The sentence is: $$p=\forall x\forall e\forall u(u\in w(x,e)\leftrightarrow(u\in x \vee u=e)).$$
The preconditions are:
- $\forall v$ $\neg v\in \emptyset$
- $\forall e\forall u(u\in w(\emptyset, e)\leftrightarrow u=e)$
- $\forall x \forall y \exists d ((d\in x \leftrightarrow d\in y)\rightarrow x=y) $
If we consider a theory $\Sigma$ made out of 1. 2. and 3., is $\Sigma$ complete?
To answer this we need to prove the first part: $\Sigma \not\vdash p$, then if we prove $\Sigma\not\vdash\neg p$ we are done. How do we begin?
Some further hints. (If the task is just to show that $\Sigma$ is not complete, then looking at $p$ in particular is already a hint ...)
Showing that $\Sigma\not\vdash\neg p$ is probably the conceptually most straightforward part. You'd need to show a model of $\Sigma$ where $\neg p$ is false -- that is, simply, a model of $\{1,2,3,p\}$. The symbols chosen ought to give you some idea what you might try to use for a model ...
Next, you can modify that model such that it doesn't satisfy $p$ anymore, but still satisfies $\Sigma$. The key observation here is that neither of 1,2,3 even speak about $w(x,y)$ when $x$ is something different from $\emptyset$, so you have great latitude to mess with what the interpretation of $w$ does, and make $p$ fail.