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Chapter 7 • Theory & Derivations

Unit 7: Orderings, Morphisms & Fuzzy Relational Equations

Advanced mathematical study of ordered fuzzy systems, relational mappings, and relational equations: fuzzy partial orderings, quasi-orderings, fuzzy dominance relations, relational morphisms (homomorphisms, strong homomorphisms, isomorphisms), fuzzy relational equations P ∘ R = Q on finite and infinite universes, the fundamental solvability criterion, Sanchez's Theorem for the greatest relational solution R̂ = P α Q via Gödel residuated implication, minimal solutions, and inverse diagnostic systems.

§7.1 Fuzzy Partial Orderings, Quasi-Orderings & Dominance

1. Fuzzy Partial Orderings

In classical order theory, a partial ordering is a reflexive, antisymmetric, and transitive relation. In fuzzy mathematics, this concept generalizes to model gradual precedence and preference:

Definition 7.1 (Fuzzy Partial Ordering): A fuzzy relation $R$ on $X \times X$ is called a fuzzy partial ordering if it satisfies:

  1. Reflexivity: $\mu_R(x, x) = 1$ for all $x \in X$.
  2. Antisymmetry: For all $x \ne y \in X$:
$$\mu_R(x, y) > 0 \implies \mu_R(y, x) = 0$$

(Strong antisymmetry: $\min(\mu_R(x, y), \mu_R(y, x)) = 0$ for $x \ne y$).

  1. Max-Min Transitivity: $R \circ R \subseteq R$.

A set $X$ equipped with a fuzzy partial ordering is called a fuzzy partially ordered set (fuzzy poset).


2. Fuzzy Pre-Orderings (Quasi-Orderings)

If the relation is reflexive and max-min transitive, but NOT antisymmetric, it is called a fuzzy pre-ordering (quasi-ordering).

  • Quasi-orderings are used extensively in social choice theory, consumer preference modeling, and multi-criteria decision ranking.

§7.2 Fuzzy Morphisms: Homomorphisms and Isomorphisms of Fuzzy Relational Systems

1. Relational Systems

Let $(X, R)$ and $(Y, S)$ be two fuzzy relational systems, where $R$ is a fuzzy relation on $X \times X$ and $S$ is a fuzzy relation on $Y \times Y$. A crisp mapping $h: X \to Y$ is called a homomorphism if it preserves the relational structure:

Definition 7.2 (Fuzzy Homomorphism): A mapping $h: X \to Y$ is a homomorphism from $(X, R)$ to $(Y, S)$ if for all $x_1, x_2 \in X$:

$$\mu_R(x_1, x_2) \le \mu_S(h(x_1), h(x_2))$$

That is, if two elements are related with strength $\alpha$ in $X$, their images under $h$ must be related with at least strength $\alpha$ in $Y$.


2. Strong Homomorphisms and Isomorphisms

1. Strong Homomorphism: $h$ is surjective and for all $y_1, y_2 \in Y$:

$$\mu_S(y_1, y_2) = \sup_{x_1 \in h^{-1}(y_1), x_2 \in h^{-1}(y_2)} \mu_R(x_1, x_2)$$

2. Fuzzy Isomorphism: $h$ is a bijective mapping and for all $x_1, x_2 \in X$:

$$\mu_S(h(x_1), h(x_2)) = \mu_R(x_1, x_2)$$

An isomorphism means the systems $(X, R)$ and $(Y, S)$ have identical relational structures up to relabeling of elements.

§7.3 Fuzzy Relational Equations of the Form $P \circ R = Q$

1. Formulation of the Problem

Let $X = \{x_1, \dots, x_m\}$, $Y = \{y_1, \dots, y_n\}$, and $Z = \{z_1, \dots, z_p\}$ be finite universes of discourse. Consider the composition:

$$P \circ R = Q$$

where:

  • $P = (p_1, \dots, p_m)$ is a known input fuzzy set (or relation on $U \times X$).
  • $Q = (q_1, \dots, q_n)$ is a known observed output fuzzy set (or relation on $U \times Y$).
  • $R \in [0, 1]^{m \times n}$ is an unknown fuzzy relation (representing the transfer matrix or system model).

Under Max-Min composition:

$$q_j = \max_{i=1}^m \min(p_i, r_{ij}) \quad (\forall j = 1, \dots, n)$$

Each column of $R$ can be solved independently:

$$\max_{i=1}^m \min(p_i, r_{ij}) = q_j$$

2. Solvability Dilemma

Unlike linear equations $A x = b$ in vector spaces, fuzzy relational equations do not form a field or ring. Instead, they are governed by idempotent semi-ring lattice theory.

  • There may be no solution at all.
  • When solutions exist, the solution set $\mathcal{S}(P, Q) = \{R \mid P \circ R = Q\}$ is generally not convex.
  • However, if solutions exist, there is always a unique GREATEST solution $\hat{R}$, and a finite set of minimal solutions!

§7.4 Solvability Criteria, Greatest and Smallest Solutions via Sanchez's Theorem

1. The $\alpha$-Operator (Gödel Residuated Implication)

To solve $P \circ R = Q$, Elie Sanchez (1976) introduced the $\alpha$-operator (the residuum of the minimum t-norm):

Definition 7.3 (Sanchez $\alpha$-Operator): For $a, b \in [0, 1]$:

$$a \mathbin{\alpha} b \equiv \begin{cases} 1, & \text{if } a \le b \\ b, & \text{if } a > b \end{cases}$$

Note: $a \mathbin{\alpha} b$ is the largest number $x \in [0, 1]$ satisfying $\min(a, x) \le b$.


2. Sanchez's Theorem

Theorem 7.1 (Sanchez's Theorem, 1976): Let $P$ and $Q$ be fuzzy vectors. Define the candidate relation $\hat{R} = P \mathbin{\alpha} Q$ by:

$$\hat{r}_{ij} = p_i \mathbin{\alpha} q_j = \begin{cases} 1, & p_i \le q_j \\ q_j, & p_i > q_j \end{cases}$$

Then:

  1. The equation $P \circ R = Q$ has a solution if and only if $\hat{R}$ is itself a solution:
$$\mathcal{S}(P, Q) \ne \emptyset \iff P \circ \hat{R} = Q$$
  1. Whenever $\mathcal{S}(P, Q) \ne \emptyset$, $\hat{R}$ is the unique greatest solution:
$$\forall R \in \mathcal{S}(P, Q), \quad R \subseteq \hat{R}$$
Proof of Optimality:

For any $R \in \mathcal{S}(P, Q)$, we have $\min(p_i, r_{ij}) \le \max_k \min(p_k, r_{kj}) = q_j$ for every $i$. By definition of the $\alpha$-operator, $r_{ij} \le p_i \mathbin{\alpha} q_j = \hat{r}_{ij}$. Thus $R \subseteq \hat{R}$ for every valid solution! $\blacksquare$


3. Structure of the Solution Set

When solvable, the solution set forms a upper semi-lattice:

$$\mathcal{S}(P, Q) = \bigcup_{k=1}^s [ \check{R}_k, \; \hat{R} ]$$

where $\check{R}_1, \dots, \check{R}_s$ are the finitely many minimal solutions.

§7.5 Inverse Relational Problems and Diagnostic Modeling

1. Medical and Fault Diagnostic Systems

In fault diagnosis and medical expert systems:

  • $P$: Fuzzy set of diseases / root causes.
  • $R$: Known symptom-disease relationship matrix (knowledge base).
  • $Q$: Observed fuzzy set of symptoms exhibited by a patient or machine.

The forward problem is $P \circ R = Q$. The inverse diagnostic problem is: Given symptoms $Q$ and relationship $R$, find the disease profile $P$!


2. Solving for the Unknown Input ($X \circ R = Q$)

By transposition, $X \circ R = Q \iff R^T \circ X^T = Q^T$. Applying Sanchez's Theorem to the dual formulation:

$$\hat{X} = (R \mathbin{\alpha} Q^T)^T \implies \hat{x}_i = \min_{j=1}^n (r_{ij} \mathbin{\alpha} q_j)$$

If $\hat{X} \circ R = Q$, then $\hat{X}$ is the largest possible cause profile that can account for symptoms $Q$ without exceeding them!

Rigorous Tiered Solved Examination Problems

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

Fundamentals Example 7.1: Problem 7.1: Application of the Gödel $\alpha$-Operator

The Gödel residuum operator is defined by $a \mathbin{\alpha} b = 1$ if $a \le b$, and $b$ if $a > b$.

  1. Compute the following scalar values:
  • $0.3 \mathbin{\alpha} 0.7$
  • $0.8 \mathbin{\alpha} 0.5$
  • $0.6 \mathbin{\alpha} 0.6$
  • $0.9 \mathbin{\alpha} 0.0$
  1. For $a = 0.8$ and $b = 0.5$, verify that $\min(a, a \mathbin{\alpha} b) = b$.
  2. Prove that for any $a, b \in [0, 1]$, $\min(a, a \mathbin{\alpha} b) \le b$ holds unconditionally.

1. Scalar Evaluations

  • $0.3 \mathbin{\alpha} 0.7$: Since $0.3 \le 0.7$, the value is $1.0$.
  • $0.8 \mathbin{\alpha} 0.5$: Since $0.8 > 0.5$, the value is $0.5$.
  • $0.6 \mathbin{\alpha} 0.6$: Since $0.6 \le 0.6$, the value is $1.0$.
  • $0.9 \mathbin{\alpha} 0.0$: Since $0.9 > 0.0$, the value is $0.0$. $\blacksquare$

2. Verification for $a = 0.8, b = 0.5$

Since $a > b$, $a \mathbin{\alpha} b = 0.5$. Then:

$$\min(a, a \mathbin{\alpha} b) = \min(0.8, 0.5) = 0.5 = b \qquad \blacksquare$$

3. General Proof that $\min(a, a \mathbin{\alpha} b) \le b$

Let $a, b \in [0, 1]$.

  • Case 1 ($a \le b$):

By definition, $a \mathbin{\alpha} b = 1$. Then:

$$\min(a, a \mathbin{\alpha} b) = \min(a, 1) = a$$

Since $a \le b$, we have $\min(a, a \mathbin{\alpha} b) = a \le b$.

  • Case 2 ($a > b$):

By definition, $a \mathbin{\alpha} b = b$. Then:

$$\min(a, a \mathbin{\alpha} b) = \min(a, b) = b \le b$$

In both exhaustive cases, $\min(a, a \mathbin{\alpha} b) \le b$ holds identically. $\blacksquare$

Computational Triad Example 7.2: Problem 7.2: Complete Solution of a Fuzzy Relational Equation $P \circ R = Q$

Consider the fuzzy relational equation $P \circ R = Q$, where:

$$P = \begin{pmatrix} 0.5 & 0.7 & 0.3 \end{pmatrix}$$

and the observed output is:

$$Q = \begin{pmatrix} 0.6 & 0.4 \end{pmatrix}$$

The unknown relation is a $3 \times 2$ matrix $R = (r_{ij}) \in [0, 1]^{3 \times 2}$.

  1. Use Sanchez's Theorem to construct the candidate greatest relation $\hat{R} = P \mathbin{\alpha} Q$.
  2. Compute the composition $P \circ \hat{R}$ and verify whether $\hat{R}$ solves the equation.
  3. Determine all minimal solutions $\check{R}$ for this system.
  4. Give a non-trivial solution matrix $R \in \mathcal{S}(P, Q)$ strictly between a minimal solution and $\hat{R}$.

1. Construction of $\hat{R} = P \mathbin{\alpha} Q$

Input: $P = (0.5, 0.7, 0.3)$. Output: $Q = (0.6, 0.4)$. Formula: $\hat{r}_{ij} = p_i \mathbin{\alpha} q_j$.

Column 1 ($q_1 = 0.6$):
  • $\hat{r}_{11} = p_1 \mathbin{\alpha} q_1 = 0.5 \mathbin{\alpha} 0.6 = 1.0$ (since $0.5 \le 0.6$)
  • $\hat{r}_{21} = p_2 \mathbin{\alpha} q_1 = 0.7 \mathbin{\alpha} 0.6 = 0.6$ (since $0.7 > 0.6$)
  • $\hat{r}_{31} = p_3 \mathbin{\alpha} q_1 = 0.3 \mathbin{\alpha} 0.6 = 1.0$ (since $0.3 \le 0.6$)
Column 2 ($q_2 = 0.4$):
  • $\hat{r}_{12} = p_1 \mathbin{\alpha} q_2 = 0.5 \mathbin{\alpha} 0.4 = 0.4$ (since $0.5 > 0.4$)
  • $\hat{r}_{22} = p_2 \mathbin{\alpha} q_2 = 0.7 \mathbin{\alpha} 0.4 = 0.4$ (since $0.7 > 0.4$)
  • $\hat{r}_{32} = p_3 \mathbin{\alpha} q_2 = 0.3 \mathbin{\alpha} 0.4 = 1.0$ (since $0.3 \le 0.4$)

Therefore, the greatest solution matrix is:

$$\hat{R} = \begin{pmatrix} 1.0 & 0.4 \\ 0.6 & 0.4 \\ 1.0 & 1.0 \end{pmatrix} \qquad \blacksquare$$

2. Verification of Solvability ($P \circ \hat{R}$)

  • Column 1:
$$q_1' = \max(\min(0.5, 1.0), \min(0.7, 0.6), \min(0.3, 1.0)) = \max(0.5, 0.6, 0.3) = 0.6 = q_1$$
  • Column 2:
$$q_2' = \max(\min(0.5, 0.4), \min(0.7, 0.4), \min(0.3, 1.0)) = \max(0.4, 0.4, 0.3) = 0.4 = q_2$$

Since $P \circ \hat{R} = (0.6, 0.4) = Q$, the equation is SOLVABLE, and $\hat{R}$ is indeed the greatest solution! $\blacksquare$


3. Finding Minimal Solutions $\check{R}$

To achieve $q_1 = 0.6$: Notice that $p_1 = 0.5 < 0.6$ and $p_3 = 0.3 < 0.6$. The term $\min(p_1, r_{11}) \le 0.5$ and $\min(p_3, r_{31}) \le 0.3$. Thus, only the second term $\min(p_2, r_{21}) = \min(0.7, r_{21})$ can reach $0.6$! We MUST have $r_{21} = 0.6$. The other entries in column 1 can be 0. Minimal column 1: $(0, 0.6, 0)^T$.

To achieve $q_2 = 0.4$: Notice that $p_3 = 0.3 < 0.4$, so $\min(p_3, r_{32}) \le 0.3$. We can achieve $0.4$ through EITHER index 1 or index 2:

  • Option A: $r_{12} = 0.4$ (since $\min(0.5, 0.4) = 0.4$).
  • Option B: $r_{22} = 0.4$ (since $\min(0.7, 0.4) = 0.4$).

Thus, there are two distinct minimal solutions:

$$\check{R}_1 = \begin{pmatrix} 0.0 & 0.4 \\ 0.6 & 0.0 \\ 0.0 & 0.0 \end{pmatrix}, \qquad \check{R}_2 = \begin{pmatrix} 0.0 & 0.0 \\ 0.6 & 0.4 \\ 0.0 & 0.0 \end{pmatrix} \qquad \blacksquare$$

4. Intermediate Solution

Any matrix $R$ satisfying $\check{R}_k \le R \le \hat{R}$ is an exact solution. For instance:

$$R_{\text{mid}} = \begin{pmatrix} 0.5 & 0.4 \\ 0.6 & 0.2 \\ 0.0 & 0.5 \end{pmatrix}$$

Check:

  • $q_1 = \max(\min(0.5, 0.5), \min(0.7, 0.6), \min(0.3, 0.0)) = \max(0.5, 0.6, 0.0) = 0.6$.
  • $q_2 = \max(\min(0.5, 0.4), \min(0.7, 0.2), \min(0.3, 0.5)) = \max(0.4, 0.2, 0.3) = 0.4$.

$P \circ R_{\text{mid}} = (0.6, 0.4) = Q$. Verified! $\blacksquare$

Honors / Proof Challenge Example 7.3: Problem 7.3: Inconsistency and Solvability Bound for Fuzzy Relational Equations

Consider the fuzzy relational equation $P \circ R = Q$ on finite sets where $P \in [0, 1]^m$ and $Q \in [0, 1]^n$.

  1. Prove that if there exists some column index $j \in \{1, \dots, n\}$ such that:
$$\max_{i=1}^m p_i < q_j$$

then the equation $P \circ R = Q$ has NO SOLUTION ($\mathcal{S}(P, Q) = \emptyset$).

  1. Prove that if $Q$ satisfies $\mathcal{S}(P, Q) \ne \emptyset$, then the greatest solution $\hat{R} = P \mathbin{\alpha} Q$ satisfies the equality:
$$P \circ (P \mathbin{\alpha} Q) = Q$$
  1. For an unsolvable equation, define the approximate Chebyshev distance error:
$$E(R) = \| P \circ R - Q \|_\infty = \max_{j=1}^n | (P \circ R)_j - q_j |$$

Show that the candidate $\hat{R} = P \mathbin{\alpha} Q$ always produces an underestimation $P \circ \hat{R} \le Q$, and evaluate the error when $P = (0.4, 0.5)$ and $Q = (0.8, 0.9)$.

1. Proof of the Necessary Solvability Bound

For any candidate matrix $R \in [0, 1]^{m \times n}$, the $j$-th component of the composed output is:

$$(P \circ R)_j = \max_{i=1}^m \min(p_i, r_{ij})$$

Since $\min(p_i, r_{ij}) \le p_i$ for all $i \in \{1, \dots, m\}$:

$$\max_{i=1}^m \min(p_i, r_{ij}) \le \max_{i=1}^m p_i$$

Therefore, for ANY relation $R$:

$$(P \circ R)_j \le \max_{i=1}^m p_i$$

Now suppose there exists some $j$ such that $\max_{i=1}^m p_i < q_j$. Then:

$$(P \circ R)_j \le \max_{i=1}^m p_i < q_j \implies (P \circ R)_j \ne q_j$$

It is impossible for $(P \circ R)_j$ to reach $q_j$. Hence no relation $R$ can satisfy the equation, proving that $\mathcal{S}(P, Q) = \emptyset$. $\blacksquare$


2. Proof of Equality for Greatest Solution

Assume $\mathcal{S}(P, Q) \ne \emptyset$. Let $R^$ be any solution in $\mathcal{S}(P, Q)$, so $P \circ R^ = Q$. By the definition of the $\alpha$-operator: For all $i, j$: $\min(p_i, r^_{ij}) \le (P \circ R^)_j = q_j$. Since $p_i \mathbin{\alpha} q_j$ is the largest value $x$ such that $\min(p_i, x) \le q_j$, we have:

$$r^*_{ij} \le p_i \mathbin{\alpha} q_j = \hat{r}_{ij} \implies R^* \subseteq \hat{R}$$

By monotonicity of the max-min composition:

$$Q = P \circ R^* \subseteq P \circ \hat{R}$$

On the other hand, by definition of $\hat{R}$:

$$\min(p_i, \hat{r}_{ij}) = \min(p_i, p_i \mathbin{\alpha} q_j) \le q_j \quad (\forall i)$$

Taking the maximum over $i$:

$$(P \circ \hat{R})_j = \max_{i=1}^m \min(p_i, \hat{r}_{ij}) \le q_j \implies P \circ \hat{R} \subseteq Q$$

Combining both inclusions:

$$Q \subseteq P \circ \hat{R} \subseteq Q \implies P \circ \hat{R} = Q \qquad \blacksquare$$

3. Approximation Analysis for Inconsistent Systems

Given $P = (0.4, 0.5)$ and $Q = (0.8, 0.9)$. Here $\max_i p_i = 0.5$. Since $q_1 = 0.8 > 0.5$ and $q_2 = 0.9 > 0.5$, the system is strictly unsolvable by Part 1!

Compute $\hat{R} = P \mathbin{\alpha} Q$:

  • For Column 1 ($q_1 = 0.8$): $p_1 = 0.4 \le 0.8 \implies 1.0$; $p_2 = 0.5 \le 0.8 \implies 1.0$.
  • For Column 2 ($q_2 = 0.9$): $p_1 = 0.4 \le 0.9 \implies 1.0$; $p_2 = 0.5 \le 0.9 \implies 1.0$.
$$\hat{R} = \begin{pmatrix} 1.0 & 1.0 \\ 1.0 & 1.0 \end{pmatrix}$$

Now compute $P \circ \hat{R}$:

  • $(P \circ \hat{R})_1 = \max(\min(0.4, 1), \min(0.5, 1)) = \max(0.4, 0.5) = 0.5 \le 0.8$.
  • $(P \circ \hat{R})_2 = \max(\min(0.4, 1), \min(0.5, 1)) = \max(0.4, 0.5) = 0.5 \le 0.9$.
$$P \circ \hat{R} = (0.5, 0.5)$$

Notice that $P \circ \hat{R} \le Q$ strictly. The Chebyshev approximation error is:

$$\| P \circ \hat{R} - Q \|_\infty = \max(|0.5 - 0.8|, |0.5 - 0.9|) = \max(0.3, 0.4) = 0.40 \qquad \blacksquare$$