A fuzzy conjunction operator denoted as \(𝑑(π‘₯,𝑦)\) and a fuzzy disjunction…

2019

A fuzzy conjunction operator denoted asΒ \(𝑑(π‘₯,𝑦)\)Β andΒ a fuzzy disjunction operator denoted asΒ \(𝑠(π‘₯,𝑦)\)Β form a dual pair if they satisfy the condition:

  1. A.

    \(𝑑(π‘₯,𝑦)=1βˆ’π‘ (π‘₯,𝑦) \)

  2. B.

    \(𝑑(π‘₯,𝑦)=𝑠(1βˆ’π‘₯,1βˆ’π‘¦) \)

  3. C.

    \(𝑑(π‘₯,𝑦)=1βˆ’π‘ (1βˆ’π‘₯,1βˆ’π‘¦) \)

  4. D.

    \(𝑑(π‘₯,𝑦)=𝑠(1+π‘₯,1+𝑦)\)

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Correct answer: C

Correct relation: t(x,y) = 1 βˆ’ s(1 βˆ’ x, 1 βˆ’ y)

Reasoning:

  • Define the standard fuzzy negation n(x) = 1 βˆ’ 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)=1βˆ’x to obtain t(x,y) = 1 βˆ’ s(1 βˆ’ x, 1 βˆ’ y).

Why the other forms fail:

  • Expression 1 βˆ’ 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 βˆ’ x, 1 βˆ’ 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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