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Summary

Irreflexivity occurs where nothing is related to itself. Anti-symmetry provides that whenever 2 elements are related "in both directions" it is because they are equal. These two concepts appear mutually exclusive but it is possible for an irreflexive relation to also be anti-symmetric.

Put another way: why does irreflexivity not preclude anti-symmetry?

More information

These are the definitions I have in my lecture slides that I am basing my question on:

Irreflexivity

$\forall x \in X : (x, x) \notin R$

Or in plain English "no elements of $X$ satisfy the conditions of $R$" i.e. no elements are related to themselves.

Anti-symmetry

$\forall x, y \in A ((xR y \land yRx) \rightarrow x = y)$

We were told that this is essentially saying that if two elements of $A$ are related in both directions (i.e. $xRy$ and $yRx$), this can only be the case where these two elements are equal.

These concepts appear mutually exclusive: anti-symmetry proposes that the bidirectionality comes from the elements being equal, but irreflexivity says that no element can be related to itself. And yet there are irreflexive and anti-symmetric relations.

I have read through a few of the related posts on this forum but from what I saw, they did not answer this question.

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    $\begingroup$ Things might become more clear if you think of antisymmetry as the rule that $x\neq y\implies\neg xRy\vee\neg yRx$ $\endgroup$ – drhab May 13 at 9:09
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Well,consider the ''less than'' relation $<$ on the set of natural numbers, i.e., $x<y$ if there exists a natural number $z>0$ such that $x+z=y$.

This relation is irreflexive, but it is also anti-symmetric. To see this, note that in $x<y \wedge y<x\Rightarrow x=y$, the premise is never satisfied and so the formula is logically true.

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  • $\begingroup$ "the premise is never satisfied and so the formula is logically true." This is exactly what I missed. I glazed over the fact that we were dealing with a logical implication and focused too much on the "plain English" translation we were given. My mistake. I'll accept this answer in 10 minutes. $\endgroup$ – daviegravee May 13 at 9:09

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