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Previous Year Question (PYQs)

Which of the following is not a valid property over two fuzzy relations $\tilde R$ and $\tilde S$ for performing $\lambda$-cut defuzzifications?

1. $(\tilde R \cup \tilde S)\lambda=\tilde R\lambda\cup \tilde S_\lambda$ 
2. $(\tilde R \cap \tilde S)\lambda=\tilde R\lambda\cap \tilde S_\lambda$ 
3. $(\overline{\tilde R})\lambda\neq \tilde R\lambda\ \text{except when }\lambda=1$
4. For any $\lambda\le \beta$, where $0\le \beta\le1$, it is true that $R_\beta\subseteq R_\lambda$




Solution

For a fuzzy set $A$, the $\lambda$-cut is defined as: $$A_\lambda = \{x \mid \mu_A(x) \ge \lambda\}$$

  • Union Property: $(A \cup B)_\lambda = A_\lambda \cup B_\lambda$ ✅
  • Intersection Property: $(A \cap B)_\lambda = A_\lambda \cap B_\lambda$ ✅
  • Monotonicity: If $\lambda \le \beta$, then $A_\beta \subseteq A_\lambda$ ✅
  • Complement: $\mu_{\overline{A}}(x) = 1 - \mu_A(x)$ ⇒ $(\overline{A})_\lambda = \{x \mid 1 - \mu_A(x) \ge \lambda\} = \{x \mid \mu_A(x) \le 1 - \lambda\}$ ❌

Hence, $(\overline{A})_\lambda$ is not equal to $A_\lambda$, even when $\lambda = 1$. Therefore, the statement “except when $\lambda = 1$” is incorrect.

✅ Final Answer: Option 3 is not valid.


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