A fuzzy conjunction operator denoted as \(π‘(π₯,π¦)\) and a fuzzy disjunctionβ¦
2019
A fuzzy conjunction operator denoted asΒ \(π‘(π₯,π¦)\)Β andΒ a fuzzy disjunction operator denoted asΒ \(π (π₯,π¦)\)Β form a dual pair if they satisfy the condition:
- A.
\(π‘(π₯,π¦)=1βπ (π₯,π¦) \) - B.
\(π‘(π₯,π¦)=π (1βπ₯,1βπ¦) \) - C.
\(π‘(π₯,π¦)=1βπ (1βπ₯,1βπ¦) \) - D.
\(π‘(π₯,π¦)=π (1+π₯,1+π¦)\)
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Correct answer: C
Correct relation: t(x,y) = 1 β s(1 β x, 1 β y)
Reasoning:
Define the standard fuzzy negation n(x) = 1 β x. This negation is involutive, so n^{-1} = n.
The general duality condition links a t-norm and s-norm by applying negation to inputs and output: t(x,y) = n( s( n(x), n(y) ) )
Because n is its own inverse, substitute n(x)=1βx to obtain t(x,y) = 1 β s(1 β x, 1 β y).
Why the other forms fail:
Expression 1 β s(x,y) only complements the output; it does not complement the inputs, so it does not satisfy the negation-input-output structure required for duality.
Expression s(1 β x, 1 β y) complements only the inputs but omits the outer complement; the final negation is required to yield the dual t.
Expression s(1 + x, 1 + y) uses inputs outside the [0,1] domain and does not follow the negation-based formula, so it is invalid as a dual relation.
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