I'm working on proving the case above thusly:
Note: I should point out that i'm using the $\lnot$ symbol in place of the complement bar that should go over the letters, as i can't seem to get that formatted on this site.
In the first case of $\lnot(A \cup B)$, we know that for every item $x, x \in \lnot (A \cup B)$, which is to say >$x \notin A \cup B$.
From this, we know that $x \notin A$ and $x \notin B$. By the definition of complements, we can conclude that $x\in \lnot A$ and $x \in \lnot B$. As such, $x \in \lnot A \cap \lnot B$, which proves that $\lnot (A \cup B) \subseteq \lnot A \cap \lnot B$.
In the second case of $\lnot A \cap \lnot B$, we know that for every item $x, x \in \lnot A \cap \lnot B$, which is to say that $x \notin A \cap B$.
From this, we can conclude that $x \notin A$ or $x \notin B$. As such, its clear that $x \in \notin A$ or $x \in \lnot B$. Further, from the definition of a union, we can conclude that $x \in \lnot (A \cup B)$, which proves that $\lnot A \cap \lnot B \subseteq \lnot (A \cup B)$.
As such, we have proven that $\lnot (A \cup B) = \lnot A \cap \lnot B$ by proving that $\lnot (A \cup B) \subseteq \lnot A \cap \lnot B$ and that $\lnot A \cap \lnot B \subseteq \lnot (A \cup B)$.
According to my instructor, the part where i say "in the second case" is where i start to go wrong, however i don't really see what i'm doing incorrectly even after having reviewed the textbook and searched around these forums.
help?