# How can a relation be both irreflexive and antisymmetric?

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?

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.

• Things might become more clear if you think of antisymmetry as the rule that $x\neq y\implies\neg xRy\vee\neg yRx$ – drhab May 13 at 9:09

Well,consider the ''less than'' relation $$<$$ on the set of natural numbers, i.e., $$x if there exists a natural number $$z>0$$ such that $$x+z=y$$.
This relation is irreflexive, but it is also anti-symmetric. To see this, note that in $$x, the premise is never satisfied and so the formula is logically true.