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.
Best Answer
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.