(v)
if I₁, I₂ satisfy (R-CP(N)), N_I₁, N_I₂, N are strict, then (E9) holds;
(vi)
if I₂ satisfies (IP), then (E6) holds;
(vii)
if I₁ satisfies (R-CP(N)) and I₂ satisfies (IP), then $E_{I_{1}, I_{2}, N} (x, y) = N_{I_{1}} (min (I_{2} (x, y), I_{2} (y, x))), x, y \in 0, 1] .$ {E_{{I_1},{I_2},N}}\left( {x,y} \right) = {N_{{I_1}}}\left( {\min \;\left( {{I_2}\left( {x,y} \right),{I_2}\left( {y,x} \right)} \right)} \right), \;\;\;x,y \in \left[ {0,1} \right].

Proof

(i)
We observe that for x ∈ [0, 1] we have $\begin{array}{l} E_{I_{1}, I_{2}, N} (x, 0) & = I_{1} (I_{2} (x, 0), N (I_{2} (0, x))) = I_{1} (N_{I_{2}} (x), N (1)) \\ = I_{1} (N_{I_{2}} (x), 0) = (N_{I_{1}} \circ N_{I_{2}}) (x), \\ E_{I_{1}, I_{2}, N} (0, x) & = I_{1} (I_{2} (0, x), N (I_{2} (x, 0))) = I_{1} (1, N (N_{I_{2}} (x))) \\ = (g_{I_{1}} \circ N \circ N_{I_{2}}) (x), \\ E_{I_{1}, I_{2}, N} (x, 1) & = I_{1} (I_{2} (x, 1), N (I_{2} (1, x))) = I_{1} (1, N (I_{2} (1, x))) \\ = (g_{I_{1}} \circ N \circ g_{I_{2}}) (x), \\ E_{I_{1}, I_{2}, N} (1, x) & = I_{1} (I_{2} (1, x), N (I_{2} (x, 1))) = I_{1} (I_{2} (1, x), N (1)) \\ = I_{1} (I_{2} (1, x), 0) = (N_{I_{1}} \circ g_{I_{2}}) (x), \end{array}$ \matrix{ {{E_{{I_1},{I_2},N}}\left( {x,0} \right)} \hfill & { = {I_1}\left( {{I_2}\left( {x,0} \right),N\left( {{I_2}\left( {0,x} \right)} \right)} \right)\; = {I_1}\left( {{N_{{I_2}}}\left( x \right),N\left( 1 \right)} \right)} \hfill \cr {} \hfill & { = {I_1}\left( {{N_{{I_2}}}\left( x \right),0} \right) = \left( {{N_{{I_1}}} \circ {N_{{I_2}}}} \right)\left( x \right),} \hfill \cr {{E_{{I_1},{I_2},N}}\left( {0,x} \right)} \hfill & { = {I_1}\left( {{I_2}\left( {0,x} \right),N\left( {{I_2}\left( {x,0} \right)} \right)\;} \right) = {I_1}\left( {1,N\left( {{N_{{I_2}}}\left( x \right)} \right)} \right)} \hfill \cr {} \hfill & { = \left( {{g_{{I_1}}} \circ N \circ {N_{{I_2}}}} \right)\left( x \right),} \hfill \cr {{E_{{I_1},{I_2},N}}\left( {x,1} \right)} \hfill & { = {I_1}\left( {{I_2}\left( {x,1} \right),N\left( {{I_2}\left( {1,x} \right)} \right)} \right) = {I_1}\left( {1,N\left( {{I_2}\left( {1,x} \right)} \right)} \right)} \hfill \cr {} \hfill & { = \left( {{g_{{I_1}}} \circ N \circ {g_{{I_2}}}} \right)\left( x \right),} \hfill \cr {{E_{{I_1},{I_2},N}}\left( {1,x} \right)} \hfill & { = {I_1}\left( {{I_2}\left( {1,x} \right),N\left( {{I_2}\left( {x,1} \right)} \right)} \right)\; = {I_1}\left( {{I_2}\left( {1,x} \right),N\left( 1 \right)} \right)} \hfill \cr {} \hfill & { = {I_1}\left( {{I_2}\left( {1,x} \right),0} \right) = \left( {{N_{{I_1}}} \circ {g_{{I_2}}}} \right)\left( x \right),} \hfill \cr } so we get (E0), (E2a), (E2b).
(ii)
Assume that I₁ satisfies (R-CP(N)). Hence, using the above, we get $\begin{array}{l} E_{I_{1}, I_{2}, N} (y, x) & = I_{1} (I_{2} (y, x), N (I_{2} (x, y))) \\ \overset{(R-CP (N))}{=} I_{1} (I_{2} (x, y), N (I_{2} (y, x))) \\ = E_{I_{1}, I_{2}, N} (x, y), x, y \in 0, 1] . \end{array}$ \matrix{ {{E_{{I_1},{I_2},N}}\left( {y,x} \right)} \hfill & { = {I_1}\left( {{I_2}\left( {y,x} \right),N\left( {{I_2}\left( {x,y} \right)} \right)} \right)} \hfill \cr {} \hfill & {{\buildrel {\left( {{\rm{R - CP}}\left( {\rm{N}} \right)} \right)} \over = } = {I_1}\left( {{I_2}\left( {x,y} \right),N\left( {{I_2}\left( {y,x} \right)} \right)} \right)} \hfill \cr {} \hfill & { = {E_{{I_1},{I_2},N}}\left( {x,y} \right), \;\;\;\;x,y \in \left[ {0,1} \right].} \hfill \cr }

Using Theorem 2.8 we get that (E7) does not hold.
(iii)
It follows from i.
(iv)
Assume that I₂ satisfies (R-CP(N)). Then we have $\begin{array}{l} N_{E_{I_{1}},_{I_{2}, N}} (N (x)) & \overset{ii}{=} N_{I_{1}} (I_{2} (1, N (x))) \overset{(R-CP (N))}{=} N_{I_{1}} (I_{2} (x, N (1))) \\ = N_{I_{1}} (I_{2} (x, 0)) = (N_{I_{1}} \circ N_{I_{2}}) (x), x \in 0, 1] . \end{array}$ \matrix{ {{N_{{E_{{I_1}}}{,_{{I_2}}}{,_N}}}\left( {N\left( x \right)} \right)} \hfill & {{\buildrel {{{ii}}} \over =}\, {N_{{I_1}}}\left( {{I_2}\left( {1,N\left( x \right)} \right)} \right){\buildrel {\left( {{\rm{R - CP}}\left( {\rm{N}} \right)} \right)} \over = } {N_{{I_1}}}\left( {{I_2}\left( {x,N\left( 1 \right)} \right)} \right)} \hfill \cr {} \hfill & { = {N_{{I_1}}}\left( {{I_2}\left( {x,0} \right)} \right) = \left( {{N_{{I_1}}} \circ {N_{{I_2}}}} \right)\left( x \right), \;\;\;\;x \in \left[ {0,1} \right].} \hfill \cr }
(v)
It follows from iv.
(vi)
Assume that I₂ satisfies (IP). Then $\begin{array}{l} E_{I_{1}, I_{2}, N} (x, x) & = I_{1} (I_{2} (x, x), N (I_{2} (x, x))) \\ \overset{(IP)}{=} I_{1} (1, N (1)) = I_{1} (1, 0) = 0, x \in 0, 1], \end{array}$ \matrix{ {{E_{{I_1},{I_2},N}}\left( {x,x} \right)} \hfill & { = {I_1}\left( {{I_2}\left( {x,x} \right),N\left( {{I_2}\left( {x,x} \right)} \right)} \right)} \hfill \cr {} \hfill & {{\buildrel {\left( {{\rm{IP}}} \right)}\over =} {I_1}\left( {1,N\left( 1 \right)} \right) = {I_1}\left( {1,0} \right) = 0, \;\;\;\;x \in \left[ {0,1} \right],} \hfill \cr } so (E6) holds.
(vii)
Let x, y ∈ [0, 1]. From ii we have that E_I₁,I₂,N is symmetric, so we can assume that y ≤ x whence from (IP) for I₂ we get I₂(y, x) = 1 and $\begin{array}{l} E_{I_{1}, I_{2}, N} (x, y) & = I_{1} (I_{2} (x, y), N (I_{2} (y, x))) \\ = I_{1} (I_{2} (x, y), N (1)) = I_{1} (I_{2} (x, y), 0) = N_{I_{1}} (I_{2} (x, y)) \\ = N_{I_{1}} (min (I_{2} (x, y), I_{2} (y, x))) . \end{array}$ \matrix{ {{E_{{I_1},{I_2},N}}\left( {x,y} \right)} \hfill & { = {I_1}\left( {{I_2}\left( {x,y} \right),N\left( {{I_2}\left( {y,x} \right)} \right)} \right)} \hfill \cr {} \hfill & { = {I_1}\left( {{I_2}\left( {x,y} \right),N\left( 1 \right)} \right) = {I_1}\left( {{I_2}\left( {x,y} \right),0} \right) = {N_{{I_1}}}\left( {{I_2}\left( {x,y} \right)} \right)} \hfill \cr {} \hfill & { = {N_{{I_1}}}\left( {\min \;\left( {{I_2}\left( {x,y} \right),{I_2}\left( {y,x} \right)} \right)} \right).} \hfill \cr }

Lemma 3.5.

Let I : [0, 1]² → [0, 1] be a fuzzy implication, N : [0, 1] → [0, 1] be a fuzzy negation. Assume that E_I,N is given by (1). Then (E7) does not hold.

Proof

Suppose that (E7) holds. First, observe that N_I is strong. Let x ∈ [0, 1]. Using Theorem 2.8 (iv) we have $x = E_{I, N} (x, 0) \overset{(3)}{=} N_{I} (N_{I} (x)),$ x = {E_{I,N}}\left( {x,0} \right){\buildrel {\left( 3 \right)}\over =} {N_I}\left( {{N_I}\left( x \right)} \right), so N_I is strong.

Next, for x ∈ [0, 1] we notice that $N_{E_{I, N}} (x) = E_{I, N} (x, 1) \overset{(5)}{=} (g_{I} \circ N \circ g_{I}) (x) .$ {N_{{E_{I,N}}}}\left( x \right) = {E_{I,N}}\left( {x,1} \right){\buildrel {\left( 5 \right)}\over =} \left( {{g_I} \circ N \circ {g_I}} \right)\left( x \right). Since g_I is non-decreasing, N_{E_I,N} is a strong fuzzy negation, then g_I is a bijection and $N = g_{I}^{- 1} \circ N_{E_{I, N}} \circ g_{I}^{- 1}$ N = g_I^{ - 1} \circ {N_{{E_{I,N}}}} \circ g_I^{ - 1} is strict.

The function $g_{I}^{- 1} \circ N_{I} \circ g_{I} \circ N \circ N_{I}$ g_I^{ - 1} \circ {N_I} \circ {g_I} \circ N \circ {N_I} is a strict fuzzy negation, so it has an equilibrium point e ∈ (0, 1). Hence $(g_{I} \circ N \circ N_{I}) (e) = (N_{I} \circ g_{I}) (e) .$ \left( {{g_I} \circ N \circ {N_I}} \right)\left( e \right) = \left( {{N_I} \circ {g_I}} \right)\left( e \right). Using the previous theorem we get $\begin{array}{l} 1 \overset{(E 7)}{=} E (E (1, e), e) \overset{(6)}{=} E ((N_{I} \circ g_{I}) (e), e) \\ = E ((g_{I} \circ N \circ N_{I}) (e), e) \overset{(4)}{=} E (E (0, e), e) \overset{(E 7)}{=} 0, \end{array}$ \matrix{ {1{\buildrel {\left( {{\rm{E}}7} \right)}\over =} E\left( {E\left( {1,e} \right),e} \right){\buildrel {\left( 6 \right)}\over =} E\left( {\left( {{N_I} \circ {g_I}} \right)\left( e \right),e} \right)} \hfill \cr \;\;\;{ = E\left( {\left( {{g_I} \circ N \circ {N_I}} \right)\left( e \right),e} \right){\buildrel {\left( 4 \right)}\over =} E\left( {E\left( {0,e} \right),e} \right){\buildrel {\left( {{\rm{E}}7} \right)}\over =} 0,} \hfill \cr } which gives us a contradiction.

Problem 3.6.

Do exist fuzzy implications I₁,I₂ and a negation N such that E_I₁,I₂,N satisfies (E7)?

3.2.

Fuzzy Xor generated from t-norm, fuzzy implication and negation

In a classical logic we can obtain a Xor in the following way $α \oplus β = ((α \Rightarrow \neg β) \land (\neg α \Rightarrow β)) .$ \alpha \oplus \beta = \left( {\left( {\alpha \Rightarrow \neg \beta } \right) \wedge \left( {\neg \alpha \Rightarrow \beta } \right)} \right). Hence, we can define the following class of functions.

Definition 3.7.

Let T : [0, 1]² → [0, 1] be a t-norm, I : [0, 1]² → [0, 1] be a fuzzy implication, N : [0, 1] → [0, 1] be a fuzzy negation. We define the function E_T,I,N : [0, 1]² → [0, 1] by the formula (7) $E_{T, I, N} (x, y) = T (I (x, N (y)), I (N (x), y)), x, y \in 0, 1] .$ {E_{T,I,N}}\left( {x,y} \right) = T\left( {I\left( {x,N\left( y \right)} \right),I\left( {N\left( x \right),y} \right)} \right), \;\;\;x,y \in \left[ {0,1} \right].

Remark 3.8.

Let S : [0, 1]² → [0, 1] be a t-conorm, T : [0, 1]² → [0, 1] be a t-norm, N : [0, 1] → [0, 1] be a strong fuzzy negation. We define $I (x, y) : = S (N (x), y), x, y \in 0, 1] .$ I\left( {x,y} \right): = S\left( {N\left( x \right),y} \right), \;\;\;x,y \in \left[ {0,1} \right]. Then $\begin{array}{l} E_{T, I, N} (x, y) & = T (I (x, N (y)), I (N (x), y)) \\ = T (S (N (x), N (y)), S (N^{2} (x), y)) \\ = T (S (N (x), N (y)), S (x, y)) = E_{T, S, N} (x, y), x, y \in 0, 1] . \end{array}$ \matrix{ {{E_{T,I,N}}\left( {x,y} \right)} \hfill & { = T\left( {I\left( {x,N\left( y \right)} \right),I\left( {N\left( x \right),y} \right)} \right)} \hfill \cr {} \hfill & { = T\left( {S\left( {N\left( x \right),N\left( y \right)} \right),S\left( {{N^2}\left( x \right),y} \right)} \right)} \hfill \cr {} \hfill & { = T\left( {S\left( {N\left( x \right),N\left( y \right)} \right),S\left( {x,y} \right)} \right) = {E_{T,S,N}}\left( {x,y} \right), \;\;\;x,y \in \left[ {0,1} \right].} \hfill \cr }

Theorem 3.9.

Let T : [0, 1]² → [0, 1] be a t-norm, I : [0, 1]² → [0, 1] be a fuzzy implication, N : [0, 1] → [0, 1] be a fuzzy negation. Then for E_T,I,N given by (7) we have:

(i)
(E0), (E2a), (E2b) hold and (8) $E_{T, I, N} (x, 0) = (N_{I} \circ N) (x), x \in 0, 1],$ {E_{T,I,N}}\left( {x,0} \right) = \left( {{N_I} \circ N} \right)\left( x \right), \;\;\;x \in \left[ {0,1} \right], (9) $E_{T, I, N} (0, x) = g_{I} (x), x \in 0, 1],$ {E_{T,I,N}}\left( {0,x} \right) = {g_I}\left( x \right), \;\;\;x \in \left[ {0,1} \right], (10) $E_{T, I, N} (x, 1) = N_{I} (x), x \in 0, 1],$ {E_{T,I,N}}\left( {x,1} \right) = {N_I}\left( x \right), \;\;\;x \in \left[ {0,1} \right], (11) $E_{T, I, N} (1, x) = (g_{I} \circ N) (x), x \in 0, 1],$ {E_{T,I,N}}\left( {1,x} \right) = \left( {{g_I} \circ N} \right)\left( x \right), \;\;\;x \in \left[ {0,1} \right], where g_I (x) := I(1, x) for x ∈ [0, 1];
(ii)
if I satisfies (R-CP(N)) and (L-CP(N)), then (E1) holds;
(iii)
I satisfies (NP) iff (E4) holds;
(iv)
N_I is strict (strong) iff (E9) ((E10)) holds;
(v)
if I satisfies (IP) and N has an equilibrium point, then (E3) does not hold;
(vi)
(E7) does not hold.

Proof

(i)
We observe that for x ∈ [0, 1] we have $\begin{array}{l} E_{T, I, N} (x, 0) & = T (I (x, N (0)), I (N (x), 0)) = T (I (x, 1), (N_{I} \circ N) (x)) \\ = T (1, (N_{I} \circ N) (x)) = (N_{I} \circ N) (x), \\ E_{T, I, N} (0, x) & = T (I (0, N (x)), I (N (0), x))) = T (1, I (1, x))) \\ = I (1, x) = g_{I} (x), \\ E_{T, I, N} (x, 1) & = T (I (x, N (1)), I (N (x), 1)) \\ = T (I (x, 0), 1) = I (x, 0) = N_{I} (x), \\ E_{T, I, N} (1, x) & = T (I (1, N (x)), I (N (1), x)) = T ((g_{I} \circ N) (x), I (0, x)) \\ = T ((g_{I} \circ N) (x), 1) = (g_{I} \circ N) (x), \end{array}$ \matrix{ {{E_{T,I,N}}\left( {x,0} \right)} \hfill & { = T\left( {I\left( {x,N\left( 0 \right)} \right),I\left( {N\left( x \right),0} \right)} \right) = T\left( {I\left( {x,1} \right),\left( {{N_I} \circ N} \right)\left( x \right)} \right)} \hfill \cr {} \hfill & { = T\left( {1,\left( {{N_I} \circ N} \right)\left( x \right)} \right) = \left( {{N_I} \circ N} \right)\left( x \right),} \hfill \cr {{E_{T,I,N}}\left( {0,x} \right)} \hfill & { = T\left( {I\left( {0,N\left( x \right)} \right),I\left( {N\left( 0 \right),x)} \right)} \right) = T\left( {1,I\left( {1,x)} \right)} \right)} \hfill \cr {} \hfill & { = I\left( {1,x} \right) = {g_I}\left( x \right),} \hfill \cr {{E_{T,I,N}}\left( {x,1} \right)} \hfill & { = T\left( {I\left( {x,N\left( 1 \right)} \right),I\left( {N\left( x \right),1} \right)} \right)} \hfill \cr {} \hfill & { = T\left( {I\left( {x,0} \right),1} \right) = I\left( {x,0} \right) = {N_I}\left( x \right),} \hfill \cr {{E_{T,I,N}}\left( {1,x} \right)} \hfill & { = T\left( {I\left( {1,N\left( x \right)} \right),I\left( {N\left( 1 \right),x} \right)} \right) = T\left( {\left( {{g_I} \circ N} \right)\left( x \right),I\left( {0,x} \right)} \right)} \hfill \cr {} \hfill & { = T\left( {\left( {{g_I} \circ N} \right)\left( x \right),1} \right) = \left( {{g_I} \circ N} \right)\left( x \right),} \hfill \cr } so we get (E0), (E2a), (E2b).
(ii)
Assume that I satisfies (R-CP(N)) and (L-CP(N)). Hence $\begin{array}{l} E_{T, I, N} (y, x) = T (I (y, N (x)), I (N (y), x)) \\ \overset{(R-CP (N))}{=} T (I (x, N (y)), I (N (y), x)) \\ \overset{(L-CP (N))}{=} T (I (x, N (y)), I (N (x), y)) = E_{T, I, N} (x, y), x, y \in 0, 1] . \end{array}$ \matrix{ {{E_{T,I,N}}\left( {y,x} \right) = T\left( {I\left( {y,N\left( x \right)} \right),I\left( {N\left( y \right),x} \right)} \right)} \hfill \cr {\;{\buildrel {\left( {{\rm{R - CP}}\left( {\rm{N}} \right)} \right)} \over = } T\left( {I\left( {x,N\left( y \right)} \right),I\left( {N\left( y \right),x} \right)} \right)} \hfill \cr {\;{\buildrel {\left( {{\rm{L - CP}}\left( {\rm{N}} \right)} \right)} \over = } T\left( {I\left( {x,N\left( y \right)} \right),I\left( {N\left( x \right),y} \right)} \right) = {E_{T,I,N}}\left( {x,y} \right), \;\;\;x,y \in \left[ {0,1} \right].} \hfill \cr }
(iii)
It follows from (9).
(iv)
It follows from (10).
(v)
Assume that I satisfies (IP) and N has an equilibrium point e ∈ (0, 1). Then $\begin{array}{l} E_{T, I, N} (e, e) & = T (I (e, N (e)), I (N (e), e)) = T (I (e, e), I (e, e)) \\ \overset{(IP)}{=} T (1, 1) = 1, x \in 0, 1] . \end{array}$ \matrix{ {{E_{T,I,N}}\left( {e,e} \right)} \hfill & { = T\left( {I\left( {e,N\left( e \right)} \right),I\left( {N\left( e \right),e} \right)} \right) = T\left( {I\left( {e,e} \right),I\left( {e,e} \right)} \right)} \hfill \cr {} \hfill & {{\buildrel {\left( {{\rm{IP}}} \right)}\over =} T\left( {1,1} \right) = 1, \;\;\;\;x \in \left[ {0,1} \right].} \hfill \cr }
(vi)
Suppose that (E7) holds. From Theorem 2.8 N_{E_T,I,N} is strong, so from iv N_I is strong. From (8) we get $x \overset{(E 7)}{=} E_{T, I, N} (E_{T, I, N} (x, 0), 0) \overset{(8)}{=} {(N_{I} \circ N)}^{2} (x), x \in 0, 1] .$ x{\buildrel {\left( {{\rm{E}}7} \right)}\over =} {E_{T,I,N}}\left( {{E_{T,I,N}}\left( {x,0} \right),0} \right) {\buildrel {\left( 8 \right)}\over =} {\left( {{N_I} \circ N} \right)^2}\left( x \right), \;\;\;x \in \left[ {0,1} \right]. Since N_I ◦N is non-decreasing (N_I ◦N)(x) = x for x ∈ [0, 1], so N = N_I. Let e ∈ (0, 1) be an equilibrium point of N_I. Using i we have $\begin{array}{l} 1 \overset{(E 7)}{=} E_{T, I, N} (E_{T, I, N} (1, e), e) \overset{(9)}{=} E_{T, I, N} ((g_{I} \circ N) (e), e) = E_{T, I, N} (g_{I} (e), e), \\ 0 \overset{(E 7)}{=} E_{T, I, N} (E_{T, I, N} (0, e), e) \overset{(11)}{=} E_{T, I, N} (g_{I} (e), e), \end{array}$ \matrix{ {1{\buildrel {\left( {{\rm{E}}7} \right)}\over =} {E_{T,I,N}}\left( {{E_{T,I,N}}\left( {1,e} \right),e} \right){\buildrel {\left( 9 \right)}\over =} {E_{T,I,N}}\left( {\left( {{g_I} \circ N} \right)\left( e \right),e} \right) = {E_{T,I,N}}\left( {{g_I}\left( e \right),e} \right),} \hfill \cr {0{\buildrel {\left( {{\rm{E}}7} \right)}\over =} {E_{T,I,N}}\left( {{E_{T,I,N}}\left( {0,e} \right),e} \right){\buildrel {\left( {11} \right)}\over =} {E_{T,I,N}}\left( {{g_I}\left( e \right),e} \right),} \hfill \cr } , which gives us a contradiction.

4.

Examples

In [3, Table 2] we can find classification of fuzzy Xor connectives. Adding new classes of fuzzy Xor we can show that all these classes are different.

	E_I,N	E_T,I,N	E_S,T,N	E_T,S,N
E_C (x, y)	(I_RS, N)	-	-	-
E₁(x, y)	-	(T, I_LK, N_C)	(S_LK, T_M, N_C)	(T, S_LK, N_C)
E_⊥(x, y)	(I₁, N)	(T, I₀, N)	-	-
E₂(x, y)	-	-	(S_D, T, N_D₂)	(T, S_D, N_D₂)

where N is an arbitrary fuzzy negation, T is an arbitrary t-norm,

\begin{matrix} \begin{array}{l} E_{C} (x, y) = \begin{array}{l} 0, & x = y, \\ 1, & x \neq y, \end{array} & E_{1} (x, y) = \begin{array}{l} x + y, & x + y \leq 1, \\ 2 - x - y, & x + y > 1, \end{array} \\ E_{⊥} (x, y) = \begin{array}{l} 0, & x - y| = 1, \\ 1, & x - y| \neq 1, \end{array} & E_{2} (x, y) = \begin{array}{l} 0, & x = y = 1, \\ max (x, y), & x = 0 \lor y = 0, \\ 1, & otherwise, \end{array} \\ I_{R S} (x, y) = \begin{array}{l} 0, & x > y, \\ 1, & x \leq y, \end{array} & I_{L K} (x, y) = \begin{array}{l} 1 - x + y, & x > y, \\ 1, & x \leq y, \end{array} \\ I_{1} (x, y) = \begin{array}{l} 0, & x = 1 \land y = 0, \\ 1, & x < 1 \lor y > 0, \end{array} & I_{0} (x, y) = \begin{array}{l} 0, & x > 0 \land y < 1, \\ 1, & x = 0 \lor y = 1, \end{array} \end{array} \\ S_{L K} (x, y) = min (1, x + y), x, y \in 0, 1], \\ S_{D} (x, y) = \begin{array}{l} max (x, y), & x = 0 \lor y = 0, \\ 1, & x > 0 \land y > 0, \end{array} \\ T_{M} (x, y) = min (x, y), x, y \in 0, 1], \\ N_{C} (x) = 1 - x, x \in 0, 1], N_{D 2} (x) = \begin{array}{l} 0, & x = 1, \\ 1, & x < 1 . \end{array} \end{matrix}

\matrix{ {\matrix{ {{E_C}\left( {x,y} \right) = \left\{ {\matrix{ {0,} \hfill & {\;x = y,} \hfill \cr {1,} \hfill & {\;x \ne y,} \hfill \cr } } \right.} \hfill & {{E_1}\left( {x,y} \right) = \left\{ {\matrix{ {x + y,} \hfill & {x + y \le 1,} \hfill \cr {2 - x - y,} \hfill & {x + y > 1,} \hfill \cr } } \right.} \hfill \cr {{E_ \bot }\left( {x,y} \right) = \left\{ {\matrix{ {0,} \hfill & {\left| {x - y} \right| = 1,} \hfill \cr {1,} \hfill & {\left| {x - y} \right| \ne 1,} \hfill \cr } } \right.} \hfill & {{E_2}\left( {x,y} \right) = \left\{ {\matrix{ {0,} \hfill & {x = y = 1,} \hfill \cr {{\rm{max\;}}\left( {x,y} \right),} \hfill & {x = 0 \vee y = 0,} \hfill \cr {1,} \hfill & {{\rm{otherwise}},} \hfill \cr } } \right.} \hfill \cr {{I_{RS}}\left( {x,y} \right) = \left\{ {\matrix{ {0,} \hfill & {\;x > y,} \hfill \cr {1,} \hfill & {\;x \le y,} \hfill \cr } } \right.} \hfill & {{I_{LK}}\left( {x,y} \right) = \left\{ {\matrix{ {1 - x + y,} \hfill & {x > y,} \hfill \cr {1,} \hfill & {x \le y,} \hfill \cr } } \right.} \hfill \cr {{I_1}\left( {x,y} \right) = \left\{ {\matrix{ {0,} \hfill & {\;x = 1 \wedge y = 0,} \hfill \cr {1,} \hfill & {\;x < 1 \vee y > 0,} \hfill \cr } } \right.} \hfill & {{I_0}\left( {x,y} \right) = \left\{ {\matrix{ {0,} \hfill & {x > 0 \wedge y < 1,} \hfill \cr {1,} \hfill & {x = 0 \vee y = 1,} \hfill \cr } } \right.} \hfill \cr } } \cr {{S_{LK}}\left( {x,y} \right) = {\rm{min\;}}\left( {1,x + y} \right), \;\;\;\;x,y \in \left[ {0,1} \right],} \cr {{S_D}\left( {x,y} \right) = \left\{ {\matrix{ {{\rm{max\;}}\left( {x,y} \right),} \hfill & {x = 0 \vee y = 0,} \hfill \cr {1,} \hfill & {x > 0 \wedge y > 0,} \hfill \cr } } \right.} \cr {{T_M}\left( {x,y} \right) = {\rm{\;min\;}}\left( {x,y} \right), \;\;\;\;x,y \in \left[ {0,1} \right],} \cr {{N_C}\left( x \right) = 1 - x, \;\;\;x \in \left[ {0,1} \right], \;\;\;{N_{D2}}\left( x \right) = \left\{ {\matrix{ {0,} \hfill & {x = 1,} \hfill \cr {1,} \hfill & {x < 1.} \hfill \cr } } \right.} \cr }

It is easy to show the form of E if we know the generators, so we show only that the above fuzzy Xors do not belong to some classes. From [3, Table 2] we know that E_C and E_⊥ do not belong to classes E_S,T,N, E_T,S,N.

Suppose that E_C = E_T,I,N. Using (8)–(11) we have $\begin{matrix} (N_{I} \circ N) (x) = g_{I} (x) = \begin{matrix} 0, & x = 0, \\ 1, & x > 0, \end{matrix} & N_{I} (x) = (g_{I} \circ N) (x) = \begin{array}{l} 0, & x = 1, \\ 1, & x < 1. \end{array} \end{matrix}$ \matrix{ {\left( {{N_I} \circ N} \right)\left( x \right) = {g_I}\left( x \right) = \left\{ {\matrix{ {0,} & {\;x = 0,} \cr {1,} & {x > 0,} \cr } } \right.} & {{N_I}\left( x \right) = \left( {{g_I} \circ N} \right)\left( x \right) = \left\{ {\matrix{ {0,} \hfill & {x = 1,} \hfill \cr {1,} \hfill & {x < 1.} \hfill \cr } } \right.} \cr } Hence I = I₁. Let x = y ∈ (0, 1). Then I(x,N(y)) = 1 = I(N(x), y) and $0 = E_{C} (x, y) = E_{T, I, N} (x, y) = T (1, 1) = 1,$ 0 = {E_C}\left( {x,y} \right) = {E_{T,I,N}}\left( {x,y} \right) = T\left( {1,1} \right) = 1, which give us a contradiction.
Suppose that E₁ = E_I,N. Using (3)–(6) we have $\begin{matrix} N_{I}^{2} (x) = (g_{I} \circ N \circ N_{I}) (x) = x, x \in 0, 1], \\ (N_{I} \circ g_{I}) (x) = (g_{I} \circ N \circ g_{I}) (x) = 1 - x, x \in 0, 1] . \end{matrix}$ \matrix{ {N_I^2\left( x \right) = \left( {{g_I} \circ N \circ {N_I}} \right)\left( x \right) = x, \;\;\;x \in \left[ {0,1} \right],} \cr {\left( {{N_I} \circ {g_I}} \right)\left( x \right) = \left( {{g_I} \circ N \circ {g_I}} \right)\left( x \right) = 1 - x, \;\;\;x \in \left[ {0,1} \right].} \cr } Hence N_I = g_I ◦ N is strong negation and $x = N_{I}^{2} (x) = (g_{I} \circ N \circ g_{I} \circ N) (x) = 1 - N (x), x \in 0, 1],$ x = N_I^2\left( x \right) = \left( {{g_I} \circ N \circ {g_I} \circ N} \right)\left( x \right) = 1 - N\left( x \right), \;\;\;x \in \left[ {0,1} \right], so N = N_C.

Suppose that I(x₀, y₀) = 1 for some x₀ > 0, y₀ < 1. Then $g_{I} (N_{c} (I (y, x))) = E_{1} (x, y) = \begin{array}{l} x + y, & x + y \leq 1, \\ 2 - x - y, & x + y > 1, \end{array} x < x_{0}, y > y_{0} .$ {g_I}\left( {{N_c}\left( {I\left( {y,x} \right)} \right)} \right)\; = {E_1}\left( {x,y} \right) = \left\{ {\matrix{ {x + y,} \hfill & {x + y \le 1,} \hfill \cr {2 - x - y,} \hfill & {x + y > 1,} \hfill \cr } } \right.\;\;\;x < {x_0},y > {y_0}. The left hand side of the above identity is non-increasing in x and non-decreasing in y which is impossible, so I(x₀, y₀) = 1 only if x₀ = 0 or y₀ = 1.

Now, we observe that $1 = E_{1} (0.5, 0.5) = I (I (0.5, 0.5), 1 - I (0.5, 0.5)), x \in 0, 1],$ 1 = {E_1}\left( {0.5,0.5} \right) = I\left( {I\left( {0.5,0.5} \right),1 - I\left( {0.5,0.5} \right)} \right), \;\;\;x \in \left[ {0,1} \right], so I(0.5, 0.5) = 0, whence $0 < N_{I} (0.5) = I (0.5, 0) \leq I (0.5, 0.5) = 0,$ 0 < {N_I}\left( {0.5} \right) = I\left( {0.5,0} \right) \le I\left( {0.5,0.5} \right) = 0, which give us a contradiction.
Suppose that E₂ = E_I,N. Using (3)–(6) we have $\begin{matrix} N_{I}^{2} (x) = (g_{I} \circ N \circ N_{I}) (x) = x, x \in 0, 1], \\ (N_{I} \circ g_{I}) (x) = (g_{I} \circ N \circ g_{I}) (x) = N_{D 2} (x), x \in 0, 1] . \end{matrix}$ \matrix{ {N_I^2\left( x \right) = \left( {{g_I} \circ N \circ {N_I}} \right)\left( x \right) = x, \;\;\;x \in \left[ {0,1} \right],} \cr {\left( {{N_I} \circ {g_I}} \right)\left( x \right) = \left( {{g_I} \circ N \circ {g_I}} \right)\left( x \right) = {N_{D2}}\left( x \right), \;\;\;x \in \left[ {0,1} \right].} \cr } Hence N_I is strong, g_I is surjective, so N_D₂ = N_I ◦ g_I is surjective, which give us a contradiction.

Suppose that E₂ = E_T,I,N. Using (8)–(11) we have $\begin{array}{l} (N_{I} \circ N) (x) = g_{I} (x) = x, x \in 0, 1], \\ N_{I} (x) = (g_{I} \circ N) (x) = N_{D 2} (x), x \in 0, 1] . \end{array}$ \matrix{ {\left( {{N_I} \circ N} \right)\left( x \right) = {g_I}\left( x \right) = x, \;\;\;x \in \left[ {0,1} \right],} \hfill \cr {\;\;\;\;\;\;\;\;{N_I}\left( x \right) = \left( {{g_I} \circ N} \right)\left( x \right) = {N_{D2}}\left( x \right), \;\;\;x \in \left[ {0,1} \right].} \hfill \cr } Hence N_I is surjective, so N_D₂ = N_I is surjective, which give us a contradiction.

A note on the fuzzy Xor

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