Quote:

Simplify this Boolean expression: .$\displaystyle A \;\cap \;\sim\![(\sim\!A) \cap (\sim\!B)] $

According to the truth table I did,

it should be equivalent to $\displaystyle \sim B$ . . . . no, it's equivalent to $\displaystyle {\color{blue}A}$

$\displaystyle A \;\cap \;\sim\![(\sim\!A) \cap (\sim\!B)] $

. . $\displaystyle = \;A \;\cap \;(\sim\sim\!A \;\cup \;\sim\sim\!B) \qquad\text{DeMorgan's law}$

. . $\displaystyle = \;A \;\cap\;(A \cap B)\qquad\text{Double negation}$

. . $\displaystyle = \;(A\cap A) \cup (A \cap B)\qquad\text{Distributive Property}$ . . . . All this is correct!

And this is where I get stuck.

If $\displaystyle A \cap A \:=\:A$, then you get into an infinite loop

of distrubutive properties and never reach your goal. . . . . This is true