A fuzzy conjunction operators, t(x, y), and a fuzzy disjunction operator, s(x, y), form a pair if they satisfy:
t(x, y) = 1 – s(1 – x, 1 - y).
If $t\left( {x,\;y} \right) = \frac{{xy}}{{\left( {x + y - xy} \right)}}$ then s(x, y) is given by

Question

A fuzzy conjunction operators, t(x, y), and a fuzzy disjunction operator, s(x, y), form a pair if they satisfy:

t(x, y) = 1 – s(1 – x, 1 - y).

If $t\left( {x,\;y} \right) = \frac{{xy}}{{\left( {x + y - xy} \right)}}$ then s(x, y) is given by

Prepp Team · Accepted Answer

Understanding Fuzzy Conjunction and Disjunction

In fuzzy logic, a fuzzy conjunction operator, often denoted as $t(x, y)$, represents the logical AND operation for fuzzy sets, where $x$ and $y$ are membership degrees between 0 and 1. Similarly, a fuzzy disjunction operator, denoted as $s(x, y)$, represents the logical OR operation.

These operators can form a pair that satisfies a specific relationship, often related to De Morgan's laws in classical logic. The question provides such a relationship:

\[ t(x, y) = 1 – s(1 – x, 1 - y) \]

This equation connects the fuzzy conjunction $t(x, y)$ with the fuzzy disjunction $s(x, y)$. The term $1 - x$ represents the complement of the membership degree $x$.

Deriving the Fuzzy Disjunction Operator s(x, y)

We are given the formula for the fuzzy conjunction operator $t(x, y)$:

\[ t\left( {x,\;y} \right) = \frac{{xy}}{{\left( {x + y - xy} \right)}} \]

Our goal is to find the corresponding fuzzy disjunction operator $s(x, y)$ using the given relationship. Let's start by rearranging the given relationship to solve for $s(1 - x, 1 - y)$:

\[ s(1 – x, 1 - y) = 1 – t(x, y) \]

Now, substitute the given expression for $t(x, y)$ into this equation:

\[ s(1 – x, 1 - y) = 1 – \frac{{xy}}{{\left( {x + y - xy} \right)}} \]

To simplify the right-hand side, we combine the terms over a common denominator:

\[ s(1 – x, 1 - y) = \frac{{\left( {x + y - xy} \right) - xy}}{{\left( {x + y - xy} \right)}} \]

\[ s(1 – x, 1 - y) = \frac{{x + y - 2xy}}{{\left( {x + y - xy} \right)}} \]

This equation gives us the formula for $s$ when its inputs are $1-x$ and $1-y$. To find $s(a, b)$ where $a$ and $b$ are arbitrary membership degrees, let's substitute $a = 1 - x$ and $b = 1 - y$. From these substitutions, we can express $x$ and $y$ in terms of $a$ and $b$:

From $a = 1 - x$, we get $x = 1 - a$.
From $b = 1 - y$, we get $y = 1 - b$.

Now, substitute these expressions for $x$ and $y$ into the formula for $s(1 - x, 1 - y)$. The left side becomes $s(a, b)$:

\[ s(a, b) = \frac{{(1 - a) + (1 - b) - 2(1 - a)(1 - b)}}{{(1 - a) + (1 - b) - (1 - a)(1 - b)}} \]

Let's expand the terms in the numerator and the denominator:

Numerator:

\[ (1 - a) + (1 - b) - 2(1 - a)(1 - b) \]

\[ = 1 - a + 1 - b - 2(1 - b - a + ab) \]

\[ = 2 - a - b - 2 + 2b + 2a - 2ab \]

\[ = a + b - 2ab \]

Denominator:

\[ (1 - a) + (1 - b) - (1 - a)(1 - b) \]

\[ = 1 - a + 1 - b - (1 - b - a + ab) \]

\[ = 2 - a - b - 1 + b + a - ab \]

\[ = 1 - ab \]

So, the formula for $s(a, b)$ is:

\[ s(a, b) = \frac{{a + b - 2ab}}{{1 - ab}} \]

Finally, replacing the variables $a$ and $b$ with the standard $x$ and $y$, we get the formula for $s(x, y)$:

\[ s(x, y) = \frac{{x + y - 2xy}}{{1 - xy}} \]

Comparing with Options

Let's compare our derived formula for the fuzzy disjunction operator $s(x, y)$ with the given options:

Option 1: $\frac{{x\; + \;y}}{{1\; -\; xy}}$
Option 2: $\frac{{x\; + \;y\; - \;2xy}}{{1\; - \;xy}}$
Option 3: $\frac{{x\; + \;y\; - \;xy}}{{1\; - \;xy}}$
Option 4: $\frac{{x\; + \;y\; - \;xy}}{{1\; + \;xy}}$

Our derived formula, $s(x, y) = \frac{{x + y - 2xy}}{{1 - xy}}$, exactly matches Option 2.

Step-by-Step Solution Summary

Here's a summary of the steps taken to find the fuzzy disjunction operator $s(x, y)$:

Start with the given relationship: $t(x, y) = 1 – s(1 – x, 1 - y)$.
Rearrange the relationship to isolate $s(1 – x, 1 - y)$: $s(1 – x, 1 - y) = 1 – t(x, y)$.
Substitute the given formula for $t(x, y)$: $s(1 – x, 1 - y) = 1 – \frac{{xy}}{{\left( {x + y - xy} \right)}}$.
Simplify the right-hand side by combining terms: $s(1 – x, 1 - y) = \frac{{x + y - 2xy}}{{\left( {x + y - xy} \right)}}$.
Let $u = 1 - x$ and $v = 1 - y$, which implies $x = 1 - u$ and $y = 1 - v$. Substitute these into the equation.
Simplify the resulting expression for $s(u, v)$ by expanding the terms.
The simplified expression for $s(u, v)$ is $\frac{{u + v - 2uv}}{{1 - uv}}$.
Replace $u$ and $v$ with $x$ and $y$ to get the final formula for $s(x, y)$: $s(x, y) = \frac{{x + y - 2xy}}{{1 - xy}}$.
Match the derived formula with the given options.

Revision Table: Key Concepts

Concept	Description	Relevance to Problem
Fuzzy Set	A set where elements have degrees of membership between 0 and 1.	Foundation of fuzzy logic where operators are applied to membership degrees.
Fuzzy Conjunction (t-norm)	A function $t(x, y)$ mapping $[0, 1] \times [0, 1]$ to $[0, 1]$ representing fuzzy AND. Must satisfy properties like commutativity, associativity, monotonicity, and boundary conditions $t(x, 1) = x$.	The given $t(x, y)$ is a specific t-norm.
Fuzzy Disjunction (t-conorm or s-norm)	A function $s(x, y)$ mapping $[0, 1] \times [0, 1]$ to $[0, 1]$ representing fuzzy OR. Must satisfy properties like commutativity, associativity, monotonicity, and boundary conditions $s(x, 0) = x$.	The operator we are asked to find.
De Morgan's Law (in Fuzzy Logic)	Relationships between t-norms and t-conorms via negation (complement). The given relationship $t(x, y) = 1 – s(1 – x, 1 - y)$ is one form of De Morgan's law in fuzzy logic.	The core principle used to derive $s(x, y)$ from $t(x, y)$.

Additional Information: Types of Fuzzy Operators

The fuzzy conjunction and disjunction operators are also known as t-norms and t-conorms (or s-norms), respectively. There are many different pairs of t-norms and s-norms that satisfy De Morgan's laws. The specific pair given in this problem, with $t(x, y) = \frac{{xy}}{{x + y - xy}}$ and $s(x, y) = \frac{{x + y - 2xy}}{{1 - xy}}$, is known as the Hamacher t-norm and Hamacher t-conorm. They are defined for $x, y \in [0, 1]$.

Some common examples of t-norms and their corresponding s-norms (using the standard negation $n(x) = 1-x$) include:

Minimum t-norm and Maximum s-norm (Drastic Operators):
- $t_{\min}(x, y) = \min(x, y)$
- $s_{\max}(x, y) = \max(x, y)$
- These satisfy $t_{\min}(x, y) = 1 - s_{\max}(1-x, 1-y)$.
Product t-norm and Probabilistic Sum s-norm:
- $t_{\text{prod}}(x, y) = xy$
- $s_{\text{psum}}(x, y) = x + y - xy$
- These satisfy $t_{\text{prod}}(x, y) = 1 - s_{\text{psum}}(1-x, 1-y)$.
Lukasiewicz t-norm and s-norm:
- $t_{luk}(x, y) = \max(0, x+y-1)$
- $s_{luk}(x, y) = \min(1, x+y)$
- These satisfy $t_{luk}(x, y) = 1 - s_{luk}(1-x, 1-y)$.

The Hamacher operators are another family of such pairs, parameterized by a parameter $\gamma $. The given operators correspond to a specific value of $\gamma$.