I managed to prove existence for the following theorem: $\forall A\in\mathcal{P}(U)\ \exists!B\in\mathcal{P}(U)\ \forall C\in\mathcal{P}(U)\ (C\setminus A=C\cap B)$ where U is any set. My assumption is that $B=U\setminus A$, and it works for existence, but I'm stuck with proving uniqueness part with $B$ being defined this way.
For uniqueness we need to prove $\forall B'\in\mathcal{P}(U)\ \forall C\in\mathcal{P}(U)\ (\ (C\setminus A=C\cap B')\rightarrow B'=B)$ where A is arbitrary element of $\mathcal{P}(U)$ but I don't how to connect $x\in B'$ or $x\in B$ to the assusmed identitiy $C\setminus A=C\cap B'$.
Any pointers are much appreciated.
EDIT
Here is my attempt of proof.
Proof: Let $A$ be an arbitrary element of $\mathcal{P}(U)$ and let $B=U\setminus A$.
Existence: Let $C$ be an arbitrary element of $\mathcal{P}(U)$. $(\rightarrow)$ Let $x$ be an arbitrary element of $C\setminus A$. Since $C\subseteq U$, then $x\in U\setminus A$. Therefore $x\in C\cap B$. $(\leftarrow)$ Let x be an arbitrary element of $C\cap B$. Then $x\in C\cap (U\setminus A)$, so we can conclude $x\in C\setminus A$.
Uniqueness: Let $B'$ be an arbitrary element of $\mathcal{P}(U)$ and suppose $\forall C\in\mathcal{P}(U)(C\setminus A=C\cap B')$.
$(B'\subseteq B)$ Since $B'\in\mathcal{P}(U)$, then in particular $B'\setminus A=B'\cap B'=B'$, so clearly $B'\cap A=B'\cap (U\setminus B)=\varnothing$. Then $\forall x(x\not\in B'\lor x\not\in U\lor x\in B)$, which is equivalent to $\forall x(x\in B'\cap U\rightarrow x\in B$). Since $B'\subseteq U$, $B'\cap U=B'$, we now have $\forall x(x\in B'\rightarrow x\in B)$, and therefore $B'\subseteq B$.
$(B\subseteq B')$ Let $C=B$. Then $B\setminus A=B\cap B'$, and because $B\cap A=\varnothing$, we have $B=B\cap B'$, so we can conclude $B\subseteq B'$.