# Boolean simplication

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.

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Are you sure you copied everything correctly? Are you using $+$ for OR and $\cdot$ or juxtaposition for AND? Then the truth conditions for your last displayed equation are not the same as for $Y=ABCD$ (e.g. they differ for A and C true, B and D false), so it seems either your copying or the module must be wrong? – joriki Apr 1 '11 at 10:02

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.