I wanted to show that both sets are equal. My Textbook says following:
$\in$ means "is element of", $\land$ is the and operator, $\lnot$ is the not operator, $\notin$ means "is not element of", $\lor$ is the or operator
$$\def\-{\setminus}\begin{split} x \in A\-A\-B &\iff\\ (x \in A) \land (\lnot(x \in (A\-B)) &\iff\\ (x \in A) \land (\lnot(x \in A \land (\lnot(x \in B))) &\iff\\ (x \in A) \land (\lnot(x \in A \land x \notin B)) &\iff\\ (x \in A) \land (x \notin A \lor x \in B)\end{split}$$
Let $(x \in A)$ be $C$ and $(x \in B)$ be $D$
- Case 1 $C$ True $D$ True -> Contradiction $x \in A \land x \notin A$ cannot be true
- Case 2 $C$ True $D$ True -> Contradiction $x \in A \land x \notin A$ cannot be true
- Case 3 $C$ False $D$ True -> True
Thus $$(x \in A) \land (x \notin A \lor x \in B) \iff\\ (x \in A) \land (x \in B) \iff\\ x \in (A \cap B)$$
What is the subsetproof? Because I have not fully understood the negation of $\land$
The equivalences, $\neg(a\land b)\equiv (\neg a\lor\neg b)$ and $\neg(a\lor b)\equiv (\neg a\land\neg b)$, are known as deMorgan's Laws. The justification is as follows:
$\neg(a\land b)$ is true when it is not the case that both $a$ and $b$ are true. This happens exactly when at least one of $a$ or $b$ is false, that is $(\neg a\lor\neg b)$ is true.
$\neg(a\lor b)$ is true when it is not the case that at least one of $a$ or $b$ are true. This happens exactly when at both of $a$ and $b$ are false, that is $(\neg a\land\neg b)$ is true.
In a similar vein, $a\land(\lnot a\lor b)$ is true exactly when $a$ is true and at least one of $\lnot a$ or $b$ is true. So it means $a$ is true, and since $\lnot a$ cannot also be true, therefore $b$ must be. Conversely $a\land b$ is true when $a$ and $b$ are both true. Now $b$ being true implies that at least one of $b$ or $\neg a$ is true. Therefore $a\land(\lnot a\lor b)\equiv (a\land b)$