empty set confusion $\neg\left (\neg{\left (A\setminus A  \right )}\setminus A  \right )$
$A\setminus A $ is simply empty set and $\neg$ of that is again empty set. Empty set $\setminus$ A is empty set or? But every empty set is included in every set?
I am confused when it comes to this..
 A: If by 'negation' you mean complement, then the 'negation' of the empty set is not the empty set, but the universal set ... or at least: the set of all things you are talking about: the Universe of Discourse. If we call that $U$, we get:
$((A\setminus A)'\setminus A)'=(\emptyset'\setminus A)'=(U\setminus A)'=A$
To go into that last step a little more:
$A \setminus B$ is the same as $A \cap B'$, and so:
$(U\setminus A)'=(U \cap A')'=U' \cup A''=\emptyset \cup A = A$
A: The emptyset is subset of every set $ E $ because you cannot find an element which is in the emptyset and not in the set $ E$.
$$\text{ Proof}$$
Let $ x$ be an arbitrary element,
$ p, q$  the propositions
 $$p \;\; : \; x\notin \emptyset$$
and
$$ q \;\; : \;\; x\in E.$$
By the emptyset Axiom, the proposition $ p$ is always true.
By the argument of Addition,
the proposition $  p \vee q$ is also true,
and by material implication,
$$\neg p \;\; \implies q \;\; \text{ is true}$$
or
$$x\in \emptyset\;\; \implies \;\;  x\in E$$
thus
$$\emptyset \;\; \subset\;\;  E$$
A: Just use the laws of operator precedence to evaluate the result of each binary or unary operation in the proper order:
\begin{align}
A\setminus A&=\emptyset\\
\therefore \neg(A\setminus A)&=\Omega\ \ \ \text{(universal set)}\\
\therefore \neg(A\setminus A)\setminus A&=\Omega\setminus A\\
\therefore \neg(\neg(A\setminus A)\setminus A)&=\neg(\Omega\setminus A)\\
&=\neg\Omega\cup A\ \ \ \text{(De Morgan)}\\
&=\emptyset\cup A\\
&=A
\end{align}
A: $A$ = everthing that is is $A$.
$A\setminus A$ = everything in $A$ that is not in $A$ = nothing = $\emptyset$
$\lnot(A\setminus A) = \lnot \emptyset$ = everything that is not in the emptyset = everything = The Universal Set. I'll call it $U$.
$\lnot(A\setminus A) \setminus A$ = $U\setminus A$= everything in the universal Set that is not in $A$ = $\lnot A$.
$\lnot (\lnot A\setminus A)\setminus A) =\lnot(\lnot A)$= everything that is not not in A = everything that is in $A$ = $A$.
.....
Or $\neg\left (\neg{\left (A\setminus A  \right )}\setminus A  \right )=$
$\{x| x\not \in  (\neg{\left (A\setminus A  \right )}\setminus A   )\}=$
$\{x|x\not \in \{x\in \neg(A\setminus A)|x\not \in A\}\}=$
$\{x|x\not \in \{x\not \in (A\setminus A)|x\not \in A\}\}=$
$\{x|x\not \in \{x\not \in \{x \in A|x\not \in A\}|x\not \in A\}\}=$
$\{x|x\not\in\{x\not \in \emptyset|x\not \in A\}\}=$
$\{x|x\not \in \{x\not \in A\}\}=$
$\{x|x \in A\}=$
$A$
