I saw the definition of a Wiener pair the other day and was curious why the sets were wrapped in an extra level of singleton. It might be to handle urelements? I'm not sure.
$$ (a,b) \;\;\text{is defined as}\;\; \{\{\{a\}, \varnothing\}, \{\{b\}\}\} $$
Does $\{\{a\}, \{b, \varnothing\}\}$ work as a definition of an ordered pair? This is a shorter version of a Wiener pair with the $\varnothing$ moved to the second element for convenience.
Here is my attempt to prove it.
Let $\pi_2(p)$ be equal to $\cup \{\cup z : z \in p \land |z| = 2\}$. Note that if the second element of $(a, b)$ is $\varnothing$, then $\{\{a\}, \{\varnothing, \varnothing\}\}$ will contain no 2-element sets.
Let $\pi_1(p)$ be equal to $\cup \{\cup z : z \in p \land |z| = 1\}$. If the second and first element are both $\varnothing$, then $\pi_1$ returns $\varnothing$. If just the second element is $\varnothing$, then there are two singletons, but $ (\cup \{a\}) \cup (\cup \{ \varnothing, \varnothing \}) $ is $a$. If neither $a$ nor $b$ are $\varnothing$, then there's only one singleton.
Since we can defined both $\pi_1$ and $\pi_2$, this means that this definition of an ordered pair satisfies the required properties.
