Boolean Algebra - Demorgan Laws

Question

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.

Answer 1

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)

Answer 2

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.