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Can anyone give me a counterexample for a relation $R\subset M\times M$ for the statement $$R\text{ antisymmetric} \wedge R\text{ not reflexive}\implies R\text{ asymmetric}$$

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No, because a relation is asymmetric if and only if it is antisymmetric and not reflexive.

To see that your implication is always true, we could check the contrapositive statement: If R is symmetric then R is not antisymmetric or R is reflexive. This is easily seen to be true since if R is symmetric and anti-symmetric, it is a sub-relation of the equality relation, in which case it is obviously reflexive.

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    Thank you for the second part - i was totally confused here.2012-12-10
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    I just got a question right now. What aboout $M=\{1,2\}$ and $R=\{(1,1),(1,2)\}$?2012-12-10
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    @ChristianIvicevic What about it?2012-12-10
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    Isn't this example antisymmetric, not reflexive and **not** asymmetric? If not, why?2012-12-10
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    @ChristianIvicevic But it *is* asymmetric, since it contains (1,2) but not (2,1).2012-12-10
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    I see it now. Let's forget what I was talking about.2012-12-10
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Actually the answer is yes, since a relation is asymmetric if and only if it is antisymmetric and "irreflexive". There are three distinct properties of a relation, reflexive, irreflexive, and neither reflexive nor irreflexive. Plus, M = {1, 2} and R = {(1, 1), (1, 2)} is a correct counterexample. This R is antisymmetric and not reflexive but it is not asymmetric.