Expressing boolean operators using logical operators From my limited understanding of logical operators, it is possible to express the more complex logical operators such as $\operatorname{xnor}$ and $\operatorname{iff}$ as a combination of just a few basic logical operators ($\operatorname{and}$, $\operatorname{or}$, $\operatorname{not}$).
Is it possible to express boolean operators (e.g. equality, subset) as a combination of logical operators too, but that yield only $true$ (tautology) or $false$ (contradiction)?
For example, I could define the $A = B$ operator (set $A$ contains the exact same elements as set $B$) as $(A \land B) \lor (\overline{A \lor B})$, which would yield $true$ when it is true. But it yields neither $true$ (set of all things) nor $false$ (empty set) when it is false, and I want it to yield $false$. Is this at all possible? And what about the boolean subset operator?
It's possible I'm combining incompatible concepts (sets, logic, boolean algebra) but please bear with me.
 A: Yes, you're combining different concepts in ways they don't really want to combine.
Your basic problem is that the logical operators (or "connectives" in the jargon of mathematical logic) such as $\land$ can only be applied between things that have truth values, and if $A$ and $B$ are sets, then "$A \land B$" doesn't make sense. ("$A\cap B$" does make sense, but produces a set). If you want to start with sets and get a truth value out, at some point in the expression you need to have a symbol that applies to sets and gives you a truth value -- such as $\in$ -- and this means your plan of making do with logical operations alone is dead in the water as written.
This is not to say there isn't a close connection between set algebra and logic -- indeed, the collection of subsets of some given base sets with operations $\cup$, $\cap$ and complement, and {true,false} with operations $\lor$, $\land$ and $\neg$ are the two prototypical examples of a boolean algebra, and you can translate from a formula in propositional logic to an expression in set algebra, and the same laws will hold in each case. However, the set expression will produce a set rather than a truth value.
A significant difference between the two settings is that in logic, every function from a number of truth-valued variables to a truth value can be realized as a Boolean expression, whereas in set algebra there are function from sets to sets that cannot be written in this way. One of these is the one you seem to want:
$$ f(A,B) = \begin{cases} U & \text{if $A$ is the same set as $B$} \\
\varnothing & \text{otherwise} \end{cases} $$
The ones you can write using Boolean operators are exactly the one where you can determine whether $x\in f(A,B)$ by knowing only whether $x\in A$ and whether $x\in B$ (but not, for example, which particular element $x$ is or whether any other elements are in $A$ and/or $B$).
A: Just consider the definition of $x\in (A \star B)$ for any of the operators $\star$ we are considering.  You can notice that they all are defined in the form
$A \star B \iff \forall x\in U,  L$
Where U is theUniverse , and L is some logical expression defined in terms of the logical primitives $(x\in A)$ and $(x\in B)$.
for example
$A \subseteq B\iff ((x\in A) \implies (x\in B))$
Where in this case $L=((x\in A) \implies (x\in B))$.  So essentially there is a duality between logical expressions over $(x\in A)$ and $(x\in B)$, and the set relations on $A$ and $B$.  From there, we notice that we can represent $L$ "as a combination of just a few basic logical operators", and by passing through the duality, represent the set relations as the combination of a few basic dual operations.
For example
$A = B\iff ((x\in A) \iff (x\in B))$
$A = B\iff (((x\in A) \implies (x\in B)) \wedge ((x\in B) \implies (x\in A)))$
$A = B\iff (A\subseteq B \wedge B\subseteq A)$
which I think is the sort of simplification you were asking about.
