Uniform elimination of imaginaries Does the following principle follow from uniform elimination of imaginaries?
For every formula $\varphi(x;y)$ there is a formula $\vartheta(x;z)$ such that 
$$\forall y\;\exists^{=1}z\;\forall x\;\Big[\varphi(x;y)\leftrightarrow\vartheta(x;z)\Big]$$
The answer is affirmative if we restrict the question to formulas such that $\forall y\;\exists x\;\varphi(x;y)$. Is this limitation necessary?
Edit (to answer a request in the comments). Uniform elimination of imaginaries says that every definable equivalence relation is the kernel of a function (that is, $aEb\leftrightarrow fa=fb$).
Elimination of imaginaries says that for every $\varphi(x;y)$ and every parameter $a$ there is a formula $\vartheta_a(x;z)$ such that 
$$\exists^{=1}z\;\forall x\;\Big[\varphi(x;a)\leftrightarrow\vartheta_a(x;z)\Big]$$
Elimination of imaginaries is equivalent to uniform elimination under very weak hypotheses (there are two definable elements).
 A: Yes, the statement follows from uniform elimination of imaginaries, with a slight warning: you need to allow the range of the variable $z$ to be a definable set (of tuples), rather than all tuples of a given length. More on that caveat later.
Let $\varphi(x,y)$ be a formula in tuples $x$ and $y$. Define an equivalence
$$y_1 E y_2 := \forall x : (\varphi(x,y_1) \leftrightarrow \varphi(x,y_2)).$$ That is, $a E b$ iff $\varphi(x,a)$ and $\varphi(x,b)$ define the same set. By uniform elimination of imaginaries, we have a definable $f$ such that $E$ is the kernel of $f$. Moreover, we can take $f$ to be surjective by considering its codomain to be its range, which is definable.
Now, set $$\vartheta(x,z) := \exists y : f(y) = z \wedge \varphi(x,y).$$ Fix $y$. We need to show existence and uniqueness of $z$ such that $\varphi(x,y)$ and $\vartheta(x,z)$ define the same set in $x$. Existence is easy: $z := f(y)$ works. For uniqueness, assume we have $z_1,z_2$ such that $\vartheta(x,z_1)$ and $\vartheta(x,z_2)$ both define the same set (in particular, that they both define $\varphi(x,y)$). By the surjectivity of $f$, take $y_1,y_2$ such that $f(y_1) = z_1, f(y_2) = z_2$. We now need only show that $y_1 E y_2$, which we already know from the definitions of $E$ and $\vartheta$.

About the caveat. The caveat appears necessary (though I don't have a counterexample) in order to consider $f$ surjective. I don't consider the caveat a problem, though, since there's no real reason to consider definable sets of tuples as "second class citizens." And even if you do, just make the definable set explicit by modifing the statement to say there is $\vartheta(x,z)$ and a definable set $Z$ such that $$(\forall y) (\exists^{=1} z \in Z) (\forall x) \varphi(x,y) \leftrightarrow \vartheta(x,z).$$
