Let $\mathfrak{M}$ be a countable transitive model of set theory, and consider HOD (the hereditarily ordinal definable elements of $\mathfrak{M}$).


Let $x$ be an object $x \in HOD$. So $x$ is hereditarily definable from ordinals. 


Can we find a defining formula $\psi(x)$ of $x$ such that $\psi(x)$ contain only quantifiers ranging over the ordinals? 


PS: by "ranging over ordinals" I mean a quantifier $Qy$ where either we have the restriction $Qy$($y$ is an ordinal ...) or that $Qy \in \alpha$ where $\alpha$ is an ordinal.


Edit: After Emil's comment, I am going to allow quantification over elements $y$ such that $y \in x$. The reason for this change is that in the application I have in mind we can allow this. And, yes, we can allow also arbitrary bounded quantifiers.



Definability in HOD
The answer is no, not necessarily. The main reason is that if $\varphi(x,\vec\alpha)$ is a formula all of whose quantifiers range only over ordinals or are bounded, with ordinal parameters $\vec\alpha$, then the truth of $\varphi(x,\vec\alpha)$ is invariant under extensions $V\subset W$ with the same ordinals and no new elements of sets in $V$, since the interpretation of the meaning of the formula does not change. In particular, the interpretation of $\varphi$ does not change by forcing. 


Consider the following example. Start in $L$, and add a Cohen real $L[c]$, and the force to make that real $c$ definable in a forcing extension $L[c][G]$, by forcing to code its digits into the GCH pattern on the $\aleph\_n$'s, for example. So $c$ is in $\text{HOD}^{L[c][G]}$, but by further forcing, we can collapse cardinals to $L[c][G][H]$, where $\text{HOD}=L$ again. Suppose that there were a nice formula $\varphi$ defining $c$ in $L[c][G]$. It follows that $\varphi(c)$ also holds in $L[c][G][H]$. But in this case, there is some condition forcing it, and this condition specifies only finitely much of $c$. Since the forcing overall is weakly homogeneous, we may find an automorphic image $d\in L[c]$ of $c$ that also contains that finite part of $c$, and so $\varphi(d)$ will also hold in $L[c][G][H]$, and therefore also in $L[c][G]$. This contradicts that $c$ was defined by $\varphi$ in $L[c][G]$. 


