Boolean Algebra - Demorgan Laws I am given the problem: 
!((!A * B) * !((!B + C) * (!C * !D))) 
where ! = NOT, * = AND, and + = OR and I tried simplifying it using only Demorgan Laws (no absorption) and I got:
(A + !B) + ((!B + C) + (C + D))
and I was just wondering if it was correct. Any help would be greatly appreciated.
 A: If you just want to do the DeMorgan's law portion of the simplification, (using your notation)
!((!A * B) * !((!B + C) * (!C * !D)))
 <==> !(!A*B) + !!((!B + C) * (!C * !D)) (first round of DeMorgan's law usage)
 <==> (A + !B) + !!((!B + C) * (!C * !D)) (second round of DeMorgan's law usage)
Now, if we want to finish simplifying,
(A + !B) + !!((!B + C) * (!C * !D))
 <==> A + !B + (!B + C) * (!C * !D)   (simplifying the !!)
 <==> A + !B + (!B * !C * !D) + (!C * C * !D)    (distribution)
 <==> A + !B + (!B * !C * !D) + 0    (!C * C * !D is equivalent to 0)
 <==> A + !B + (!B * !C * !D) 
 <==> A + !B   (using the unavailable absorption law you mentioned)
A: If we let $\overline{A}$ signify "$\text{not}\ A$", then I get the following
$$\phantom{\iff\quad} !((!A * B) * !((!B + C) * (!C * !D)))$$
$$\iff\quad \overline{(\overline{A} * B) * \overline{(\overline{B} + C) * (\overline{C} * \overline{D}))}}$$
$$\iff\quad \overline{(\overline{A} * B)} + (\overline{B} + C) * (\overline{C} * \overline{D}))$$
$$\iff\quad (A + \overline{B}) + (\overline{B} + C) * (\overline{C} * \overline{D}))$$
$$\iff\quad (A * !B) + ((!B + C) * (!C * !D))$$
This is slightly different from the answer you got.
