Mathematics / Fuzzy Mathematics Fuzzy Sets, Logic & Relational Systems 100% Free Open Access
Chapter 2 • Theory & Derivations

Unit 2: Operations on Fuzzy Sets: Complements, Unions & Intersections

Comprehensive mathematical axiomatics of fuzzy set operations: classical Zadeh operators (min, max, 1-c), the axiomatic foundation of fuzzy complements (monotonicity, boundary conditions, continuous involutions, equilibrium points c(e) = e), triangular norms (t-norms) for intersection (minimum, algebraic product, bounded difference, drastic product, Hamacher, Frank, Schweizer-Sklar families), triangular conorms (s-norms) for union, generalized De Morgan duality, and compensatory averaging operators including ordered weighted averaging (OWA).

§2.1 Standard Zadeh Operators (Min, Max, Standard Inversion) & De Morgan's Laws

1. Zadeh's Original Set Theoretic Operations

In 1965, Lotfi A. Zadeh defined the elementary operations on fuzzy sets $A$ and $B$ in universe $X$ point-by-point through their membership functions:

Definition 2.1 (Standard Fuzzy Operations): For all $x \in X$:

  1. Fuzzy Complement ($A^c$):
$$\mu_{A^c}(x) = 1 - \mu_A(x)$$
  1. Fuzzy Intersection ($A \cap B$):
$$\mu_{A \cap B}(x) = \min(\mu_A(x), \mu_B(x)) = \mu_A(x) \wedge \mu_B(x)$$
  1. Fuzzy Union ($A \cup B$):
$$\mu_{A \cup B}(x) = \max(\mu_A(x), \mu_B(x)) = \mu_A(x) \vee \mu_B(x)$$
  1. Fuzzy Inclusion ($A \subseteq B$):
$$A \subseteq B \iff \mu_A(x) \le \mu_B(x) \quad (\forall x \in X)$$

2. Preservation of Algebraic Lattice Properties

Under standard operations $(\min, \max, 1 - \cdot)$, the set of all fuzzy subsets $\mathcal{F}(X)$ forms a distributive, bounded pseudo-complemented lattice (a de Morgan algebra / Kleene algebra).

Satisfied Properties ($\forall A, B, C \in \mathcal{F}(X)$):

1. Involution (Double Negation): $(A^c)^c = A$

2. Idempotence: $A \cup A = A$, and $A \cap A = A$

3. Commutativity: $A \cup B = B \cup A$, and $A \cap B = B \cap A$

4. Associativity: $(A \cup B) \cup C = A \cup (B \cup C)$, and $(A \cap B) \cap C = A \cap (B \cap C)$

5. Absorption: $A \cup (A \cap B) = A$, and $A \cap (A \cup B) = A$

6. Distributivity:

$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$

7. Identity Elements: $A \cup \emptyset = A$, and $A \cap X = A$


3. De Morgan's Laws for Fuzzy Sets

Theorem 2.1 (De Morgan's Laws): For any fuzzy sets $A, B \in \mathcal{F}(X)$:

$$(A \cup B)^c = A^c \cap B^c$$
$$(A \cap B)^c = A^c \cup B^c$$
Proof:

For any $x \in X$, let $a = \mu_A(x)$ and $b = \mu_B(x)$. For the first identity:

$$\mu_{(A \cup B)^c}(x) = 1 - \mu_{A \cup B}(x) = 1 - \max(a, b)$$

Notice that if $a \ge b$, then $\max(a, b) = a$, so $1 - \max(a, b) = 1 - a$. At the same time, $1 - a \le 1 - b$, so $\min(1 - a, 1 - b) = 1 - a$. In general, for any two real numbers $a, b$:

$$1 - \max(a, b) = \min(1 - a, 1 - b)$$

Therefore:

$$\mu_{(A \cup B)^c}(x) = \min(1 - \mu_A(x), 1 - \mu_B(x)) = \min(\mu_{A^c}(x), \mu_{B^c}(x)) = \mu_{A^c \cap B^c}(x)$$

Since this holds for all $x \in X$, $(A \cup B)^c = A^c \cap B^c$. The second identity follows symmetrically from $1 - \min(a, b) = \max(1 - a, 1 - b)$. $\blacksquare$

§2.2 Axiomatic Skeleton of Fuzzy Complements: Involution, Monotonicity & Equilibrium Points

1. General Axioms of Fuzzy Complements

Why must a complement be $1 - a$? In 1980, mathematicians generalized the notion of complementation by defining an axiomatic system for mapping $c: [0, 1] \to [0, 1]$.

Definition 2.2 (Axiomatic Fuzzy Complement): A function $c: [0, 1] \to [0, 1]$ is a fuzzy complement if it satisfies the following two axiomatic requirements:

  • Axiom C1 (Boundary Conditions): $c(0) = 1$ and $c(1) = 0$.
  • Axiom C2 (Monotonicity): For all $a, b \in [0, 1]$, if $a \le b$, then $c(a) \ge c(b)$ (strictly non-increasing).

A complement is called involutive if it satisfies:

  • Axiom C3 (Involution): $c(c(a)) = a$ for all $a \in [0, 1]$.
  • Axiom C4 (Continuity): $c$ is a continuous function.

(Theorem: Any involutive complement satisfying C1-C2 is strictly decreasing and continuous!)


2. Parameterized Families of Fuzzy Complements

1. Sugeno's Complement Family:

For parameter $\lambda \in (-1, \infty)$:

$$c_\lambda(a) = \frac{1 - a}{1 + \lambda a}$$
  • When $\lambda = 0$: $c_0(a) = 1 - a$ (Zadeh's standard complement).
  • When $\lambda \to \infty$: $c_\infty(a) \to 0$ for all $a > 0$.
  • When $\lambda \to -1$: $c_{-1}(a) \to 1$ for all $a < 1$.
2. Yager's Complement Family:

For parameter $w \in (0, \infty)$:

$$c_w(a) = \left( 1 - a^w \right)^{1/w}$$
  • When $w = 1$: $c_1(a) = 1 - a$ (standard complement).
  • Every member of Yager's family is strictly involutive: $c_w(c_w(a)) = (1 - (1 - a^w))^{1/w} = a$.

3. Equilibrium Points

Definition 2.3 (Equilibrium Point): An equilibrium point of a fuzzy complement $c$ is a value $e \in [0, 1]$ that is its own complement:

$$c(e) = e$$

Theorem 2.2 (Existence and Uniqueness of Equilibrium): Every continuous fuzzy complement $c$ possesses a unique equilibrium point $e \in (0, 1)$.

Proof:

Define the auxiliary function $g(a) = c(a) - a$ on the domain $[0, 1]$.

  1. At $a = 0$: $g(0) = c(0) - 0 = 1 - 0 = 1 > 0$.
  2. At $a = 1$: $g(1) = c(1) - 1 = 0 - 1 = -1 < 0$.
  3. Since $c$ is continuous, $g$ is continuous on $[0, 1]$.

By the Intermediate Value Theorem, there exists at least one $e \in (0, 1)$ such that $g(e) = 0 \implies c(e) = e$. Because $c$ is strictly decreasing, $g(a) = c(a) - a$ is strictly decreasing, so the root $e$ is strictly unique! $\blacksquare$

  • For Zadeh's complement $c(a) = 1 - a$: $1 - e = e \implies e = 0.5$.
  • For Sugeno's complement: $\frac{1 - e}{1 + \lambda e} = e \implies \lambda e^2 + 2e - 1 = 0 \implies e = \frac{\sqrt{1 + \lambda} - 1}{\lambda}$.

§2.3 Triangular Norms (t-Norms): Product, Łukasiewicz, Drastic, Hamacher & Frank Families

1. The Axiomatization of Fuzzy Intersections

The concept of a triangular norm ($t$-norm) was introduced by Karl Menger (1942) in the study of probabilistic metric spaces and later adopted by Schweizer and Sklar to generalize fuzzy intersections.

Definition 2.4 (Triangular Norm / t-Norm): A binary operator $i: [0, 1] \times [0, 1] \to [0, 1]$ (often denoted $T(a, b)$ or $a \top b$) is a t-norm if it satisfies for all $a, b, c, d \in [0, 1]$:

  1. Boundary Condition: $T(a, 1) = a$ (1 is the neutral identity element).
  2. Monotonicity: If $a \le c$ and $b \le d$, then $T(a, b) \le T(c, d)$.
  3. Commutativity: $T(a, b) = T(b, a)$.
  4. Associativity: $T(T(a, b), c) = T(a, T(b, c))$.

Note: From boundary and monotonicity: $T(a, 0) \le T(1, 0) = T(0, 1) = 0 \implies T(a, 0) = 0$.


2. The Four Fundamental Archetypal t-Norms

1. Minimum (Standard Zadeh Intersection):
$$T_{\min}(a, b) = \min(a, b)$$

This is the largest possible t-norm: for any t-norm $T$, $T(a, b) \le \min(a, b)$. It is the unique idempotent t-norm: $T(a, a) = a \iff T = T_{\min}$.

2. Algebraic Product:
$$T_{\text{prod}}(a, b) = a \cdot b$$

Strictly positive for all $a, b > 0$.

3. Bounded Difference (Łukasiewicz t-Norm):
$$T_{\text{Luk}}(a, b) = \max(0, a + b - 1)$$

Nilpotent t-norm: $T(a, b) = 0$ can occur even when $a > 0$ and $b > 0$.

4. Drastic Product:
$$T_D(a, b) = \begin{cases} a, & b = 1 \\ b, & a = 1 \\ 0, & \text{otherwise} \end{cases}$$

This is the smallest possible t-norm: for any t-norm $T$, $T_D(a, b) \le T(a, b)$.


3. Universal Ordering Chain of Archetypal t-Norms

Theorem 2.3 (Ordering of Fundamental t-Norms): For all $a, b \in [0, 1]$:

$$T_D(a, b) \le T_{\text{Luk}}(a, b) \le T_{\text{prod}}(a, b) \le T_{\min}(a, b)$$
Proof:
  1. $T_D \le T$ is immediate from Definition 2.4.
  2. For $T_{\text{Luk}} \le T_{\text{prod}}$:

Notice that $(1 - a)(1 - b) \ge 0 \implies 1 - a - b + ab \ge 0 \implies ab \ge a + b - 1$. Since $ab \ge 0$, we have $ab \ge \max(0, a + b - 1) = T_{\text{Luk}}(a, b)$.

  1. For $T_{\text{prod}} \le T_{\min}$:

Since $b \le 1$, $ab \le a$. Since $a \le 1$, $ab \le b$. Thus $ab \le \min(a, b) = T_{\min}(a, b)$. $\blacksquare$

§2.4 Triangular Conorms (t-Conorms / s-Norms): Algebraic Sum, Bounded Sum, Drastic Sum

1. The Axiomatization of Fuzzy Unions

The dual operation to a $t$-norm is a triangular conorm ($t$-conorm or $s$-norm).

Definition 2.5 (Triangular Conorm / s-Norm): A binary operator $u: [0, 1] \times [0, 1] \to [0, 1]$ (denoted $S(a, b)$ or $a \bot b$) is an s-norm if it satisfies for all $a, b, c, d \in [0, 1]$:

  1. Boundary Condition: $S(a, 0) = a$ (0 is the neutral identity element).
  2. Monotonicity: If $a \le c$ and $b \le d$, then $S(a, b) \le S(c, d)$.
  3. Commutativity: $S(a, b) = S(b, a)$.
  4. Associativity: $S(S(a, b), c) = S(a, S(b, c))$.

Note: $S(a, 1) = 1$ for all $a \in [0, 1]$.


2. The Four Fundamental Archetypal s-Norms

1. Maximum (Standard Zadeh Union):
$$S_{\max}(a, b) = \max(a, b)$$

This is the smallest possible s-norm: for any s-norm $S$, $\max(a, b) \le S(a, b)$. It is the unique idempotent s-norm: $S(a, a) = a \iff S = S_{\max}$.

2. Algebraic Sum (Probabilistic Sum):
$$S_{\text{sum}}(a, b) = a + b - ab$$
3. Bounded Sum (Łukasiewicz s-Norm):
$$S_{\text{Luk}}(a, b) = \min(1, a + b)$$
4. Drastic Sum:
$$S_D(a, b) = \begin{cases} a, & b = 0 \\ b, & a = 0 \\ 1, & \text{otherwise} \end{cases}$$

This is the largest possible s-norm: for any s-norm $S$, $S(a, b) \le S_D(a, b)$.


3. Generalized De Morgan Duality

Theorem 2.4 (Duality Theorem): Let $c$ be an involutive fuzzy complement. For every t-norm $T$, the dual operator defined by:

$$S(a, b) = c\left( T(c(a), c(b)) \right)$$

is an s-norm. Conversely, for every s-norm $S$:

$$T(a, b) = c\left( S(c(a), c(b)) \right)$$

is a t-norm. The pair $(T, S, c)$ satisfies generalized De Morgan's laws.

Universal Ordering Chain of Archetypal s-Norms:
$$S_{\max}(a, b) \le S_{\text{sum}}(a, b) \le S_{\text{Luk}}(a, b) \le S_D(a, b)$$

§2.5 Averaging Operators, Generalized Means & Ordered Weighted Averaging (OWA)

1. The Spectrum Between Intersection and Union

Notice the strict bounds governing norm operations:

$$T(a, b) \le \min(a, b) \le \max(a, b) \le S(a, b)$$

$t$-norms represent strict conjunction ("AND"), while $s$-norms represent full disjunction ("OR"). However, human decision-making frequently requires compensatory aggregation (a trade-off where a high score in one criterion compensates for a lower score in another). This motivates averaging operators $M(a, b)$ that lie strictly between min and max:

$$\min(a, b) \le M(a, b) \le \max(a, b)$$

2. Generalized Means (Power Means)

For elements $a_1, a_2, \dots, a_n \in [0, 1]$ and weights $w_i \ge 0$ with $\sum w_i = 1$:

$$M_p(a; w) = \left( \sum_{i=1}^n w_i a_i^p \right)^{1/p}, \qquad p \in \mathbb{R} \setminus \{0\}$$
  • $p \to -\infty$: $M_{-\infty} = \min(a_1, \dots, a_n)$ (pure intersection).
  • $p = -1$: Harmonic Mean $M_{-1} = \frac{1}{\sum \frac{w_i}{a_i}}$.
  • $p \to 0$: Geometric Mean $M_0 = \prod_{i=1}^n a_i^{w_i}$.
  • $p = 1$: Arithmetic Mean $M_1 = \sum_{i=1}^n w_i a_i$.
  • $p = 2$: Quadratic (RMS) Mean $M_2 = \sqrt{\sum w_i a_i^2}$.
  • $p \to +\infty$: $M_{+\infty} = \max(a_1, \dots, a_n)$ (pure union).

3. Ordered Weighted Averaging (OWA) Operators

Introduced by Ronald R. Yager (1988), the OWA operator decouples weights from individual criteria and associates them with magnitudes:

Definition 2.6 (OWA Operator): An OWA operator of dimension $n$ is a mapping $F_w: [0, 1]^n \to [0, 1]$ associated with weighting vector $w = (w_1, \dots, w_n)$ where $w_i \in [0, 1]$ and $\sum w_i = 1$:

$$F_w(a_1, \dots, a_n) = \sum_{j=1}^n w_j b_j$$

where $b_j$ is the $j$-th largest element of the collection $\{a_1, \dots, a_n\}$ ($b_1 \ge b_2 \ge \dots \ge b_n$).

Extreme Cases:
  • $w = (1, 0, \dots, 0) \implies F_w(a) = b_1 = \max(a_i)$ (Pure OR).
  • $w = (0, 0, \dots, 1) \implies F_w(a) = b_n = \min(a_i)$ (Pure AND).
  • $w = (1/n, 1/n, \dots, 1/n) \implies F_w(a) = \frac{1}{n} \sum a_i$ (Standard Arithmetic Mean).

The degree of "orness" of an OWA operator is measured by:

$$\text{orness}(w) = \frac{1}{n - 1} \sum_{j=1}^n (n - j) w_j \in [0, 1]$$
  • $\text{orness} = 1$ for Max; $\text{orness} = 0$ for Min; $\text{orness} = 0.5$ for Arithmetic Mean.

Rigorous Tiered Solved Examination Problems

Step-by-step unskipped derivations, complete proofs, and verification across Foundational, Advanced, and Honors tiers.

Fundamentals Example 2.1: Problem 2.1: Comparison of t-Norms and s-Norms for Specific Truth Grades

Let two fuzzy propositions have truth values $a = 0.7$ and $b = 0.4$.

  1. Compute their intersection using the four fundamental t-norms:
  • Minimum $T_{\min}(a, b)$
  • Algebraic Product $T_{\text{prod}}(a, b)$
  • Łukasiewicz Bounded Difference $T_{\text{Luk}}(a, b)$
  • Drastic Product $T_D(a, b)$

and verify the ordering chain $T_D \le T_{\text{Luk}} \le T_{\text{prod}} \le T_{\min}$.

  1. Compute their union using the four fundamental s-norms:
  • Maximum $S_{\max}(a, b)$
  • Algebraic Sum $S_{\text{sum}}(a, b)$
  • Łukasiewicz Bounded Sum $S_{\text{Luk}}(a, b)$
  • Drastic Sum $S_D(a, b)$

and verify the ordering chain $S_{\max} \le S_{\text{sum}} \le S_{\text{Luk}} \le S_D$.

  1. Compute the standard Yager complement with parameter $w = 2$ for $a = 0.7$.

1. Calculation of t-Norms ($a = 0.7, b = 0.4$)

1. Minimum:

$$T_{\min}(0.7, 0.4) = \min(0.7, 0.4) = 0.40$$

2. Algebraic Product:

$$T_{\text{prod}}(0.7, 0.4) = 0.7 \times 0.4 = 0.28$$

3. Łukasiewicz Bounded Difference:

$$T_{\text{Luk}}(0.7, 0.4) = \max(0, 0.7 + 0.4 - 1) = \max(0, 0.10) = 0.10$$

4. Drastic Product:

Since neither $a = 1$ nor $b = 1$:

$$T_D(0.7, 0.4) = 0.00$$
Verification of Ordering:
$$0.00 \le 0.10 \le 0.28 \le 0.40 \iff T_D \le T_{\text{Luk}} \le T_{\text{prod}} \le T_{\min}$$

Holds with strict inequalities! $\blacksquare$


2. Calculation of s-Norms ($a = 0.7, b = 0.4$)

1. Maximum:

$$S_{\max}(0.7, 0.4) = \max(0.7, 0.4) = 0.70$$

2. Algebraic Sum:

$$S_{\text{sum}}(0.7, 0.4) = 0.7 + 0.4 - (0.7)(0.4) = 1.10 - 0.28 = 0.82$$

3. Łukasiewicz Bounded Sum:

$$S_{\text{Luk}}(0.7, 0.4) = \min(1, 0.7 + 0.4) = \min(1, 1.10) = 1.00$$

4. Drastic Sum:

Since neither $a = 0$ nor $b = 0$:

$$S_D(0.7, 0.4) = 1.00$$
Verification of Ordering:
$$0.70 \le 0.82 \le 1.00 \le 1.00 \iff S_{\max} \le S_{\text{sum}} \le S_{\text{Luk}} \le S_D$$

Holds identically! $\blacksquare$


3. Yager Complement for $w = 2, a = 0.7$

$$c_2(0.7) = \left( 1 - 0.7^2 \right)^{1/2} = \sqrt{1 - 0.49} = \sqrt{0.51} \approx 0.7141 \qquad \blacksquare$$
Computational Triad Example 2.2: Problem 2.2: Equilibrium Points of Sugeno and Yager Complements

An equilibrium point $e \in [0, 1]$ of a fuzzy complement satisfies $c(e) = e$.

  1. For the Sugeno complement $c_\lambda(a) = \frac{1 - a}{1 + \lambda a}$ with parameter $\lambda \in (-1, \infty)$:
  • Show that the equilibrium point is given by $e_\lambda = \frac{\sqrt{1 + \lambda} - 1}{\lambda}$ for $\lambda \ne 0$.
  • Evaluate $\lim_{\lambda \to 0} e_\lambda$ and show it matches Zadeh's standard equilibrium point $0.5$.
  • Calculate $e_\lambda$ for $\lambda = 3$ and $\lambda = -0.75$.
  1. For the Yager complement $c_w(a) = (1 - a^w)^{1/w}$ with parameter $w \in (0, \infty)$:
  • Find the exact analytical expression for the equilibrium point $e_w$.
  • Evaluate $e_w$ for $w = 1, 2, 3$.

1. Sugeno Complement Equilibrium Point

Step A: Quadratic Derivation

Set $c_\lambda(e) = e$:

$$\frac{1 - e}{1 + \lambda e} = e \implies 1 - e = e(1 + \lambda e) \implies 1 - e = e + \lambda e^2$$

Rearranging into standard quadratic form:

$$\lambda e^2 + 2e - 1 = 0$$

Using the quadratic formula (for $\lambda \ne 0$):

$$e = \frac{-2 \pm \sqrt{4 - 4(\lambda)(-1)}}{2\lambda} = \frac{-2 \pm \sqrt{4 + 4\lambda}}{2\lambda} = \frac{-1 \pm \sqrt{1 + \lambda}}{\lambda}$$

Since $\lambda > -1$, $\sqrt{1 + \lambda} > 0$. To ensure $e \in [0, 1]$, we take the positive root:

$$e_\lambda = \frac{\sqrt{1 + \lambda} - 1}{\lambda} \qquad \blacksquare$$
Step B: Limit as $\lambda \to 0$

Using rationalization:

$$e_\lambda = \frac{(\sqrt{1 + \lambda} - 1)(\sqrt{1 + \lambda} + 1)}{\lambda (\sqrt{1 + \lambda} + 1)} = \frac{(1 + \lambda) - 1}{\lambda (\sqrt{1 + \lambda} + 1)} = \frac{\lambda}{\lambda (\sqrt{1 + \lambda} + 1)} = \frac{1}{\sqrt{1 + \lambda} + 1}$$

Taking the limit as $\lambda \to 0$:

$$\lim_{\lambda \to 0} e_\lambda = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} = 0.5 \qquad \blacksquare$$
Step C: Specific Evaluations
  • For $\lambda = 3$:
$$e_3 = \frac{\sqrt{1 + 3} - 1}{3} = \frac{2 - 1}{3} = \frac{1}{3} \approx 0.3333$$
  • For $\lambda = -0.75$:
$$e_{-0.75} = \frac{\sqrt{1 - 0.75} - 1}{-0.75} = \frac{\sqrt{0.25} - 1}{-0.75} = \frac{0.5 - 1}{-0.75} = \frac{-0.5}{-0.75} = \frac{2}{3} \approx 0.6667 \qquad \blacksquare$$

2. Yager Complement Equilibrium Point

Set $c_w(e) = e$:

$$\left( 1 - e^w \right)^{1/w} = e$$

Raise both sides to power $w$:

$$1 - e^w = e^w \implies 2e^w = 1 \implies e^w = \frac{1}{2}$$

Taking the $w$-th root:

$$e_w = \left(\frac{1}{2}\right)^{1/w} = 2^{-1/w} \qquad \blacksquare$$
Evaluations:
  • For $w = 1$:
$$e_1 = 2^{-1} = 0.5$$
  • For $w = 2$:
$$e_2 = 2^{-1/2} = \frac{1}{\sqrt{2}} \approx 0.7071$$
  • For $w = 3$:
$$e_3 = 2^{-1/3} \approx 0.7937 \qquad \blacksquare$$
Honors / Proof Challenge Example 2.3: Problem 2.3: Uniqueness of Min/Max via Idempotence and OWA Optimization
  1. Prove that the minimum operator $T_{\min}(a, b) = \min(a, b)$ is the ONLY idempotent t-norm:
$$\forall a \in [0, 1], \; T(a, a) = a \iff T(a, b) = \min(a, b)$$
  1. Let $F_w(a_1, \dots, a_n) = \sum_{j=1}^n w_j b_j$ be an OWA operator with inputs $a = (0.9, 0.4, 0.8, 0.2)$.
  • Evaluate $F_w$ with weights $w = (0.4, 0.3, 0.2, 0.1)$.
  • Compute the degree of orness: $\text{orness}(w) = \frac{1}{n-1} \sum_{j=1}^n (n-j) w_j$.
  • Compute the dispersion (entropy) of the weights: $H(w) = -\sum_{j=1}^n w_j \ln w_j$.
  1. Show that as $\text{orness}(w) \to 1$, $H(w) \to 0$.

1. Proof of the Uniqueness of $T_{\min}$ via Idempotence

Step A: Verification that $\min$ is Idempotent

$\min(a, a) = a$ holds trivially for all $a \in [0, 1]$.

Step B: Uniqueness Proof

Assume $T$ is an arbitrary t-norm satisfying the idempotence axiom:

$$T(x, x) = x \quad (\forall x \in [0, 1])$$

Let $a, b \in [0, 1]$. Without loss of generality, assume $a \le b$. Then $\min(a, b) = a$. Now use the axioms of t-norms:

  1. By monotonicity (Axiom 2), since $a \le b$:
$$T(a, a) \le T(a, b)$$

By idempotence, $T(a, a) = a$, so:

$$a \le T(a, b)$$
  1. On the other hand, since $b \le 1$, by monotonicity:
$$T(a, b) \le T(a, 1)$$

By the boundary condition (Axiom 1), $T(a, 1) = a$, so:

$$T(a, b) \le a$$

Combining (1) and (2):

$$a \le T(a, b) \le a \implies T(a, b) = a = \min(a, b)$$

If $b \le a$, by commutativity $T(a, b) = T(b, a) = b = \min(a, b)$. Therefore, $T(a, b) = \min(a, b)$ for all $a, b \in [0, 1]$. $T_{\min}$ is the unique idempotent t-norm! $\blacksquare$


2. OWA Operator Calculations

Inputs: $a = (0.9, 0.4, 0.8, 0.2)$. Dimension $n = 4$. Weights: $w = (0.4, 0.3, 0.2, 0.1)$.

Step A: Sort in Descending Order
$$b_1 = 0.9, \quad b_2 = 0.8, \quad b_3 = 0.4, \quad b_4 = 0.2$$
$$b = (0.9, 0.8, 0.4, 0.2)$$
Step B: Evaluate OWA Aggregation
$$\begin{aligned} F_w(a) &= w_1 b_1 + w_2 b_2 + w_3 b_3 + w_4 b_4 \\ &= (0.4)(0.9) + (0.3)(0.8) + (0.2)(0.4) + (0.1)(0.2) \\ &= 0.36 + 0.24 + 0.08 + 0.02 \\ &= 0.70 \qquad \blacksquare \end{aligned}$$
Step C: Degree of Orness

With $n = 4$:

$$\begin{aligned} \text{orness}(w) &= \frac{1}{4 - 1} \sum_{j=1}^4 (4 - j) w_j \\ &= \frac{1}{3} \left[ 3 w_1 + 2 w_2 + 1 w_3 + 0 w_4 \right] \\ &= \frac{1}{3} \left[ 3(0.4) + 2(0.3) + 1(0.2) + 0(0.1) \right] \\ &= \frac{1}{3} \left[ 1.2 + 0.6 + 0.2 + 0 \right] = \frac{2.0}{3} \approx 0.6667 \qquad \blacksquare \end{aligned}$$
Step D: Dispersion (Entropy)
$$H(w) = -\sum_{j=1}^4 w_j \ln w_j$$
  • $w_1 = 0.4 \implies 0.4 \ln(0.4) \approx 0.4(-0.9163) = -0.3665$
  • $w_2 = 0.3 \implies 0.3 \ln(0.3) \approx 0.3(-1.2040) = -0.3612$
  • $w_3 = 0.2 \implies 0.2 \ln(0.2) \approx 0.2(-1.6094) = -0.3219$
  • $w_4 = 0.1 \implies 0.1 \ln(0.1) \approx 0.1(-2.3026) = -0.2303$

Sum:

$$H(w) = -(-0.3665 - 0.3612 - 0.3219 - 0.2303) = 1.2799 \text{ nats} \qquad \blacksquare$$

3. Asymptotic Behavior as $\text{orness} \to 1$

When $\text{orness}(w) = 1$, all weight is concentrated on the first component:

$$w^* = (1, 0, \dots, 0)$$

Then:

$$H(w^*) = -1 \ln(1) - \lim_{p \to 0^+} \sum_{j=2}^n p \ln p = 0 - 0 = 0$$

This demonstrates that pure disjunction (Max) has zero entropy, representing maximum certainty of relying strictly on the best score, whereas equal weights $w = (1/n, \dots, 1/n)$ achieve maximum entropy $\ln n$ with neutral $\text{orness} = 0.5$. $\blacksquare$