why does the axiom of pair has 2 different claims on Wiki? I was reading ZFC Set theory on wiki, and here is the definition of the axiom of pair on its page:

If $x$ and $y$ are sets, then there exists a set which contains $x$ and $y$ as elements.
$$\forall x\forall y\exists z\Bigl( (x\in z)\wedge (y\in z)\Bigr).$$

Which only states that both $x$ and $y$ are elements of some $z$.
But when I look into the ordered pair axiom page of the wiki
(or from Herbert Enderton's Elements of Set Theory), it states the following:

For any $A$ and $B$ sets, there is a set $C$ such that
$$(x\in C)\iff (x=A\vee x=B).$$

In my understanding, this means there exists set $C$ that any set $A$ or $B$ would be an element of $C$.
So it seems like the axiom of order pair has stated 2 different things: the first statement must have 2 of any element belonging to some set $z$. but the second statement says that is ok that only 1 of any element belongs to some set $C$.
Therefore I'm having a hard time understanding which is more correct.
Thank you!
 A: You will see this issue appear for other axioms such as the power set and the union axiom. The reason why it doesn't matter is that you have the comprehension axiom scheme so if I have a $z$ containing $x$ and $y$ then the set $\{x,y\}$ exists by comprehension on $z$ with the formula $\varphi(u,x,y)=u=x\vee u=y$. For union and the power set axiom you can the axiom state that there exists a set containing the union or the powerset and by comprehension you obtain the union or power set. What is interesting is that the axiom of pairing is not necessary and follows from the other $ZFC$ axioms.
One case where the weaker form of these axioms can be useful is for example when proving that the Boolean universe in forcing values all of the $ZFC$ axioms as $1$ in such a case veryfying the claim for a weaker axiom makes the proof a bit easier.
Edit: as Andreas pointed out I did not address one of the concerns which was that of understanding the notation of $(x\in c)\leftrightarrow (x=a\vee x=b)$. Something I may add to his comment is that this would be equivalent in set builder notation to say $c=\{x:x=a\vee x=b\}=\{a,b\}$ which makes sense in a context with extensionality. I hope this helps
