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Let S be a set, and R an antisymmetric relation on S. Prove that R^c is...

Let S be a set, and R an antisymmetric relation on S. Prove that R^c is trichotomous.

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Answer #1

I don't thing the given statement is true because you just consider the antisymmetric relation divisibility on natural numbers. Then compliment of this relation is

a is related to b iff a does not divde b . This is not a trichotomous relation because

7 R^c 3\; and \; 3 R^c\; 7 \quad 7\neq 3

Hence R^c is not trichotomous

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