Is


$$U\_{\omega}=\Big\{x\mid\forall z\Big(\big(\emptyset\in z\wedge \forall u, v\;(u,v\in z\rightarrow\{w\mid w\in u\vee w=v\}\in z)\big)\rightarrow x\in z\Big)\Big\}$$


identical with the set $V\_{\omega}$ of hereditarily finite sets, i.e. the level $\omega$ of the cumulative hierarchy?



Adjunction, infinity and hereditarily finite sets
I assume that the definition of $U\_\omega$ has a typo: adjunction operation usually means the following binary operation:
$$u;v:=u\cup\{v\}$$


If you just assume $z$ in the definition of $U\_\omega$ is closed under successor operator $u\mapsto u\cup\{u\}$, then $U\_\omega$ would be $\omega$.




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It suffices to show that every set which contains $\varnothing$ and is closed under adjunction contains $V\_\omega$.


Let $z$ be a set such that $\varnothing\in z$ and is closed under adjunction.
Assume inductively that $V\_n\subseteq z$. For $x\in V\_{n+1}=\mathcal{P}(V\_n)$, we have $x=\{a\_0,\cdots, a\_k\}$ for some $a\_0,\cdots,a\_k\in V\_n$. By induction on $i\le k$, you can see that $\{a\_0,\cdots, a\_i\}\in z$ (In the case of $i=0$, $\{a\_0\}$ is the adjunction of $\varnothing$ and $a\_0\in z$.) Hence $x\in z$. This shows $V\_{n+1}\subseteq z$.


