Let us consider some consistent subset of naïve set theory, in which a universal set $U$ exists, the power set $\mathcal{P}(A)$ and complement $A'=U-A$ exists of any set $A$, on top of the usual binary operations $\cup$, $\cap$, $-$, $\Delta$ (symmetric difference), etc.
Considering that $\mathcal{P}(U)=U$ (as $U$ is the universe, rather than some arbitrary domain of discourse), since all subsets of $U$ are elements of $U$ and vice-versa,
is there a "closed-form" expression for $\mathcal{P}(A')$?
By "closed-form" here, I mean an expression only in terms of $A$ and $U$ under the given operations (so, no set-builder notation).
An immediate answer evades me, partly because naïve set theory is often unintuitive, but also because I am sleep deprived (so I apologize If I have overlooked something trivial).
We may first ask the question of what exactly $\mathcal{P}(A')$ represents. It is not so hard to deduce that $\mathcal{P}(A')=U-\Gamma$, where $\Gamma$ is the family of all sets which are not disjoint with $A$. I.e; $$\Gamma=\{x\in U:x\cap A\neq\emptyset\}.$$ If we were allowed set-builder notation, this would indeed be our answer! However, since we don't, the question becomes whether or not $\Gamma$ can be represented in some closed-form.
Equivalently, $$\mathcal{P}(A')=\{x\in U:x\cap A=\emptyset\}$$ can be seen to be the family of all sets which are disjoint with $A$ (which is, IMO, an interesting property!). Whether this is easier to be shown to have or not have a "closed form" is not obvious to me, but worth mentioning.
Any and all advice to this recreational problem would be greatly appreciated.
Suppose the universe is uncountable, and that $A$ is countable. Then any set you can build using the given list of operations is either countable or has countable complement. But $\mathcal{P}(A^c)$ has neither of these properties.