I'm trying to simplify the following booleans:
$$Y=[\overline{ \overline{(A+B)} \quad \overline{(C+D)}}]$$
My solution is:
$$Y=[\overline{ \overline{(A+B)} \quad\overline{(C+D)}}]$$ $$ = [\overline{ \overline{(A+B)}} \quad \overline{ \overline{(C+D)}}]$$
After this step my module is saying that the next step would be:
$$Y = [(AB)(CD)]$$
But I am not getting how,since I believe the next step is:$$Y = [(A+B)(C+D)]$$ which is just be canceling the negation both times,am I wrong?
This is a part of entire problem which is simplify: $$Y= [(A+B)(C+D)] \cdot [\overline{ \overline{(A+B)} \quad\overline{(C+D)}}]$$ my solution for which is $Y = [(A+B)(C+D)]$ but they are showing that $Y=ABCD$ should be the correct solution.
A rule you have undoubtedly seen is $\overline{XY}=\overline{X}+\overline{Y}$. If you apply that, with $X=\overline{A+B}$ and $Y=\overline{C+D}$ then the first thing we get is $$\overline{\overline{A+B}}+\overline{\overline{C+D}}$$
Then, using the ``double negation'' rule, we obtain $$(A+B)+(C+D)$$ (here the brackets are unnecessary). You can verify, using truth tables, that the starting expression you gave and the one obtained above are equivalent.